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Success in Mathematics is just a click away… With WileyPLUS, students and instructors will experience success in the classroom. When students succeed in your course—when they stay on-task and make the breakthrough that turns confusion into confidence—they are empowered to realize the possibilities for greatness that lie within each of them. We know your goal is to create an environment where students reach their full potential and experience the exhilaration of academic success that will last them a lifetime. WileyPLUS can help you reach that goal.
WileyPLUS is an online suite of resources—including the complete text—that will help your students: • come to class better prepared for your lectures • get immediate feedback and context-sensitive help on assignments and quizzes • track their progress throughout the course
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FOR INSTRUCTORS WileyPLUS is built around the activities you perform in your class each day. With WileyPLUS you can: Prepare & Present
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Create outstanding class presentations using a wealth of resources such as PowerPoint™ slides, image galleries, interactive simulations, and more. You can even add materials you have created yourself.
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FOR STUDENTS You have the potential to make a difference! WileyPLUS is a powerful online system packed with features to help you make the most of your potential and get the best grade you can!
With WileyPLUS you get: • A complete online version of your text and other study resources.
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*Based upon 7,000 responses to student surveys in academic year 2006-2007
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FOURTH EDITION
EXPLORATIONS IN COLLEGE ALGEBRA LINDA ALMGREN KIME JUDITH CLARK University of Massachusetts, Boston, Retired
B E V E R LY K . M I C H A E L University of Pittsburgh in collaboration with Norma M. Agras Miami Dade College Robert F. Almgren Courant Institute, New York University Linda Falstein University of Massachusetts, Boston, Retired Meg Hickey Massachusetts College of Art John A. Lutts University of Massachusetts, Boston Peg Kem McPartland Golden Gate University, Retired Jeremiah V. Russell University of Massachusetts, Boston; Boston Public Schools software developed by Hubert Hohn Massachusetts College of Art Funded by a National Science Foundation Grant
JOHN WILEY & SONS, INC.
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Publisher
Laurie Rosatone
Acquisitions Editor
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This book was set in 10/12 Times Roman by Publication Services, and printed and bound by Courier (Westford). The cover was printed by Courier (Westford). This book is printed on acid-free paper. ` Copyright © 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, website http://www.wiley.com/go/permission. To order books or for customer service call 1-800-CALL-WILEY(225-5945). ISBN 978-0471-91688-8 Printed in the United States of America. 10
9
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To our students, who inspired us.
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A Letter from a Student
My name is Lexi Fo urnier and semester I am enro I am a fre ll shman he ed in the A in College re at Pitt. p p lied Algeb Algebra.” This ra B course us efore com courses v ing “Explo ing to Pitt, arying fro rations I had take m algebra frustration n numero to calculu , stress, an u s s , m a ll ath of which d a detesta that I was produced tion for m required to ath as a su take a ma and creati bject. Wh th course ve writing en I was to h e re , m I ld ajor; why was livid. by inform do I need I am a pre ing me of math? My -law this new m non-math adviser ca ath class /science m lm a im e d e me ajors the d at teach At first I basic skil ing was skep ls th e y w tical, but ill need in recomme I’m writin everyday nd this co g life. u to r s y e . What I h ou now to been reali ave learn emphatic stic math e ally d thus far skills pre confidenc sented in in this co e and mo u a r “ s le e f h ti t ave vation. Fo brain” me relatable. r once in thod that The conc m fosters y career a epts are c amorphou s a studen lear and r s topics a t, e m a li ath is s tic (as op ddressed to this cla posed to in my ear ss. I enjoy th lier math e abstract, doing my classes). the lesson homewor I lo s are appli o k forward k and pro cable to m empower jects beca y life and ed by my u s e I feel that m y future a understan This cour n d d in b g e . c a u se is a vit se I feel al additio my view n to the m on the su ath depar bject and call “ever tment. It stimulate yday math has altere d a n apprecia .” d tion for w It is my b hat I like elief that to many stu and helpf dents wil ul as I ha l find the ve. Thank class as e you for y ncouragin our attenti g on. Sincerely , Lexi Fou rnier Student, U niversity of P
ittsburgh
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PREFACE This text was born from a desire to reshape the college algebra course, to make it relevant and accessible to all of our students. Our goal is to shift the focus from learning a set of discrete mechanical rules to exploring how algebra is used in the social and physical sciences. Through connecting mathematics to real-life situations, we hope students come to appreciate its power and beauty.
Guiding Principles The following principles guided our work. • Develop mathematical concepts using real-world data and questions. • Pose a wide variety of problems designed to promote mathematical reasoning in different contexts. • Make connections among the multiple representations of functions. • Emphasize communication skills, both written and oral. • Facilitate the use of technology. • Provide sufficient practice in skill building to enhance problem solving.
Evolution of Explorations in College Algebra The fourth edition of Explorations is the result of a 15-year long process. Funding by the National Science Foundation enabled us to develop and publish the first edition, and to work collaboratively with a nationwide consortium of schools. Faculty from selected schools in the consortium continued to work with us on the second, third, and now the fourth editions. During each stage of revision we solicited extensive feedback from our colleagues, reviewers and students. Throughout the text, families of functions are used to model real-world phenomena. After an introductory chapter on data and functions, we first focus on linear and exponential functions, since these are the two most commonly used mathematical models. We then discuss logarithmic, power, quadratic, and polynomial functions and finally turn to ways to extend and combine all the types of functions we’ve studied to create new functions. The text adopts a problem-solving approach, where examples and exercises lie on a continuum from open-ended, nonroutine questions to problems on algebraic skills. The materials are designed for flexibility of use and offer multiple options for a wide range of skill levels and departmental needs. The text is currently used in small classes, laboratory settings, and large lectures, and in both two- and four-year institutions.
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Special Features and Supplements An instructor is free to choose among a number of special features. The Instructor’s Teaching Manual provides support for using these features and includes sample test questions. The Instructor’s Solutions Manual contains answers to the even exercises and even problems in the Chapter Reviews. Both manuals are available free for adopters either online at www.wiley.com/college/kimeclark or in hardcopy by contacting your local Wiley representative.
Exploring Mathematical Ideas Explorations These are open-ended investigations designed to be used in parallel
with the text. They appear at the end of each chapter and in two chapter-length Extended Explorations. New! Chapter Review: Putting It All Together Each review contains problems that apply all of the basic concepts in the chapter. The answers to the odd-numbered problems are in the back of the text. Check Your Understanding A set of mostly true/false questions at the end of each
chapter (with answers in the back of the text) offer students a chance to assess their understanding of that chapter’s mathematical ideas.
?
SOMETHING TO THINK ABOUT
Something to Think About Provocative questions, posed throughout the text, can
be used to generate class discussion or for independent inquiry. 60-Second Summaries Short writing assignments in the exercises and Explorations ask students to succinctly summarize their findings. Readings A variety of articles related to topics covered in the text are available on the
course website at www.wiley.com/college/kimeclark.
Using Technology COURSE WEBSITE
Technology is not required to teach this course. However, we provide numerous resources, described below, for teaching with technology at the course website www.wiley.com /college/kimeclark. Graphing Calculator Manual The manual offers step-by-step instructions for using
the TI83/TI84 family of calculators that are coordinated with the chapters in the text. It is free on the course website or at a discount when packaged in hardcopy with the text. Interactive Software for Mac and PC Programs for visualizing mathematical
concepts, simulations, and practice in skill building are available on the course website. They may be used in classroom demonstrations or a computer lab, or downloaded for student use at home. DATA
Excel and TI83/TI84 Graph Link Files Data files containing all the major data sets used in the text are available on the course website.
Practice in Skill Building Algebra Aerobics
These collections of skill-building practice problems are integrated throughout each chapter. Answers for all Algebra Aerobics problems are in the back of the text. WileyPLUS This is a powerful online tool that provides a completely integrated suite
of teaching and learning resources in one easy-to-use website. It offers an online assessment system with full gradebook capabilities, which contains algorithmically generated skill-building questions from the Algebra Aerobics problems and the exercises in each chapter. Faculty can view the online demo at www.wiley.com/college/wileyplus.
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Acknowledgments
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The Fourth Edition Overall Changes Extensive faculty reviews guided our work on the fourth edition. The sequence of the chapters remains the same as in the third edition, but we have included • New chapter reviews, called “Putting It All Together,” with problems that bring together the major concepts of the chapter. • A relocation of exercises from the end of the chapter to the end of each section. • Expanded coverage of several topics, including function notation, range and domain, piecewise linear functions (including absolute value and step functions), rational functions, composition, and inverse functions. • Extensive updates of the data sets. • Revisions to many chapters for greater clarity. • Many new problems and exercises, ranging from basic algebraic manipulations to real-world applications.
Detailed Changes CHAPTER REVIEWS: “PUTTING IT ALL TOGETHER” appear at the end of each chapter. CHAPTER 1: Making Sense of Data and Functions has a new section on the language of functions, with expanded coverage of function notation, domain, and range. Boxes have been added to highlight important concepts. CHAPTER 2: Rates of Change and Linear Functions has a new subsection on piecewise linear functions, including the absolute value function and step functions.
EXTENDED EXPLORATION: Looking for Links between Education and Earnings uses an updated data set from the U.S. Census about 1000 individuals. CHAPTER 5: Growth and Decay: An Introduction to Exponential Functions has an expanded discussion on constructing an exponential function given its doubling time or half-life. CHAPTER 7: Power Functions has an added discussion of asymptotes for negative integer power functions. CHAPTER 8: Quadratics, Polynomials, and Beyond has changed the most. The old Section 8.6 has been expanded and broken up into three sections. Section 8.6, “New Functions from Old,” discusses the effect of stretching, compressing, shifting, reflecting, or rotating a function. Section 8.7, “Combining Two Functions,” includes the algebra of functions and an expanded subsection on rational functions. Section 8.8, “Composition and Inverse Functions,” extends the coverage of these topics.
Acknowledgments We wish to express our appreciation to all those who helped and supported us during this extensive collaborative endeavor. We are grateful for the support of the National Science Foundation, whose funding made this project possible, and for the generous help of our program officers then, Elizabeth Teles and Marjorie Enneking. Our original Advisory Board, especially Deborah Hughes Hallett and Philip Morrison, and our original editor, Ruth Baruth provided invaluable advice and encouragement. Over the last 15 years, through five printings (including a rough draft and preliminary edition), we worked with more faculty, students, TAs, staff, and administrators than we can possibly list here. We are deeply grateful for supportive colleagues at our own University. The generous and ongoing support we received from Theresa Mortimer, Patricia Davidson, Mark Pawlak, Maura Mast, Dick Cluster, Anthony Beckwith, Bob Seeley, Randy Albelda, Art MacEwan, Rachel Skvirsky, Brian Butler, among many others, helped to make this a successful project.
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PREFACE
We are deeply indebted to Ann Ostberg and Rebecca Hubiak for their dedicated search for mathematical errors in the text and solutions, and finding (we hope) all of them. A text designed around the application of real-world data would have been impossible without the time-consuming and exacting research done by Patrick Jarrett, Justin Gross, and Jie Chen. Edmond Tomastik and Karl Schaffer were gracious enough to let us adapt some of their real-world examples in the text. One of the joys of this project has been working with so many dedicated faculty who are searching for new ways to reach out to students. These faculty, and their teaching assistants and students all offered incredible support, encouragement, and a wealth of helpful suggestions. In particular, our heartfelt thanks to members of our original consortium: Sandi Athanassiou and all the wonderful TAs at University of Missouri, Columbia; Natalie Leone, University of Pittsburgh; Peggy Tibbs and John Watson, Arkansas Technical University; Josie Hamer, Robert Hoburg, and Bruce King, past and present faculty at Western Connecticut State University; Judy Stubblefield, Garden City Community College; Lida McDowell, Jan Davis, and Jeff Stuart, University of Southern Mississippi; Ann Steen, Santa Fe Community College; Leah Griffith, Rio Hondo College; Mark Mills, Central College; Tina Bond, Pensacola Junior College; and Curtis Card, Black Hills State University. The following reviewers’ thoughtful comments helped shape the fourth edition: Mark Gïnn, Appalachian State; Ernie Solheid, California State University, Fullerton; Pavlov Rameau, Florida International University; Karen Becker, Fort Lewis College; David Phillips, Georgia State University; Richard M. Aron and Beverly Reed, Kent State University; Nancy R. Johnson, Manatee Community College; Lauren Fern, University of Montana; Warren Bernard, Linda Green, and Laura Younts, Santa Fe Community College; Sarah Clifton, Southeastern Louisiana University; and Jonathan Prewett, University of Wyoming. We are especially indebted to Laurie Rosatone at Wiley, whose gracious oversight helped to keep this project on track. Particular thanks goes to our new editors Jessica Jacobs, Acquisitions Editor; John-Paul Ramin, Developmental Editor; Michael Shroff, Assistant Editor; and their invaluable assistant Jeffrey Benson. It has been a great pleasure, both professionally and personally, to work with Maddy Lesure on her creative cover design and layout of the text. “Explorations” and the accompanying media would never have been produced without the experienced help from Sandra Dumas, Dorothy Sinclair, and Stefanie Liebman. Kudos to Jan Fisher at Publication Services. Throughout the production of this text, her cheerful attitude and professional skills made her a joy to work with. Over the years many others at Wiley have been extraordinarily helpful in dealing with the myriad of endless details in producing a mathematics textbook. Our thanks to all of them. Our families couldn’t help but become caught up in this time-consuming endeavor. Linda’s husband, Milford, and her son Kristian were invaluable scientific and, more importantly, emotional resources. They offered unending encouragement and sympathetic shoulders. Judy’s husband, Gerry, become our Consortium lawyer, and her daughters, Rachel, Caroline, and Kristin provided support, understanding, laughter, editorial help and whatever was needed. Beverly’s husband, Dan, was patient and understanding about the amount of time this edition took. Her daughters Bridget and Megan would call from college to cheer her on and make sure she was not getting too stressed! All our family members ran errands, cooked meals, listened to our concerns, and gave us the time and space to work on the text. Our love and thanks. Finally, we wish to thank all of our students. It is for them that this book was written. Linda, Judy, and Bev
P.S. We’ve tried hard to write an error-free text, but we know that’s impossible. You can alert us to any errors by sending an email to
[email protected]. Be sure to reference Explorations in College Algebra. We would very much appreciate your input.
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COVERING THE CONTENTS
The following flow chart suggests some alternative paths through the chapters that have worked successfully for others.
Ch. 1: Making Sense of Data and Functions
Ch. 2: Rates of Change and Linear Functions
Extended Exploration: Looking for Links between Education and Earnings
Ch. 3: When Lines Meet: Linear Systems Ch. 4: The Laws of Exponents and Logarithms: Measuring the Universe
Ch. 5: Growth and Decay: An Introduction to Exponential Functions
Ch. 6: Logarithmic Links: Logarithmic and Exponential Functions
Ch. 7: Power Functions
Ch. 8: Quadratics, Polynomials, and Beyond
Extended Exploration: The Mathematics of Motion
The straight vertical path through Chapters 1, 2, 4, 5, and 8, covering linear, exponential, quadratic and other polynomial functions, indicates the core content of the text. You may choose to cover these chapters in depth, spending time on the explorations, readings and student discussions, writing, and presentations. Or you may pick up the pace and include as many of the other chapters and Extended Explorations as is appropriate for your department’s needs.
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TABLE OF CONTENTS CHAPTER 1
MAKING SENSE OF DATA AND FUNCTIONS
1.1 Describing Single-Variable Data
2
Visualizing Single-Variable Data 2 Mean and Median: What is “Average” Anyway? An Introduction to Algebra Aerobics 7
6
1.2 Describing Relationships between Two Variables Visualizing Two-Variable Data 13 Constructing a “60-Second Summary” Using Equations to Describe Change
13
14 16
1.3 An Introduction to Functions
22
What is a Function? 22 Representing Functions in Multiple Ways 23 Independent and Dependent Variables 24 When is a Relationship Not a Function? 24
1.4 The Language of Functions Function Notation Domain and Range
29
29 33
1.5 Visualizing Functions
39
Is There a Maximum or Minimum Value? 39 Is the Function Increasing or Decreasing? 40 Is the Graph Concave Up or Concave Down? Getting the Big Idea 42 CHAPTER SUMMARY
40
49
C H E C K Y O U R U N D E R S TA N D I N G
50
C H A P T E R 1 R E V I E W: P U T T I N G I T A L L T O G E T H E R
52
E X P L O R AT I O N 1 . 1 Collecting, Representing, and Analyzing Data E X P L O R AT I O N 1 . 2 Picturing Functions
61
E X P L O R AT I O N 1 . 3 Deducing Formulas to Describe Data
CHAPTER 2
RATES OF CHANGE AND 2.1 Average Rates of Change 66 Describing Change in the U.S. Population over Time LINEAR FUNCTIONS Defining the Average Rate of Change 67 Limitations of the Average Rate of Change
66
68
2.2 Change in the Average Rate of Change
71
2.3 The Average Rate of Change is a Slope
76
Calculating Slopes
76
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63
58
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2.4 Putting a Slant on Data
82
Slanting the Slope: Choosing Different End Points Slanting the Data with Words and Graphs 83
82
2.5 Linear Functions: When Rates of Change Are Constant What If the U.S. Population Had Grown at a Constant Rate? Real Examples of a Constant Rate of Change 87 The General Equation for a Linear Function 90
2.6 Visualizing Linear Functions The Effect of b The Effect of m
87
94
94 94
2.7 Finding Graphs and Equations of Linear Functions
99
Finding the Graph 99 Finding the Equation 100
2.8 Special Cases
108
Direct Proportionality 108 Horizontal and Vertical Lines 110 Parallel and Perpendicular Lines 112 Piecewise Linear Functions 114 The absolute value function 115 Step functions 117
2.9 Constructing Linear Models for Data
122
Fitting a Line to Data: The Kalama Study 123 Reinitializing the Independent Variable 125 Interpolation and Extrapolation: Making Predictions CHAPTER SUMMARY
126
131
C H E C K Y O U R U N D E R S TA N D I N G
132
C H A P T E R 2 R E V I E W: P U T T I N G I T A L L T O G E T H E R E X P L O R AT I O N 2 . 1 Having It Your Way
134
139
E X P L O R AT I O N 2 . 2 A Looking at Lines with the Course Software E X P L O R AT I O N 2 . 2 B Looking at Lines with a Graphing Calculator
AN EXTENDED EXPLORATION: LOOKING FOR LINKS BETWEEN EDUCATION AND EARNINGS
Using U.S. Census Data
148
Is There a Relationship between Education and Earnings? Regression Lines: How Good a Fit? 151
148
Interpreting Regression Lines: Correlation vs. Causation 154
Do Earnings Depend on Age? 155 Do Earnings Depend upon Gender? How Good are the Data? 157 How Good is the Analysis? 157 EXPLORING ON YOUR OWN EXERCISES
159
142
146
Summarizing the Data: Regression Lines
Raising More Questions
141
155
157
153
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CHAPTER 3
WHEN LINES MEET: LINEAR SYSTEMS
3.1 Systems of Linear Equations
166
An Economic Comparison of Solar vs. Conventional Heating Systems
3.2 Finding Solutions to Systems of Linear Equations Visualizing Solutions 171 Strategies for Finding Solutions 172 Linear Systems in Economics: Supply and Demand
171
176
3.3 Reading between the Lines: Linear Inequalities Above and Below the Line 183 Manipulating Inequalities 184 Reading between the Lines 185 Breakeven Points: Regions of Profit or Loss
166
183
187
3.4 Systems with Piecewise Linear Functions: Tax Plans
193
Graduated vs. Flat Income Tax 193 Comparing the Two Tax Models 195 The Case of Massachusetts 196 CHAPTER SUMMARY
201
C H E C K Y O U R U N D E R S TA N D I N G
202
C H A P T E R 3 R E V I E W: P U T T I N G I T A L L T O G E T H E R
204
E X P L O R AT I O N 3 . 1 Flat vs. Graduated Income Tax: Who Benefits?
209
CHAPTER 4
THE LAWS OF EXPONENTS AND LOGARITHMS: MEASURING THE UNIVERSE
4.1 The Numbers of Science: Measuring Time and Space Powers of 10 and the Metric System Scientific notation 214
212
4.2 Positive Integer Exponents
218
Exponent Rules 219 Common Errors 221 Estimating Answers 223
4.3 Negative Integer Exponents Evaluating AabB 2n 227 4.4 Converting Units
226
230
Converting Units within the Metric System 230 Converting between the Metric and English Systems Using Multiple Conversion Factors 231
4.5 Fractional Exponents
235 1
Square Roots: Expressions of the Form a /2 235 1 nth Roots: Expressions of the Form a /n 237 Rules for Radicals 238 m Fractional Powers: Expressions of the Form a /n
4.6 Orders of Magnitude
239
242
Comparing Numbers of Widely Differing Sizes
242
231
212
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Orders of Magnitude 242 Graphing Numbers of Widely Differing Sizes: Log Scales
4.7 Logarithms Base 10
248
Finding the Logarithms of Powers of 10 248 Finding the Logarithm of Any Positive Number Plotting Numbers on a Logarithmic Scale 251 CHAPTER SUMMARY
244
250
255
C H E C K Y O U R U N D E R S TA N D I N G
256
C H A P T E R 4 R E V I E W: P U T T I N G I T A L L T O G E T H E R
257
E X P L O R AT I O N 4 . 1 The Scale and the Tale of the Universe
260
E X P L O R AT I O N 4 . 2 Patterns in the Positions and Motions of the Planets
CHAPTER 5
GROWTH AND DECAY: AN INTRODUCTION TO EXPONENTIAL FUNCTIONS
5.1 Exponential Growth
266
The Growth of E. coli Bacteria 266 The General Exponential Growth Function 267 Looking at Real Growth Data for E. coli Bacteria
268
5.2 Linear vs. Exponential Growth Functions
271
Linear vs. Exponential Growth 271 Comparing the Average Rates of Change 273 A Linear vs. an Exponential Model through Two Points 274 Identifying Linear vs. Exponential Functions in a Data Table
5.3 Exponential Decay
275
279
The Decay of Iodine-131 279 The General Exponential Decay Function
279
5.4 Visualizing Exponential Functions The Effect of the Base a 284 The Effect of the Initial Value C Horizontal Asymptotes 287
284
285
5.5 Exponential Functions: A Constant Percent Change Exponential Growth: Increasing by a Constant Percent Exponential Decay: Decreasing by a Constant Percent Revisiting Linear vs. Exponential Functions 293
290 291
5.6 Examples of Exponential Growth and Decay
298
Half-Life and Doubling Time 299 The “rule of 70” 301 Compound Interest Rates 304 The Malthusian Dilemma 308 Forming a Fractal Tree 309
5.7 Semi-log Plots of Exponential Functions CHAPTER SUMMARY
316
320
C H E C K Y O U R U N D E R S TA N D I N G
321
C H A P T E R 5 R E V I E W: P U T T I N G I T A L L T O G E T H E R E X P L O R AT I O N 5 . 1 Properties of Exponential Functions
322 327
290
262
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CHAPTER 6
LOGARITHMIC LINKS: LOGARITHMIC AND EXPONENTIAL FUNCTIONS
6.1 Using Logarithms to Solve Exponential Equations Estimating Solutions to Exponential Equations Rules for Logarithms 331 Solving Exponential Equations 336
330
330
6.2 Base e and Continuous Compounding
340
What is e? 340 Continuous Compounding 341 Exponential Functions Base e 344
6.3 The Natural Logarithm 6.4 Logarithmic Functions
349 352
The Graphs of Logarithmic Functions 353 The Relationship between Logarithmic and Exponential Functions Logarithmic vs. exponential growth 354 Logarithmic and exponential functions are inverses of each other Applications of Logarithmic Functions 357 Measuring acidity: The pH scale 357 Measuring noise: The decibel scale 359
6.5 Transforming Exponential Functions to Base e Converting a to ek
354 355
363
364
6.6 Using Semi-log Plots to Construct Exponential Models for Data Why Do Semi-Log Plots of Exponential Functions Produce Straight Lines? CHAPTER SUMMARY
369
369
374
C H E C K Y O U R U N D E R S TA N D I N G
375
C H A P T E R 6 R E V I E W: P U T T I N G I T A L L T O G E T H E R
377
E X P L O R AT I O N 6 . 1 Properties of Logarithmic Functions
380
CHAPTER 7
POWER FUNCTIONS
7.1 The Tension between Surface Area and Volume
384
Scaling Up a Cube 384 Size and Shape 386
7.2 Direct Proportionality: Power Functions with Positive Powers Direct Proportionality 390 Properties of Direct Proportionality 390 Direct Proportionality with More Than One Variable
7.3 Visualizing Positive Integer Powers The Graphs of ƒ(x) 5 x and g(x) 5 x Odd vs. Even Powers 399 2
3
389
393
397
397
Symmetry 400
The Effect of the Coefficient k
400
7.4 Comparing Power and Exponential Functions
405
Which Eventually Grows Faster, a Power Function or an Exponential Function?
405
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7.5 Inverse Proportionality: Power Functions with Negative Integer Powers 409 Inverse Proportionality 410 Properties of Inverse Proportionality Inverse Square Laws 415
411
7.6 Visualizing Negative Integer Power Functions The Graphs of ƒ(x) 5 x and g(x) 5 x Odd vs. Even Powers 422 21
22
420
420
Asymptotes 423 Symmetry 423
The Effect of the Coefficient k
423
7.7 Using Logarithmic Scales to Find the Best Functional Model Looking for Lines 429 Why is a Log-Log Plot of a Power Function a Straight Line? 430 Translating Power Functions into Equivalent Logarithmic Functions Analyzing Weight and Height Data 431
429
430
Using a standard plot 431 Using a semi-log plot 431 Using a log-log plot 432
Allometry: The Effect of Scale CHAPTER SUMMARY
434
442
C H E C K Y O U R U N D E R S TA N D I N G
443
C H A P T E R 7 R E V I E W: P U T T I N G I T A L L T O G E T H E R E X P L O R AT I O N 7 . 1 Scaling Objects
444
448
E X P L O R AT I O N 7 . 2 Predicting Properties of Power Functions
450
E X P L O R AT I O N 7 . 3 Visualizing Power Functions with Negative Integer Powers
CHAPTER 8
QUADRATICS, POLYNOMIALS, AND BEYOND
8.1 An Introduction to Quadratic Functions The Simplest Quadratic 454 Designing parabolic devices 455 The General Quadratic 456 Properties of Quadratic Functions 457 Estimating the Vertex and Horizontal Intercepts
454
459
8.2 Finding the Vertex: Transformations of y = x2 Stretching and Compressing Vertically 464 Reflections across the Horizontal Axis 464 Shifting Vertically and Horizontally 465 Using Transformations to Get the Vertex Form 468 Finding the Vertex from the Standard Form 470 Converting between Standard and Vertex Forms 472
8.3 Finding the Horizontal Intercepts
480
Using Factoring to Find the Horizontal Intercepts Factoring Quadratics 482
481
463
451
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Using the Quadratic Formula to Find the Horizontal Intercepts The discriminant 485 Imaginary and complex numbers 487 The Factored Form 488
484
8.4 The Average Rate of Change of a Quadratic Function 8.5 An Introduction to Polynomial Functions
xix
493
498
Defining a Polynomial Function 498 Visualizing Polynomial Functions 500 Finding the Vertical Intercept 502 Finding the Horizontal Intercepts 503
8.6 New Functions from Old
510
Transforming a Function 510 Stretching, compressing and shifting Reflections 511 Symmetry 512
8.7 Combining Two Functions
510
521
The Algebra of Functions 521 Rational Functions: The Quotient of Two Polynomials Visualizing Rational Functions 525
8.8 Composition and Inverse Functions Composing Two Functions 531 Composing More Than Two Functions 533 Inverse Functions: Returning the Original Value A Final Example 540 CHAPTER SUMMARY
524
531
534
547
C H E C K Y O U R U N D E R S TA N D I N G
548
C H A P T E R 8 R E V I E W: P U T T I N G I T A L L T O G E T H E R
550
E X P L O R AT I O N 8 . 1 How Fast Are You? Using a Ruler to Make a Reaction Timer
AN EXTENDED EXPLORATION: THE MATHEMATICS OF MOTION
The Scientific Method
560
The Free-Fall Experiment
560
Interpreting Data from a Free-Fall Experiment 561 Deriving an Equation Relating Distance and Time 563 Returning to Galileo’s Question 565 Velocity: Change in Distance over Time 565 Acceleration: Change in Velocity over Time 566 Deriving an Equation for the Height of an Object in Free Fall Working with an Initial Upward Velocity 569
568
C O L L E C T I N G A N D A N A L Y Z I N G D ATA F R O M A F R E E FA L L E X P E R I M E N T EXERCISES
573
555
570
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Student Data Tables for Exploration 2.1
SOLUTIONS
For Algebra Aerobics, odd-numbered Exercises, Check Your Understanding, and odd-numbered Problems in the Chapter Review: Putting It All Together. (All solutions are grouped by chapter.) 583
INDEX
691
579
See www.wiley.com/college/kimeclark for Course Software, Anthology of Readings, Excel and Graph Link data files, and the Graphing Calculator Manual. The Instructor’s Teaching Manual and Instructor’s Solutions Manual are also available on the site, but password protected to restrict access to Instructors.
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MAKING SENSE OF DATA AND FUNCTIONS OVERVIEW How can you describe patterns in data? In this chapter we explore how to use graphs to visualize the shape of single-variable data and to show changes in two-variable data. Functions, a fundamental concept in mathematics, are introduced and used to model change. After reading this chapter, you should be able to • describe patterns in single- and two-variable data • construct a “60-second summary” • define a function and represent it in multiple ways • identify properties of functions • use the language of functions to describe and create graphs
1
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1.1 Describing Single-Variable Data
Visualizing Single-Variable Data Humans are visual creatures. Converting data to an image can make it much easier to recognize patterns. Bar charts: How well educated are Americans? Categorical data are usually displayed with a bar chart. Typically the categories are listed on the horizontal axis. The height of the bar above a single category tells you either the frequency count (the number of observations that fall into that category) or the relative frequency (the percentage of total observations). Since the relative size of the bars is the same using either frequency or relative frequency counts, we often put the two scales on different vertical axes of the same chart. For example, look at the vertical scales on the left- and right-hand sides of Figure 1.1, a bar chart of the educational attainment of Americans age 25 or older in 2004. American Educational Attainment (2004) 60,000,000
30%
50,000,000 40,000,000
20%
30,000,000 20,000,000
10%
Doctoral degree
Professional degree
Master’s degree
Bachelor’s degree
Associates degree
Some college(no degree)
High school degree
Grades 9–11
0
Grades K–8
10,000,000
Percentage of Americans
70,000,000 Number of Americans
Exploration 1.1 provides an opportunity to collect your own data and to think about issues related to classifying and interpreting data.
This course starts with you. How would you describe yourself to others? Are you a 5-foot 6-inch, black, 26-year-old female studying biology? Or perhaps you are a 5-foot 10-inch, Chinese, 18-year-old male English major. In statistical terms, characteristics such as height, race, age, and major that vary from person to person are called variables. Information collected about a variable is called data.1 Some variables, such as age, height, or number of people in your household, can be represented by a number and a unit of measure (such as 18 years, 6 feet, or 3 people). These are called quantitative variables. For other variables, such as gender or college major, we use categories (such as male and female or biology and English) to classify information. These are called categorical (or qualitative) data. The dividing line between classifying a variable as categorical or quantitative is not always clear-cut. For example, you could ask individuals to list their years of education (making education a quantitative variable) or ask for their highest educational category, such as college or graduate school (making education a categorical variable). Many of the controversies in the social sciences have centered on how particular variables are defined and measured. For nearly two centuries, the categories used by the U.S. Census Bureau to classify race and ethnicity have been the subject of debate. For example, Hispanic used to be considered a racial classification. It is now considered an ethnic classification, since Hispanics can be black, or white, or any other race.
0%
Education level
Figure 1.1 Bar chart showing the education levels for Americans age 25 or older. Source: U.S. Bureau of the Census, www.census.gov. 1
Data is the plural of the Latin word datum (meaning “something given”)—hence one datum, two data.
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The vertical scale on the left tells us the number (the frequency count) of Americans who fell into each educational category. For example, in 2004 approximately 60 million Americans age 25 or older had a high school degree but never went on to college. It’s often more useful to know the percentage (the relative frequency) of all Americans who have only a high school degree. Given that in 2004 the number of people 25 years or older was approximately 186,877,000 and the number who had only a high school degree was approximately 59,810,000, then the percentage of those with only a high school degree was Number with only a high school degree 59,810,000 5 Total number of people age 25 or older 186,877,000 < 0.32 sin decimal formd or 32% The vertical scale on the right tells us the percentage (relative frequency). Using this scale, the percentage of Americans with only a high school degree was about 32%, which is consistent with our calculation. E X A M P L E
1
What does the bar chart tell us? a. Using Figure 1.1, estimate the number and percentage of people age 25 or older who have bachelor’s degrees, but no further advanced education. b. Estimate the total number of people and the percentage of the total population age 25 or older who have at least a high school education. c. What doesn’t the bar chart tell us? d. Write a brief summary of educational attainment in the United States.
S O L U T I O N
a. Those with bachelor’s degrees but no further education number about 34 million, or 18%. b. Those who have completed a high school education include everyone with a high school degree up to a Ph.D. We could add up all the numbers (or percentages) for each of those seven categories. But it’s easier to subtract from the whole those who do not meet the conditions, that is, subtract those with either a grade school or only some high school education from the total population (people age 25 or older) of about 187 million. Grade School 1 Some High School 5 Total without High School Degree Number (approx.) Percentage (approx.)
12 million 1 16 million 6% 1 9%
5 28 million 5 15%
The number of Americans (age 25 or over) with a high school degree is about 187 million 2 28 million 5 159 million. The corresponding percentage is about 100% 2 15% 5 85%. So more than four out of five Americans 25 years or older have completed high school. c. The bar chart does not tell us the total size of the population or the total number (or percentage) of Americans who have a high school degree. For example, if we include younger Americans between age 18 and age 25, we would expect the percentage with a high school degree to be higher. d. About 85% of adult Americans (age 25 or older) have at least a high school education. The breakdown for the 85% includes 32% who completed high school but did not go on, 43% who have some college (up to a bachelor’s degree), and about 10% who have graduate degrees. This is not surprising, since the United States population ranks among the mostly highly educated in the world.
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An important aside: What a good graph should contain When you encounter a graph in an article or you produce one for a class, there are three elements that should always be present: 1. An informative title that succinctly describes the graph 2. Clearly labeled axes (or a legend) including the units of measurement (e.g., indicating whether age is measured in months or years) 3. The source of the data cited in the data table, in the text, or on the graph
See the program “F1: Histograms.”
Histograms: What is the distribution of ages in the U.S. population? A histogram is a specialized form of a bar chart that is used to visualize single-variable quantitative data. Typically, the horizontal axis on a histogram is a subset of the real numbers with the unit (representing, for example, number of years) and the size of each interval marked. The intervals are usually evenly spaced to facilitate comparisons (e.g., placed every 10 years). The size of the interval can reveal or obscure patterns in the data. As with a bar chart, the vertical axis can be labeled with a frequency or a relative frequency count. For example, the histogram in Figure 1.2 shows the distribution of ages in the United States in 2005.
U.S. Population Age Distribution (2005) 25,000,000 Number of people
20,000,000 15,000,000 10,000,000
95– 99 years
90– 94 years
85– 89 years
100 years and over
Age interval
80– 84 years
75– 79 years
70– 74 years
65– 69 years
60– 64 years
55– 59 years
50– 54 years
45– 49 years
40– 44 years
35– 39 years
30– 34 years
25– 29 years
20– 24 years
15– 19 years
10– 14 years
0
5– 9 years
5,000,000
Under 5 years
4
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Figure 1.2 Age distribution of the U.S. population in 5-year intervals. Source: U.S. Bureau of the Census, www.census.gov.
E X A M P L E
2
What does the histogram tell us? a. What 5-year age interval contains the most Americans? Roughly how many are in that interval? (Refer to Figure 1.2.) b. Estimate the number of people under age 20. c. Construct a topic sentence for a report about the U.S. population.
S O L U T I O N
a. The interval from 40 to 44 years contains the largest number of Americans, about 23 million. b. The sum of the frequency counts for the four intervals below age 20 is about 80 million. c. According to the U.S. Census Bureau 2005 data, the number of Americans in each 5-year age interval remained fairly flat up to age 40, peaked between ages 40 to 50, then fell in a gradual decline.
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5
Describe the age distribution for Tanzania, one of the poorest countries in the world (Figure 1.3). Tanzanian Population Age Distribution (2005) 7,000,000 6,000,000
Number of people
5,000,000 4,000,000 3,000,000 2,000,000
80 +
75– 79
70– 74
65– 69
60– 64
55– 59
50– 54
45– 49
40– 44
35– 39
30– 34
25– 29
20– 24
15– 19
10– 14
5– 9
0
Under 5
1,000,000
Age interval
Figure 1.3 The age distribution in 2005 of the Tanzanian population in 5-year intervals. Source: U.S. Bureau of the Census, International Data Base, April 2005.
S O L U T I O N
?
SOMETHING TO THINK ABOUT
What are some trade-offs in using pie charts versus histograms?
The age distributions in Tanzania and the United States are quite different. Tanzania is a much smaller country and has a profile typical of a developing country; that is, each subsequent 5-year interval has fewer people. For example, there are about 6 million children 0 to 4 years old, but only about 5.3 million children age 5–9 years, a drop of over 10%. For ages 35 to 39 years, there are only about 1.7 million people, less than a third of the number of children between 0 and 4 years. Although the histogram gives a static picture of the Tanzanian population, the shape suggests that mortality rates are much higher than in the United States. Pie charts: Who gets the biggest piece? Both histograms and bar charts can be transformed into pie charts. For example, Figure 1.4 shows two pie charts of the U.S. and Tanzanian age distributions (both now divided into 20-year intervals). One advantage of using a pie chart is that it clearly shows the size of each piece relative to the whole. Hence, they are usually labeled with percentages rather than frequency counts. Ages of U.S. Population (2005)
Ages of Tanzanian Population (2005) 0.3% 4%
4% 13%
11%
28% Under 20 20–39 40–59
55%
60–79 28%
80+
29%
28 %
Figure 1.4 Two pie charts displaying information about the U.S. and Tanzanian
age distributions. Source: U.S. Bureau of the Census, www.census.gov.
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In the United States the first three 20-year age intervals (under 20, 20–39, and 40–59 years) are all approximately equal in size and together make up about 84% of the population. Those 60 and older represent 17% of the population. Note that the percentages add up to more than 100% due to rounding. In Tanzania, the proportions are entirely different. Over half of the population are under 20 years and more than 80% are under 40 years old. Those 60 and older make up less than 5% of Tanzania’s population.
Mean and Median: What Is “Average” Anyway? In 2005 the U.S. Bureau of the Census reported that the mean age for Americans was 37.2 and the median age was 36.7 years. The Mean and Median The mean is the sum of a list of numbers divided by the number of terms in the list. The median is the middle value of an ordered numerical list; half the numbers lie at or below the median and half at or above it.
?
SOMETHING TO THINK ABOUT
If someone tells you that in his town “all of the children are above average,” you might be skeptical. (This is called the “Lake Wobegon effect.”) But could most (more than half) of the children be above average? Explain.
See the reading “The Median Isn’t the Message” to find out how an understanding of the median gave renewed hope to the renowned scientist Stephen J. Gould when he was diagnosed with cancer.
The mean age of 37.2 represents the sum of the ages of every American divided by the total number of Americans. The median age of 36.7 means that if you placed all the ages in order, 36.7 would lie right in the middle; that is, half of Americans are younger than or equal to 36.7 and half are 36.7 or older. In the press you will most likely encounter the word “average” rather than the term “mean” or “median.”2 The term “average” is used very loosely. It usually represents the mean, but it could also represent the median or something much more vague, such as the “average” American household. For example, the media reported that: • The average American home now has more television sets than people. . . . There are 2.73 TV sets in the typical home and 2.55 people.3 • The average American family now owes more than $9,000 in credit debt. . . . and is averaging about seven cards.4 The significance of the mean and median The median divides the number of entries in a data set into two equal halves. If the median age in a large urban housing project is 17, then half the population is 17 or under. Hence, issues such as day care, recreation, and education should be high priorities with the management. If the median age is 55, then issues such as health care and wheelchair accessibility might dominate the management’s concerns. The median is unchanged by changes in values above and below it. For example, as long as the median income is larger than the poverty level, it will remain the same even if all poor people suddenly increase their incomes up to that level and everyone else’s income remains the same. The mean is the most commonly cited statistic in the news media. One advantage of the mean is that it can be used for calculations relating to the whole data set. Suppose a corporation wants to open a new factory similar to its other factories. If the managers know the mean cost of wages and benefits for employees, 2
The word “average” has an interesting derivation according to Klein’s etymological dictionary. It comes from the Arabic word awariyan, which means “merchandise damaged by seawater.” The idea being debated was that if your ships arrived with water-damaged merchandise, should you have to bear all the losses yourself or should they be spread around, or “averaged,” among all the other merchants? The words averia in Spanish, avaria in Italian, and avarie in French still mean “damage.” 3 Source: USA Today, www.usatoday.com/life/television/news/2006. 4 Source: Newsweek, www.msnbc.msn.com/id/14366431/site/newsweek/.
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they can make an estimate of what it will cost to employ the number of workers needed to run the new factory: total employee cost 5 smean cost for employeesd ? snumber of employeesd The mean, unlike the median, can be affected by a few extreme values called outliers. For example, suppose Bill Gates, founder of Microsoft and the richest man in the world, were to move into a town of 10,000 people, all of whom earned nothing. The median income would be $0, but the mean income would be in the millions. That’s why income studies usually use the median.
E X A M P L E
4
“Million-dollar Manhattan apartment? Just about average” According to a report cited on money.cnn.com, in 2006 the median price of purchasing an apartment in Manhattan was $880,000 and the mean price was $1.4 million. How could there be such a difference in price? Which value do you think better represents apartment prices in Manhattan?
S O L U T I O N
Apartments that sold for exorbitant prices (in the millions) could raise the mean above the median. If you want to buy an apartment in Manhattan, the median price is probably more important because it tells you that half of the apartments cost $880,000 or less.
An Introduction to Algebra Aerobics In each section of the text there are “Algebra Aerobics” with answers in the back of the book. They are intended to give you practice in the algebraic skills introduced in the section and to review skills we assume you have learned in other courses. These skills should provide a good foundation for doing the exercises at the end of each chapter. The exercises include more complex and challenging problems and have answers for only the odd-numbered ones. We recommend you work out these Algebra Aerobics practice problems and then check your solutions in the back of this book. The Algebra Aerobics are numbered according to the section of the book in which they occur.
Algebra Aerobics 1.1 1. Fill in Table 1.1. Round decimals to the nearest thousandth. Fraction
Decimal
Percent
7 12 0.025 2% 1 200 0.35 0.8% Table 1.1
2. Calculate the following: a. A survey reported that 80 people, or 16% of the group, were smokers. How many people were surveyed? b. Of the 236 students who took a test, 16.5% received a B grade. How many students received a B grade? c. Six of the 16 people present were from foreign countries. What percent were foreigners? 3. When looking through the classified ads, you found that 16 jobs had a starting salary of $20,000, 8 had a starting salary of $32,000, and 1 had a starting salary of $50,000. Find the mean and median starting salary for these jobs.5
Recall that given the list of numbers 9, 2, 22, 6, 5, the mean 5 the sum 9 1 2 1 (22) 1 6 1 5 divided by 5 (the number of items in the list) 5 20/5 5 4; the median 5 the middle number of the list in ascending order 22, 2, 5, 6, 9, which is 5. If the list had an even number of elements—for example, 22, 2, 5, 6—the median would be the mean of the middle two numbers on the ordered list, in this case (2 1 5)/2 5 7/2 5 3.5.
5
8
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4. Find the mean and median grade point average (GPA) from the data given in Table 1.2.
6. a. Fill in Table 1.3. Round your answers to the nearest whole number. Frequency Count
GPA
Frequency Count
Age
1.0 2.0 3.0 4.0
56 102 46 12
1–20 21–40 41–60 61–80
35 28
Total
137
Table 1.2
5. Figure 1.5 presents information about the Hispanic population in the United States from 2000 to 2005. Hispanic Population in the United States (2000–2005) (in millions) 42.7 35.3
2000
2005
12.5%
14.4%
Source: U.S. Bureau of the Census, www.census.gov.
a. What does the bar chart tell you that the pie chart does not? b. Using the bar and pie charts, what was the U.S. population in 2000? In 2005?
38
b. Calculate the percentage of the population who are over 40 years old. 7. Use Table 1.3 to create a histogram and pie chart. 8. From the histogram in Figure 1.6, create a frequency distribution table. Assume that the total number of people represented by the histogram is 1352. (Hint: Estimate the relative frequencies from the graph and then calculate the frequency count in each interval.) 35 30 25 20 15 10 5 0
Percentage of total population
Figure 1.5 Change in the Hispanic population in the United States.
Relative Frequency (%)
Table 1.3
Percentage
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1–20
21–40 41–60 Years
61–80
Figure 1.6 Distribution of ages (in
years).
9. Calculate the mean and median for the following data: a. $475, $250, $300, $450, $275, $300, $6000, $400, $300 b. 0.4, 0.3, 0.3, 0.7, 1.2, 0.5, 0.9, 0.4 10. Explain why the mean may be a misleading numerical summary of the data in Problem 9(a).
Exercises for Section 1.1 1. Internet use as reported by teenagers in 2006 in the United States is shown in the accompanying graph. a. What percentage of 13- to 17-year-old females spend at least 3 hours per day on the internet outside of school?
b. What percentage of 13- to 17-year-old males spend at least 3 hours per day on the internet outside of school? c. What additional information would you need in order to find out the percentage of 13 to 17 year olds who spend at least 3 hours per day on the internet?
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Where the Money Will Go in 2003
Time Spent Per Day on the Internet Outside of School 13–17 Year Olds in U.S.
other 4% give away 1% 30% save/invest
14.5%
More than five hours
6% spend/save
19.9%
More than four hours, but less than five hours
7.9% 6.8%
More than three hours, but less than four hours
21% spend 12.3% 13.2% 34% bills/debts
More than two hours, but less than three hours
Source: ABCNEWS/Washington Post poll. May, 2003.
19.5% 16.5%
More than one hour, but less than two hours
a. What is the largest category on which people say they will spend their rebates? Why does the category look so much larger than its actual relative size? b. What might make you suspicious about the numbers in this pie chart?
24.7% 25.5% 21.1%
Less than an hour
18.1%
4. The point spread in a football game is the difference between the winning team’s score and the losing team’s score. For example, in the 2004 Super Bowl game, the Patriots won with 32 points versus the Carolina Panthers’ 29 points. So the point spread was 3 points. a. In the accompanying bar chart, what is the interval with the most likely point spread in a Super Bowl? The least likely?
% of Respondents Male
Female
Chart 1 – Time Spent Online Outside of School Source: BURST Research, May 2006
2. The accompanying bar chart shows the five countries with the largest populations in 2006.
Point Spreads in 39 Super Bowl Games 16 14
Population (in millions)
1,400 1,200 1,000
12 Frequency
The Five Most Populous Countries (2006)
10 8 6 4
800
2
600
0 0–4
400 200 0
5–9
10–14 Point spread
15–19
Source: www.docsports.com/point-spreads-for-every-super-bowl.html. China
India
U.S.
Indonesia
Brazil
Source: CIA Factbook, www.cia.gov/cia.
a. What country has the largest population, and approximately what is its population, size? b. The population of India is projected in the near future to exceed the population of China. Given the current data, what is the minimum number of additional persons needed to make India’s population larger than China’s? c. The world population in 2006 was estimated to be about 6.5 billion. Approximately what percentage of the world’s population live in China? In India? In the United States? 3. In 2003 some taxpayers received $300–600 tax rebates. Congress approved this spending as a means to stimulate the economy. According to a May 2003 ABC News/Washington Post poll, the accompanying pie chart shows how people would use the money.
b. What percentage of these Super Bowl games had a point spread of 9 or less? Of 14 or less? 5. Given here is a table of salaries taken from a survey of recent graduates (with bachelor degrees) from a well-known university in Pittsburgh. Salary (in thousands)
Number of Graduates Receiving Salary
21–25 26–30 31–35 36–40 41–45 46–50
2 3 10 20 9 1
a. How many graduates were surveyed? b. Is this quantitative or qualitative data? Explain.
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5. (continued) c. What is the relative frequency of people having a salary between $26,000 and $30,000? d. Create a histogram of the data. 6. The accompanying bar chart shows the predictions of the U.S. Census Bureau about the future racial composition of American society. Hispanic origin may be of any race, so the other categories may include people of Hispanic origin. Percentage of U.S. Population by Race and Ethnicity (2001–2050 est) 90% 80%
2001 2010 2020 2050 (est)
Percent
70% 60%
a. In what single category did Americans spend the largest percentage of their income? Estimate this percentage. b. According to this chart, if an American family has an income of $35,000, how much of it would be spent on food? c. If you were to write a newspaper article to accompany this pie chart, what would your opening topic sentence be? 8. Attendance at a stadium for the last 30 games of a college baseball team is listed as follows: 5072 2341 1980 1376 3654
3582 2478 1784 978 3845
2504 3602 1493 2035 673
4834 5435 3674 1239 2745
2456 3903 4593 2456 3768
3956 4535 5108 5189 5227
Create a histogram to display these data. Decide how large the intervals should be to illustrate the data well without being overly detailed.
50% 40% 30% 20% 10% 0% American Asian and Indian, Pacific Eskimo, Aleut Islander
White
Black
Hispanic origin (of any race)
Sources: U.S. Bureau of the Census, Statistical Abstract of the United States: 2002.
a. Estimate the following percentages: i. Asian and Pacific Islanders in the year 2050 ii. Combined white and black population in the year 2020 iii. Non-Hispanic population in the year 2001 b. The U.S. Bureau of the Census has projected that there will be approximately 392,031,000 people in the United States in the year 2050. Approximately how many people will be of Hispanic origin in the year 2050? c. Write a topic sentence describing the overall trend. 7. Shown is a pie chart of America’s spending patterns at the end of 2006. How Americans Spend Their Money How the average American spends $100, as measured in late 2006. The Labor Department uses this survey-—along with price samples—to calculate the inflation rate. $4.42 $5.55
$1.12 Housing
$6.50
Transportation Food $36.16 $10.92
Personal insurance and pensions Health care Entertainment
$14.38
9. a. Compute the mean and median for the list: 5, 18, 22, 46, 80, 105, 110. b. Change one of the entries in the list in part (a) so that the median stays the same but the mean increases. 10. Suppose that a church congregation has 100 members, each of whom donates 10% of his or her income to the church. The church collected $250,000 last year from its members. a. What was the mean contribution of its members? b. What was the mean income of its members? c. Can you predict the median income of its members? Explain your answer. 11. Suppose that annual salaries in a certain corporation are as follows: Level I (30 employees) Level II (8 employees) Level III (2 employees)
$18,000 $36,000 $80,000
Find the mean and median annual salary. Suppose that an advertisement is placed in the newspaper giving the average annual salary of employees in this corporation as a way to attract applicants. Why would this be a misleading indicator of salary expectations? 12. Suppose the grades on your first four exams were 78%, 92%, 60%, and 85%. What would be the lowest possible average that your last two exams could have so that your grade in the class, based on the average of the six exams, is at least 82%? 13. Read Stephen Jay Gould’s article “The Median Isn’t the Message” and explain how an understanding of statistics brought hope to a cancer victim.
Apparel and services $20.95
Reading materials and tobacco products
Sources: U.S. Bureau of the Census,
Statistical Abstract of the United States: 2006.
14. a. On the first quiz (worth 25 points) given in a section of college algebra, one person received a score of 16, two people got 18, one got 21, three got 22, one got 23, and one got 25. What were the mean and median of the quiz scores for this group of students?
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14. (continued) b. On the second quiz (again worth 25 points), the scores for eight students were 16, 17, 18, 20, 22, 23, 25, and 25. i. If the mean of the scores for the nine students was 21, then what was the missing score? ii. If the median of the scores was 22, then what are possible scores for the missing ninth student? 15. Why is the mean age larger than the median age in the United States? What prediction would you make for your State? What predictions would you make for other countries? Check your predictions with data from the U.S. Census Bureau at www.census.gov. 16. Up to and including George W. Bush, the ages of the last 15 presidents when they first took office6 were 56, 55, 51, 54, 51, 60, 62, 43, 55, 56, 52, 69, 64, 46, 54. a. Find the mean and median ages of the past 15 presidents when they took office. b. If the mean age of the past 16 presidents is 54.94, at what age did the missing president take office? c. Beginning with age 40 and using 5-year intervals, find the frequency count for each age interval. d. Create a frequency histogram using your results from part (c). 17. Herb Caen, a Pulitzer Prize–winning columnist for the San Francisco Chronicle, remarked that a person moving from state A to state B could raise the average IQ in both states. Is he right? Explain. 18. Why do you think most researchers use median rather than mean income when studying “typical” households? 19. According to the 2000 U.S. Census, the median net worth of American families was $55,000 and the mean net worth was $282,500. How could there be such a wide discrepancy? 20. Read the CHANCE News article and explain why the author was concerned. 21. The Greek letter (called sigma) is used to represent the sum of all of the terms of a certain group. Thus, a1 1 a2 1 a3 1 · · · 1 an can be written as n
g ai i51
which means to add together all of the values of ai from a1 to an. a. Using notation, write an algebraic expression for the mean of the five numbers x 1, x 2, x 3, x 4, x 5. b. Using notation, write an algebraic expression for the mean of the n numbers t 1, t 2, t 3, . . . , t n. c. Evaluate the following sum: 5
g 2k k51
6
http://www.campvishus.org/PresAgeDadLeft.htm#AgeOffice.
Describing Single-Variable Data
11
22. The accompanying table gives the ages of students in a mathematics class. Ages of Students Age Interval
Frequency Count
15–19 20–24 25–29 30–34 35–39 40–44 45–49
2 8 4 3 2 1 1
Total
21
a. Use this information to estimate the mean age of the students in the class. Show your work. (Hint: Use the mean age of each interval.) b. What is the largest value the actual mean could have? The smallest? Why? 23. (Use of calculator or other technology recommended.) Use the following table to generate an estimate of the mean age of the U.S. population. Show your work. (Hint: Replace each age interval with an age approximately in the middle of the interval.) Ages of U.S. Population in 2004 Age (years) Under 10 10–19 20–29 30–39 40–49 50–59 60–74 75–84 85 and over Total
Population (thousands)
DA T A
USPOPAGE
39,677 41,875 40,532 41,532 45,179 35,986 31,052 12,971 4,860 293,655
Source: U.S. Bureau of the Census, Statistical Abstract of the United States: 2006.
24. An article titled “Venerable Elders” (The Economist, July 24, 1999) reported that “both Democratic and Republican images are selective snapshots of a reality in which the median net worth of households headed by Americans aged 65 or over is around double the national average—but in which a tenth of such households are also living in poverty.” What additional statistics would be useful in forming an opinion on whether elderly Americans are wealthy or poor compared with Americans as a whole? 25. Estimate the mean and median from the given histogram. (See hint in Exercise 23.) The program “F4: Measures of Central Tendency” in FAM1000 Census Graphs can help you understand the mean and median and their relationship to histograms.
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Number of students
Monthly Allowance of Junior High School Students 14 12 10 8 6 4 2 0
10–19
0–9
20–29 Dollars
30–39
27. (Computer and course software required.) Open up the program “F1: Histograms” in FAM1000 Census Graphs in the course software. The 2006 U.S. Census data on 1000 randomly selected U.S. individuals and their families are imbedded in this program. You can use it to create histograms for education, age, and different measures of income. Try using different interval sizes to see what patterns emerge. Decide on one variable (say education) and compare the histograms of this variable for different groups of people. For example, you could compare education histograms for men and women or for people living in two different regions of the country. Pick a comparison that you think is interesting. Create a possible headline for these data. Describe three key features that support your headline. 28. Population pyramids are a type of chart used to depict the overall age structure of a society. Use the accompanying population pyramids for the United States to answer the following questions. United States: 2005
Female 85+ 80–84 75–79 70–74 65–69 60–64 55–59 50–54 45–49 40–44 35–39 30–34 25–29 20–24 15–19 10–14 5–9 0–4
40–49
26. Choose a paragraph of text from any source and construct a histogram of word lengths (the number of letters in the word). If the same word appears more than once, count it as many times as it appears. You will have to make some reasonable decisions about what to do with numbers, abbreviations, and contractions. Compute the mean and median word lengths from your graph. Indicate how you would expect the graph to be different if you used: a. A children’s book c. A medical textbook b. A work of literature
Male
United States: 2050 (projected) Male
16 14 12 10 8
6
4
2 0
6
4
2
0
0
2
6
8 10 12 14 16
4
6
8 10 12 14 16
a. Estimate the number of: i. Males who were between the ages 35 and 39 years in 2005. ii. Females who were between the ages 55 and 59 years in 2005. iii. Males 85 years and older in the year 2050; females 85 years and older in the year 2050. iv. All males and females between the ages of 0 and 9 years in the year 2050. b. Describe two changes in the distribution of ages from the year 2005 to the predictions for 2050. 29. The accompanying population pyramid shows the age structure in Ghana, a developing country in Africa, for 2005. The previous exercise contains a population pyramid for the United States, an industrialized nation, for 2005. Describe three major differences in the distribution of ages in these two countries in 2005. Ghana: 2005 Male
Female
Female
4
2
Source: U.S. Bureau of the Census, International Data Base, www.census.gov.
85+ 80–84 75–79 70–74 65–69 60–64 55–59 50–54 45–49 40–44 35–39 30–34 25–29 20–24 15–19 10–14 5–9 0–4 16 14 12 10 8
0
Population (in millions)
2.0
1.5
1.0
0.5
100+ 95–99 90–94 85–89 80–84 75–79 70–74 65–69 60–64 55–59 50–54 45–49 40–44 35–39 30–34 25–29 20–24 15–19 10–14 5–9 0–4 0.0 0.0
0.5
1.0
1.5
2.0
Population (in millions)
Population (in millions)
Source: U.S. Bureau of Census, International Data Base, www.census.gov.
Source: U.S. Census Bureau, International Data Base, www.census.gov.
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1.2 Describing Relationships between Two Variables By looking at two-variable data, we can learn how change in one variable affects change in another. How does the weight of a child determine the amount of medication prescribed by a pediatrician? How does median age or income change over time? In this section we examine how to describe these changes with graphs, data tables, written descriptions, and equations.
Visualizing Two-Variable Data E X A M P L E
1
Scatter Plots Table 1.4 shows data for two variables, the year and the median age of the U.S. population. Plot the data in Table 1.4 and then use your graph to describe the changes in the U.S. median age over time. Median Age of the U.S. Population, 1850–2050*
DA T A
MEDAGE Excel and graph link files for the median age data are called MEDAGE.
Median Age
Year
Median Age
1850 1860 1870 1880 1890 1900 1910 1920 1930 1940
18.9 19.4 20.2 20.9 22.0 22.9 24.1 25.3 26.4 29.0
1950 1960 1970 1980 1990 2000 2005 2010 2025 2050
30.2 29.5 28.0 30.0 32.8 35.3 36.7 36.0 37.5 38.1
Table 1.4 *Data for 2010–2050 are projected. Source: U.S., Bureau of the Census, Statistical Abstract of the United States: 1, 2006.
In Table 1.4 we can think of a year and its associated median age as an ordered pair of the form (year, median age). For example, the first row corresponds to the ordered pair (1850, 18.9) and the second row corresponds to (1860, 19.4). Figure 1.7 shows a scatter plot of the data. The graph is called a time series because it shows changes over time. In newspapers and magazines, the time series is the most frequently used form of data graphic.7 40 35 Median age (in years)
S O L U T I O N
Year
30 25 20 15 10 5 0 1850
1890
1930
1970
2010
2050
Year
Figure 1.7 Median age of U.S. population over time. 7
Edward Tufte in The Visual Display of Information (Cheshire, Conn.: Graphics Press, 2001, p. 28) reported on a study that found that more than 75% of all graphics published were time series.
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Our graph shows that the median age of the U.S. population grew quite steadily for one hundred years, from 1850 to 1950. Although the median age decreased between 1950 and 1970, since 1970 it has continued to increase. From 1850 to the present, the median age nearly doubled, and projections for 2025 and 2050 indicate continued increases, though at a slower pace.
SOMETHING TO THINK ABOUT
What are some of the trade-offs in using the median instead of the mean age to describe changes over time?
Constructing a “60-Second Summary” To communicate effectively, you need to describe your ideas succinctly and clearly. One tool for doing this is a “60-second summary”—a brief synthesis of your thoughts that could be presented in one minute. Quantitative summaries strive to be straightforward and concise. They often start with a topic sentence that summarizes the key idea, followed by supporting quantitative evidence. After you have identified a topic you wish to write about or present orally, some recommended steps for constructing a 60-second summary are: • • • • • •
Collect relevant information (possibly from multiple sources, including the Internet). Search for patterns, taking notes. Identify a key idea (out of possibly many) that could provide a topic sentence. Select evidence and arguments that support your key idea. Examine counterevidence and arguments and decide if they should be included. Construct a 60-second summary, starting with your topic sentence.
You will probably weave back and forth among the steps in order to refine or modify your ideas. You can help your ideas take shape by putting them down on paper. Quantitative reports should not be written in the first person. For example, you might say something like “The data suggest that . . .” rather than “I found that the data . . .” E X A M P L E
2
A 60-Second Summary The annual federal surplus (1) or deficit (2) since World War II is shown in Table 1.5 and Figure 1.8 (a scatter plot where the points have been connected). Construct a 60second summary describing the changes over time.
Federal Budget: Surplus (1) or Deficit (–) Billions of Dollars
Year
Billions of Dollars
1945 1950 1955 1960 1965 1970 1971 1972 1973 1974 1975 1976 1977 1978
2$48 2$3 2$3 $0 2$1 2$3 2$23 2$23 2$15 2$6 2$53 2$74 2$54 2$59
1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992
2$41 2$74 2$79 2$128 2$208 2$185 2$212 2$221 2$150 2$155 2$152 2$221 2$269 2$290
Year
Billions of Dollars
1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005
2$255 2$203 2$164 2$108 2$22 1$69 1$126 1$236 1$127 2$159 2$378 2$412 2$427
$300 Federal budget (billions of dollars)
Year
$200 $100 $0 –$100 –$200 –$300 –$400 –$500 1945
1955
1965
1975 Year
1985
1995
2005
Figure 1.8 Annual federal budget surplus or deficit in
billions of dollars.
Table 1.5 Source: U.S. office of Management and Budget.
S O L U T I O N
Between 1945 and 2005 the annual U.S. federal deficit moved from a 30-year stable period, with as little as $0 deficit, to a period of oscillations, leading in 2005 to the
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15
largest deficit ever recorded. From 1971 to 1992, the federal budget ran an annual deficit, which generally was getting larger until it reached almost $300 billion in 1992. From 1992 to 1997, the deficit steadily decreased, and from 1998 to 2001 there were relatively large surpluses. The maximum surplus occurred in 2000, when it reached $236 billion. But by 2002 the federal government was again running large deficits. In 2005 the deficit reached $427 billion, the largest recorded up to that time.
Algebra Aerobics 1.2a 1. The net worth of a household at any given time is the difference between assets (what you own) and liabilities (what you owe). Table 1.6 and Figure 1.9 show the median net worth of U.S. households, adjusted for inflation.8 Median Net Worth of Households (adjusted for inflation using year 2000 dollars) Year
Median Net Worth ($)
1984 1988 1991 1993 1995 1998 2000
50,018 49,855 44,615 43,567 44,578 49,932 55,000
b. The year that it is projected to reach 8 billion. c. The number of years it will take to grow from 4 to 8 billion. 3. Use Figure 1.10 to estimate the following projections for the year 2150. a. The total world population. b. The total populations of all the more developed countries. c. The total populations of all the less developed countries. d. Write a topic sentence about the estimated world population in 2150. 4. Use Figure 1.10 to answer the following: a. The world population in 2000 was how many times greater than the world population in 1900? What was the difference in population size? b. The world population in 2100 is projected to be how many times greater than the world population in 2000? What is the difference in population size? c. Describe the difference in the growth in world population in the twentieth century (1900–2000) versus the projected growth in the twenty-first century (2000–2150).
Table 1.6 Source: U.S. Bureau of the Census, www.census.gov.
Median Net Worth of Households (in 2000 dollars) $60,000 $50,000 $40,000 $30,000 $20,000 $10,000
World Population with Projections to 2150
2000
8
8
Less
6
developed countries
4 2
2150
2100
2050
More developed countries
0
2000
a. Write a few sentences about the trend in U.S. median household net worth. b. What additional information might be useful in describing the trend in median net worth? 2. Use Figure 1.10 to estimate: a. The year when the world population reached 4 billion.
10
1950
Figure 1.9
1900
Year
1850
1996
1800
1992
1750
1988
Population (in billions)
$0 1984
Figure 1.10 World population growth, 1750–2150 (est.). Source: Population Reference Bureau, www.prb.org.
“Constant dollars” is a measure used by economists to compare incomes and other variables in terms of purchasing power, eliminating the effects of inflation. To say the median income in 1986 was $37,546 in “constant 2000 dollars” means that the median income in 1986 could buy an amount of goods and services that would cost $37,546 to buy in 2000. The actual median income in 1986 (measured in what economists call “current dollars”) was much lower. Income corrected for inflation is sometimes called “real” income.
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Using Equations to Describe Change The relationship between two variables can also be described with an equation. An equation gives a rule on how change in the value of one variable affects change in the value of the other. If the variable n represents the number of years of education beyond grammar school and e represents yearly median earnings (in dollars) for people living in the United States, then the following equation models the relationship between e and n: e 5 3780 1 4320n This equation provides a powerful tool for describing how earnings and education are linked and for making predictions. 9 For example, to predict the median earnings, e, for those with a high school education, we replace n with 4 (representing 4 years beyond grammar school, or a high school education) in our equation to get e 5 3780 1 4320 ? 4 5 $21,060 Thus our equation predicts that for those with a high school education, median earnings will be about $21,060. An equation that is used to describe a real-world situation is called a mathematical model. Such models offer compact, often simplified descriptions of what may be a complex situation. Thus, the accuracy of the predictions made with such models can be questioned and disciplines outside of mathematics may be needed to help answer such questions. Yet these models are valuable guides in our quest to understand social and physical phenomena in our world. Describing the relationship between abstract variables Variables can represent quantities that are not associated with real objects or events. The following equation or mathematical sentence defines a relationship between two quantities, which are named by the abstract variables x and y: y 5 x2 1 2x 2 3 By substituting various values for x and finding the associated values for y, we can generate pairs of values for x and y, called solutions to the equation, that make the sentence true. By convention, we express these solutions as ordered pairs of the form (x, y). Thus, (1, 0) would be a solution to y 5 x2 1 2x 2 3 , since 0 5 12 1 2(1) 2 3, whereas (0, 1) would not be a solution, since 1 2 02 1 2s0d 2 3. There are infinitely many possible solutions to the equation y 5 x2 1 2x 2 3, since we could substitute any real number for x and find a corresponding y. Table 1.7 lists a few solutions. We can use technology to graph the equation (see Figure 1.11). All the points on the graph represent solutions to the equation, and every solution is a point on the graph of the equation.
9
In “Extended Exploration: Looking for Links between Education and Earnings,” which follows Chapter 2, we show how such equations are derived and how they are used to analyze the relationship between education and earnings.
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10
x
y
–4 –3 –2 –1 0 1 2 3
5 0 –3 –4 –3 0 5 12
Describing Relationships between Two Variables
17
y Coordinates of point P are: (horizontal coordinate, vertical coordinate) (x, y) (2, 5)
P(2, 5)
x –10
10
–10
Figure 1.11 Graph of y 5 x2 1 2x 2 3 where one solution
Table 1.7
of infinitely many solutions is labeled.
Note that sometimes an arrow is used to show that a graph extends indefinitely in the indicated direction. In Figure 1.11, the arrows show that both arms of the graph extend indefinitely upward.
Solutions of an Equation The solutions of an equation in two variables x and y are the ordered pairs (x, y) that make the equation a true statement. Graph of an Equation The graph of an equation in two variables displays the set of points that are solutions to the equation.
E X A M P L E
3
Solutions for equations in one or two variables Describe how the solutions for the following equations are similar and how they differ. 3x 1 5 5 11 x125x12 31x5y15
S O L U T I O N
The solutions are similar in the sense that each solution for each particular statement makes the statement true. They are different because: There is only one solution (x 5 2) of the single-variable equation 3x 1 5 5 11. There are an infinite number of solutions for x of the single-variable equation x 1 2 5 x 1 2, since any real number will make the statement a true statement. There are infinitely many solutions, in the form of ordered pairs (x, y), of the two-variable equation 3 1 x 5 y 1 5. y
E X A M P L E
4
Estimating solutions from a graph The graph of the equation x2 1 4y2 5 4 is shown in Figure 1.12. a. From the graph, estimate three solutions of the equation. b. Check your solutions using the equation.
2 x –3
3 –2
Figure 1.12
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S O L U T I O N
a. The coordinates (0, 1), (22, 0), and (1, 0.8) appear to lie on the ellipse, which is the graph of the equation x2 1 4y2 5 4 . b. If substituting the ordered pair (0, 1) into the equation makes it a true statement, then (0, 1) is a solution. x2 1 4y2 5 4
Given substitute x 5 0 and y 5 1
s0d 2 1 4s1d 2 5 4 454
evaluate
We get a true statement, so (0, 1) is a solution to the equation. For the ordered pair (22, 0): x2 1 4y2 5 4
Given substitute x 5 22 and y 5 0
s22d 2 1 4s0d 2 5 4 4105 4 45 4
evaluate
Again we get a true statement, so (22, 0) is a solution to the equation. For the ordered pair (1, 0.8): x2 1 4y2 5 4
Given substitute x 5 1 and y 5 0.8
s1d 2 1 4s0.8d 2 5 4 1 1 4(0.64) 5 4
evaluate
3.56 2 4 We get a false statement, so (1, 0.8) is not a solution, although it is close to a solution.
Algebra Aerobics 1.2b Problem 4(c) requires a graphing program. 1. a. Describe in your own words how to compute the value for y, given a value for x, using the following equation: y 5 3x2 2 x 1 1 b. Which of the following ordered pairs represent solutions to the equation? (0, 0), (0, 1), (1, 0), (21, 2), (22, 3), (21, 0) c. Use x 5 0, 61, 62, 63 to generate a small table of values that represent solutions to the equation. 2. Repeat the directions in Problem 1(a), (b), and (c) using the equation y 5 sx 2 1d 2.
3. Given the equations y1 5 4 2 3x and y2 5 22x2 2 3x 1 5 , fill in Table 1.8.
x
24
22
21
0
1
2
4
y1 y2 Table 1.8
a. Use the table to create two scatter plots, one for the ordered pairs (x, y1) and the other for (x, y2). b. Draw a smooth curve through the points on each graph.
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c. Is (1, 1) a solution for equation y1? For y2? d. Is (21, 6) a solution for equation y1? For y2? e. Look at the graphs. Is the ordered pair (23, 2) a solution for either equation? Verify your answer by substituting the values into each equation.
Describing Relationships between Two Variables
19
4. Given the equation y 5 x2 2 3x 1 2, a. If x 5 3/2, find y. b. Find two points that are not solutions to this equation. c. If available, use technology to graph the equation and then confirm your results for parts (a) and (b).
Exercises for Section 1.2 Course software recommended for Exercise 20. 1. Assume you work for a newspaper and are asked to report on the following data. AIDS Cases in U.S.
c. Estimate the highest value for the Dow Jones during that period. When did it occur? d. Write a topic sentence describing the change in the Dow Jones over the given time period. 3. The accompanying table shows the number of personal and property crimes in the United States from 1995 to 2003.
79,879 (all-time high)
39,206
42,514 40,267 41,831
Year
Personal Crimes (in thousands)
Property Crimes (in thousands)
1995 1998 2000 2003
1,799 1,534 1,425 1,381
12,064 10,952 10,183 10,436
Source: U.S. Bureau of the Census, Statistical Abstract, 2006.
'93
'94
'95
'96
'97
'98
'99
'00
'01
'02
'03
'04
Source: Centers for Disease Control and Prevention, www.cdc.gov.
a. What are three important facts that emerge from this graph? b. Construct a 60-second summary that could accompany the graph in the newspaper article. 2. The following graph shows changes in the Dow Jones Industrial Average, which is based on 30 stocks that trade on the New York Stock Exchange and is the best-known index of U.S. stocks. Dow Jones Industrial Average
a. Create a scatter plot of the personal crimes over time. Connect the points with line segments. b. Approximately how many times more property crimes than personal crimes were committed in 1995? In 2003? c. Write a topic sentence that compares property and personal crime from 1995 to 2003. 4. The National Cancer Institute now estimates that after 70 years of age, 1 woman in 8 will have gotten breast cancer. Fortunately, they also estimate that 95% of breast cancer can be cured, especially if caught early. The data in the accompanying table show how many women in different age groups are likely to get breast cancer.
12,500
Lifetime Risk of Developing Breast Cancer 12,000 11,500 11,000 10,500 Jun06 Jul06
Aug06
Sep06
Oct06
Nov06
Dec06
Source: http://finance.yahoo.com.
a. What time period does the graph cover? b. Estimate the lowest Dow Jones Industrial Average. During what month did it occur?
Age Group
Chance of Developing Cancer
Chance in 1000 Women
30–39 40–49 50–59 60–69 701
1 in 229 1 in 68 1 in 37 1 in 26 1 in 8
4 per 1000 15 per 1000 27 per 1000 38 per 1000 125 per 1000
(Note: Men may get breast cancer too, but less than 1% of all breast cancer cases occur in men.) a. What is the overall relationship between age and breast cancer?
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b. Make a bar chart using the chance of breast cancer in 1000 women for the age groups given. c. Using the “chance in 1000 women” data, estimate how much more likely that women in their 40s would have had breast cancer than women in their 30s. How much more likely for women in their 50s than women in their 40s? d. It is common for women to have yearly mammograms to detect breast cancer after they turn 50, and health insurance companies routinely pay for them. Looking at these data, would you recommend an earlier start for yearly mammograms? Explain your answer in terms of the interests of the patient and the insurance company. (Note: Some research says that mammograms are not that good at detection.) 5. The National Center for Chronic Disease Prevention and Health Promotion published the following data on the chances that a man has had prostate cancer at different ages. Lifetime Risk of Developing Prostate Cancer Age
Risk
Percent Risk
45 50 55 60 65 70 75 80
1 in 25,000 1 in 476 1 in 120 1 in 43 1 in 21 1 in 13 1 in 9 1 in 6
0.004% 0.21% 0.83% 2.3% 4.8% 7.7% 11.1% 16.7%
a. What is the relationship between age and getting prostate cancer? b. Make a scatter plot of the percent risk for men of the ages given. c. Using the “percent risk” data, how much more likely are men 50 years old to have had prostate cancer than men who are 45? How much more likely are men 55 years old to have had prostate cancer than men who are 50? d. Looking at these data, when would you recommend annual prostate checkups to begin for men? Explain your answer in terms of the interests of the patient and the insurance company. 6. Birth rate data in the United States are given as the number of live births per 1000 women in each age category. Age
1950
2000
10–14 15–19 20–24 25–29 30–34 35–39 40–44 45–49
1.0 81.6 196.6 166.1 103.7 52.9 15.1 1.2
0.9 48.5 112.3 121.4 94.1 40.4 7.9 0.5
Source: National Center for Health Statistics, U.S. Dept. of Health and Human Services.
a. Construct a bar chart showing the birth rates for the year 1950. Which mother’s age category had the highest rate of live births? What percentage of women in that category delivered live babies? In which age category was the lowest rate of babies born? What percentage of women in that category delivered live babies? b. Construct a bar chart showing the birth rates for the year 2000. Which mother’s age category had the highest rate of live births? What percentage of women in that category delivered live babies? In which age category was the lowest rate of babies born? What percentage of women in that category delivered live babies? c. Write a paragraph comparing and contrasting the birth rates in 1950 and in 2000. Bear in mind that since 1950 there have been considerable medical advances in saving premature babies and in increasing the fertility of couples. 7. The National Center for Health Statistics published the accompanying chart on childhood obesity. Percentage of American Children Who Are Overweight 16
Percent
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Ages 6–11 years
12
Ages 12–19 years
10 8 6 4 2 0 1963– 1970
1971– 1974
1976– 1980
1988– 1994
1999– 2002
Years Source: National Center for Health Statistics,
www.cdc.gov/nchs.
a. How would you describe the overall trend in the weights of American children? b. Over which years did the percentage of overweight children age 6 to 11 increase? c. Over which time period was there no change in the percentage of overweight children age 6 to 11? d. During which time period were there relatively more overweight 6- to 11-year-olds than 12- to 19-yearolds? e. One of the national health objectives for the year 2010 is to reduce the prevalence of obesity among children to less than 15%. Does this seem like a reasonable goal? 8. Some years are more severe for influenza- and pneumoniarelated deaths than others. The table at the top of the next page shows data from Centers for Disease Control figures for the U.S. for selected years from 1950 to 2000.
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1.2
Death Rate per 100,000 Males
Females
1950 1960 1970 1980 1990 2000
55 66 54 42 48 29
42 44 33 25 31 21
x
Cost
Cruising speed
Range (miles)
Size
Age
B
22 21 2
a. b. c. d.
a. B
Passenger capacity
(2, 1)
11. Consider the equation R 5 2 2 5T. a. Determine which, if any, of the following points (T, R) satisfy this equation. (1, 23),
x11 . x21 Describe in words how to find the value for y given a value for x. Find the ordered pair that represents a solution to the equation when the value of x is 5. Find the ordered pair that represents a solution to the equation when the value of y is 3. Is there an ordered-pair solution to the equation when the value of x is 1? If so, find it; if not, explain why.
13. For parts (a)–(d) use the following equation: y 5
A
b. Find one additional ordered pair that is a solution to the equation above. Show how you found your solution.
(0, 4),
–5
1 . x11 Describe in words how to find the value for y given a value for x. Find the ordered pair that represents a solution to the equation when the value of x is 0. Find the ordered pair that represents a solution to the equation when the value of y is 4. Is there an ordered-pair solution to the equation when the value of x is 21? If so, find it; if not, explain why.
14. For parts (a)–(d) use the following equation: y 5
10. a. Which (if any) of the following ordered pairs (x, y) is a solution to the equation y 5 x2 2 2x 1 1 ? Show how you came to your conclusion. (1, 0),
x 4
–4
7
For parts (a)–(d), decide whether the statement is true or false. Explain your reasoning. a. The newer car is more expensive. b. The slower car is larger. c. The larger car is newer. d. The less expensive car carries more passengers. e. State two other facts you can derive from the graphs. f. Which car would you buy? Why?
(22, 7),
y
2
9. The following three graphs describe two cars, A and B.
A
15
21
a. Create a double bar chart showing the death rates both for men and for women who died of influenza and pneumonia between 1950 and 2000. b. In which year were death rates highest for both men and women? c. Were there any decades in which there was an increase in male deaths but a decrease for women? d. Write a 60-second summary about deaths due to influenza and pneumonia over the years 1950 to 2000.
B
y
23
Source: Centers for Disease Control and Prevention, www.cdc.gov.
A
21
12. Use the accompanying graph to estimate the missing values for x or y in the table.
Age-Adjusted Death Rate for Influenza and Pneumonia
Year
Describing Relationships between Two Variables
(2, 0)
b. Find two additional ordered pairs that are solutions to the equation. c. Make a scatter plot of the solution points found. d. What does the scatter plot suggest about where more solutions could be found? Check your predictions.
b. c. d.
15. For parts (a)–(d) use the following equation: y 5 22x2 . a. If x 5 0, find the value of y. b. If x is greater than zero, what can you say about the value of y? c. If x is a negative number, what can you say about the value of y? d. Can you find an ordered pair that represents a solution to the equation when y is greater than zero? If so, find it; if not, explain why. 16. Find the ordered pairs that represent solutions to each of the following equations when x 5 0, when x 5 3, and when x 5 22. a. y 5 2x2 1 5x b. y 5 2x2 1 1
c. y 5 x3 1 x2 d. y 5 3(x 2 2)(x 2 1)
17. Given the four ordered pairs (21, 3), (1, 0), (2, 3), and (1, 2), for each of the following equations, identify which points (if any) are solutions for that equation. a. y 5 2x 1 5 b. y 5 x2 2 1
c. y 5 x2 2 x 1 1 4 d. y 5 x11
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Age Distribution of the 2006 Populations (in 1000s) of China and the United States 140,000 China United States
Population in thousands
120,000 100,000 80,000 60,000 40,000 20,000
0–4 5–9 10–14 15–19 20–24 25–29 30–34 35–39 40–44 45–49 50–54 55–59 60–64 65–69 70–74 75–79 80–84 85–89 90–94 95–99 100 +
0
Age in years
Percent of 2006 Populations of China and United States by Age 10% China United States
9% 8% Percentage of population
18. The accompanying graphs (and related Excel and DA T A graph link file USCHINA) contain information about the populations of the United States and China. Write a 60-second summary comparing the two populations.
7% 6%
19. Read The New York Times op-ed article “A Fragmented War on Cancer” by Hamilton Jordan, who was President Jimmy Carter’s chief of staff. Jordan claims that we are on the verge of a cancer epidemic. a. Use what you know about the distribution of ages over time in the United States to refute his claim. (See Sec.1.1 Exercise 28 for some ideas.) Are there other arguments that refute his claim? b. Read The New York Times letter to the editor by William M. London, Director of Public Health, American Council on Science and Health. London argues that Hamilton Jordan’s assertions are misleading. What questions are raised by the arguments of William London? What additional data would you need to evaluate his arguments? c. Write a paragraph refuting Hamilton Jordan’s claim that we are on the verge of a cancer epidemic.
5% 4% 3% 2%
0%
0–4 5–9 10–14 15–19 20–24 25–29 30–34 35–39 40–44 45–49 50–54 55–59 60–64 65–69 70–74 75–79 80–84 85–89 90–94 95–99 100 +
1%
20. (Computer required.) Make a prediction about the distribution of income for males and females in the United States. Check your predictions using the course software “F1: Histograms” in FAM1000 Census Graphs and/or using data from the U.S. Census Bureau at www.census.gov. Write a 60-second summary describing your results.
Age in years
Source: U.S. Bureau of the Census, www.census.gov.
1.3 An Introduction to Functions What Is a Function? When we speak informally of one quantity being a function of some other quantity, we mean that one depends on the other. For example, someone may say that what they wear is a function of where they are going, or what they weigh is a function of what they eat, or how well a car runs is a function of how well it is maintained. In mathematics, the word “function” has a precise meaning. A function is a special relationship between two quantities. If the value of one quantity uniquely determines the value of a second quantity, then the second quantity is a function of the first. Median age and the federal deficit are functions of time since each year determines a unique (one and only one) value of median age or the federal deficit. The equation y 5 x2 1 2x 2 3 defines y as a function of x since each value of x we substitute in the equation determines a unique value of y.
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23
Definition of a Function A variable y is a function of a variable x if each value of x determines a unique value of y.
Representing Functions in Multiple Ways We can think of a function as a “rule” that takes certain inputs and assigns to each input value exactly one output value. The rule can be described using words, data tables, graphs, or equations. 1
Sales tax Eleven states have a sales tax of 6%; that is, for each dollar spent in a store in these states, the law says that you must pay a tax of 6 cents, or $0.06. Represent the sales tax as a function of purchase price using an equation, table, and graph.10
S O L U T I O N
Using an Equation We can write this relationship as an equation where T represents the amount of sales tax and P represents the price of the purchase (both measured in dollars): Amount of sales tax 5 0.06 ? price of purchase T 5 0.06P Our function rule says: “Take the given value of P and multiply it by 0.06; the result is the corresponding value of T.” The equation represents T as a function of P, since for each value of P the equation determines a unique (one and only one) value of T. The purchase price, P, is restricted to dollar amounts greater than or equal to zero. Using a Table We can use this formula to make a table of values for T determined by the different values of P (see Table 1.9). Such tables were once posted next to many cash registers. P (purchase price in $) T (sales tax in $)
0
1
2
3
4
5
6
7
8
9
10
0.00 0.06 0.12 0.18 0.24 0.30 0.36 0.42 0.48 0.54 0.60
Table 1.9
Using a Graph The points in Table 1.9 were used to create a graph of the function (Figure 1.13). The table shows the sales tax only for selected purchase prices, but we could have used any positive dollar amount for P. We connected the points on the scatter plot to suggest the many possible intermediate values for price. For example, if P 5 $2.50, then T 5 $0.15. 1.00 Sales tax (dollars)
E X A M P L E
T
0.50
0
2
4
6
8
10
P
Purchase price (dollars)
Figure 1.13 Graph of 6%
sales tax. 10
In 2007, a sales tax of 6% was the most common rate for a sales tax in the United States. See www.taxadmin.org for a listing of the sales tax rates for all of the states.
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Independent and Dependent Variables Since a function is a rule that assigns to each input a unique output, we think of the output as being dependent on the input. We call the input of a function the independent variable and the output the dependent variable. When a set of ordered pairs represents a function, then each ordered pair is written in the form
y
(independent variable, dependent variable)
Dependent variable
24
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or equivalently,
(input, output)
If x is the independent and y the dependent variable, then the ordered pairs would be of the form (x, y)
x Independent variable
The mathematical convention is for the first variable, or input of a function, to be represented on the horizontal axis and the second variable, or output, on the vertical axis. Sometimes the choice of the independent variable is arbitrary or not obvious. For example, economists argue as to whether wealth is a function of education or education is a function of wealth. As seen in the next example, there may be more than one correct choice. E X A M P L E
2
Identifying Independent and dependent variables In the sales tax example, the equation T 5 0.06P gives the sales tax, T, as a function of purchase price, P. In this case T is the dependent variable, or output, and P is the independent variable, or input. But for this equation we can see that P is also a function of T; that is, each value of T corresponds to one and only one value of P. It is easier to see the relationship if we solve for P in terms of T, to get P5
T 0.06
Now we are thinking of the purchase price, P, as the dependent variable, or output, and the sales tax, T, as the independent variable, or input. So, if you tell me how much tax you paid, I can find the purchase price.
When Is a Relationship Not a Function? Not all relationships define functions. A function is a special type of relationship, one where for each input, the rule specifies one and only one output. Examine the following examples. Function Input 1 2 3
Not a Function
Output 6 7 8
Each input has only one output.
E X A M P L E
3
Input 1 2 3
Output 6 7 8 9
The input of 1 gives two different outputs, 6 and 7, so this relationship is not a function.
Function Input 1 2 3
Output 6
Each input has only one output. Note that a function may have identical outputs for different inputs.
Does the table represent a function? Consider the set of data in Table 1.10. The first column shows the year, T, of the Olympics. The second column shows the winning distance, D (in feet), for that year for the men’s Olympic 16-pound shot put. a. Is D a function of T? b. Is T a function of D? c. What should be your choice for the dependent and independent variables?
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25
Olympic Shot Put Year, T
Winning Distance in Feet Thrown, D
1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004
65 67 67 70 70 70 70 74 71 71 70 69
Table 1.10 Source: The World Almanac and Book of Facts, 2006.
S O L U T I O N
1964 1968
67
1964 67
1968
a. D is a function of T. To determine if D is a function of T, we need to find out if each value of T (the input) determines one and only one value for D (the output). So, the ordered pair representing the relationship would be of the form (T, D). Using Table 1.10, we can verify that for each T, there is one and only one D. So D, the winning shot put distance, is a function of the year of the Olympics, T. Note that different inputs (such as 1964 and 1968) can have the same output (67 feet), and the relationship can still be a function. There are even 5 different years that have 70 feet as their output. b. T is not a function of D. To determine if T is a function of D, we need to find out if each value of D (now the input) determines one and only one value for T (the output). Now the ordered pairs representing the relationship would be of the form (D, T). Table 1.11 shows this new pairing, where D is thought of as the input and T as the output. The year, T, is not a function of D, the winning distance, since some values of D give more than one value for T. For example, when D 5 67 there are two corresponding values for T, 1964 and 1968, and this violates the condition of a unique (one and only one) output for each input. Olympic Shot Put Winning Distance in Feet Thrown, D 65 67 67 70 70 70 70 74 71 71 70 69 Table 1.11
Year, T 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004
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c. D, the winning distance, is a function of the year, T, but T is not a function of D. The distance, D, depends on the year, T, but T does not depend on D. To construct a function relating T and D, we must choose T as the independent variable and D as the dependent variable. The ordered pairs that represent the function would be written as (T, D). How would the axes be labeled for each graph of the following functions? a. Density of water is a function of temperature. b. Radiation intensity is a function of wavelength. c. A quantity Q is a function of time t.
S O L U T I O N
a.
b.
c.
Density of water
Quantity
4
Radiation intensity
E X A M P L E
Temperature
Wavelength
Time
Figure 1.14 Various labels for axes, depending on the context.
Exploration 1.2 will help you develop an intuitive sense of functions.
How to tell if a graph represents a function: The vertical line test For a graph to represent a function, each value of the input on the horizontal axis must be associated with one and only one value of the output on the vertical axis. If you can draw a vertical line that intersects a graph in more than one point, then at least one input is associated with two or more outputs, and the graph does not represent a function. The graph in Figure 1.15 represents y as a function of x. For each value of x, there is only one corresponding value of y. No vertical line intersects the curve in more than one point. The graph in Figure 1.16 does not represent a function. One can draw a vertical line (an infinite number, in fact) that intersects the graph in more than one point. Figure 1.16 shows a vertical line that intersects the graph at both (4, 2) and (4, 22). That means that the value x 5 4 does not determine one and only one value of y. It corresponds to y values of both 2 and 22. y
y
5
5
(4, 2)
–5 5
x –5
5
x
(4, –2)
–5
Figure 1.15 The graph represents y as a function of x since there is no vertical line that intersects the curve at more than one point.
–5
Figure 1.16 The graph does not
represent y as a function of x since there is at least one vertical line that intersects this curve at more than one point.
Vertical Line Test If there is a vertical line that intersects a graph more than once, the graph does not represent a function.
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An Introduction to Functions
27
Algebra Aerobics 1.3
Input
Output
Input
Output
1 2 3 4
5 8 8 10
1 2 2 4
5 7 8 10
Table A
5. Consider the scatter plot in Figure 1.19. Is weight a function of height? Is height a function of weight? Explain your answer. 160 Weight (pounds)
1. Which of the following tables represent functions? Justify your answer.
Table B
Output
1 1 3 4
5 7 8 10
130 120 110 50 54 58 Height (inches)
62
Figure 1.19 Graph of weight
versus height.
6. Consider Table 1.12. a. Is D a function of Y? b. Is Y a function of D?
3. Refer to the graph in Figure 1.17. Is y a function of x? y
140
100 46
2. Does the following table represent a function? If so, why? If not, how could you change the values in the table so it represents a function? Input
150
10
Y
1992
1993
1994
1995
1996
1997
D
$2.50
$2.70
$2.40
2$0.50
$0.70
$2.70
Table 1.12
x –10
10
Figure 1.17 Graph of an
abstract relationship between x and y.
4. Which of the graphs in Figure 1.18 represent functions and which do not? Why?
Graph A
Graph B
Graph C
Figure 1.18
7. Plot the following points with x on the horizontal and y on the vertical axis. Draw a line through the points and determine if the line represents a function. a. b. x
y
x
y
21 0 2 3
3 3 3 3
22 22 22 22
21 0 1 4
8. a. Write an equation for computing a 15% tip in a restaurant. Does your equation represent a function? If so, what are your choices for the independent and dependent variables? b. How much would the equation suggest you tip for an $8 meal? c. Compute a 15% tip on a total check of $26.42.
Exercises for Section 1.3 A graphing program is required for Exercise 12. 1. The following table gives the high temperature in Rome, Italy, for each of five days in October 2006. a. Is the temperature a function of the date? b. Is the date a function of the temperature?
Date
Rome High Temperature
Oct. 26 Oct. 27 Oct. 28 Oct. 29 Oct. 30
27C 27C 25C 26C 22C
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2. Which of the following tables describe functions? Explain your answer.
7. a. Find an equation that represents the relationship between x and y in each of the accompanying tables. i.
a. Input value Output value
22 28
21 21
0 0
1 1
2 8
b. Input value Output value
0 24
1 22
2 0
1 2
0 4
c. Input value Output value
10 3
7 6
4 9
7 12
10 15
d. Input value Output value
0 3
3 3
9 3
12 3
15 3
y
x x Graph A
x
Graph B
Graph C
5. Which of the accompanying graphs describe functions? Explain your answer.
Graph A
Graph B
Graph C
6. Consider the accompanying table, listing the weights (W) and heights (H) of five individuals. Based on this table, is height a function of weight? Is weight a function of height? Justify your answers. Weight W (lb)
Height H (in)
120 120 125 130 135
54 55 58 60 56
0 1 2 3 4
5 6 7 8 9
ii.
x
y
0 1 2 3 4
1 2 5 10 17
iii.
x
y
0 1 2 3 4
3 3 3 3 3
8. For each of the accompanying tables find a function formula that takes the x values and produces the given y values.
4. Which of the accompanying graphs describe functions? Explain your answer. y
y
b. Which of your equations represents y as function of x? Justify your answer.
3. Determine whether each set of points represents a function. (Hint: It may be helpful to plot the points.) a. (2, 6), (24, 6), (1, 23), (4, 23) b. (2, 22), (3, 22), (4, 22), (6, 22) c. (2, 23), (2, 3), (–2, 23), (22, 3) d. (21, 2), (21, 0), (21, 21), (21, 22)
y
x
a.
x
y
0 1 2 3 4
0 3 6 9 12
b.
x
y
0 1 2 3 4
–2 1 4 7 10
c.
x
y
0 1 2 3 4
0 –1 –4 –9 –16
9. The basement of a large department store features discounted merchandise. Their policy is to reduce the previous month’s price of the item by 10% each month for 5 months, and then give the unsold items to charity. a. Let S1 be the sale price for the first month and P the original price. Express S1 as a function of P. What is the price of a $100 garment on sale for the first month? b. Let S2 be the sale price for the second month and P the original price. Express S2 as a function of P. What is the price of a $100 garment on sale for the second month? c. Let S3 be the sale price for the third month and P the original price. Express S3 as a function of P. What is the price of a $100 garment on sale for the third month? d. Let S5 be the sale price for the fifth month and P the original price. Express S5 as a function of P. What is the final price of a $100 garment on sale for the fifth month? By what total percentage has the garment now been reduced from its original price? 10. Write a formula to express each of the following sentences: a. The sale price is 20% off the original price. Use S for sale price and P for original price to express S as a function of P. b. The time in Paris is 6 hours ahead of New York. Use P for Paris time and N for New York time to express P as a function of N. (Represent your answer in terms of a 12-hour clock.) How would you adjust your formula if P comes out greater than 12? c. For temperatures above 08F the wind chill effect can be estimated by subtracting two-thirds of the wind speed (in miles per hour) from the outdoor temperature. Use C for the effective wind chill temperature, W for wind speed, and T for the actual outdoor temperature to write an equation expressing C in terms of W and T. 11. Determine whether y is a function of x in each of the following equations. If the equation does not define a function, find a value of x that is associated with two different y values.
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1.4
a. y 5 x2 1 1 b. y 5 3x 2 2
c. y 5 5 d. y2 5 x
12. (Graphing program required.) For each equation below, write an equivalent equation that expresses z in terms of t. Use technology to sketch the graph of each equation. Is z a function of t? Why or why not? a. 3t 2 5z 5 10 c. 2(t 2 4) 2 (z 1 1) 5 0 b. 12t2 2 4z 5 0 13. If we let D stand for ampicillin dosage expressed in milligrams and W stand for a child’s weight in kilograms, then the equation D 5 50W
The Language of Functions
29
gives a rule for finding the safe maximum daily drug dosage of ampicillin (used to treat respiratory infections) for children who weigh less than 10 kilograms (about 22 pounds).11 a. What are logical choices for the independent and dependent variables? b. Does the equation represent a function? Why? c. Generate a small table and graph of the function. d. Think of the function D 5 50W for ampicillin dosage as an abstract mathematical equation. How will the table and graph change?
1.4 The Language of Functions As we have seen, not all equations represent functions. But functions have important qualities, so it is useful to have a way to indicate when a relationship is a function.
Function Notation When a quantity y is a function of x, we can write y is a function of x or in abbreviated form, y equals “f of x” or using function notation, y 5 f(x) The expression y 5 f(x) means that the rule f is applied to the input value x to give the output value, f(x): output 5 f(input) or
dependent variable 5 f (independent variable) The letter f is often used to denote the function, but we could use any letter, not just f. Understanding the symbols Suppose we have a function that triples the input. We could write this function as y 5 3x
(1)
or with function notation as Tsxd 5 3x
where y 5 T sxd
(2)
Equations (1) and (2) represent the same function, but with function notation we name the function—in this case T—and identify the input, x, and output, 3x. Function notation can provide considerable economy in writing and reading. For example, throughout a discussion we can use T (x) instead of the full expression to represent the function. 11
Information extracted from Anna M. Curren and Laurie D. Muntlay, Math for Meds; Dosages and Solutions, 6th ed. (San Diego: W. I. Publications, 1990), p. 198.
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The Language of Functions If y is a function, f, of x, then y f(x) where
f is the name of the function, y is the output or dependent variable, x is the input or independent variable. output f(input) dependent f(independent)
Finding output values: Evaluating a function Function notation is particularly useful when a function is being evaluated at a specific input value. Suppose we want to find the value of the previous function T(x) when our input value is 10. Using equation (1) we would say, “find the value of y when x 5 10.” With function notation, we simply write T(10). To evaluate the function T at 10 means calculating the value of the output when the value of the input is 10: T(x) 5 3x
Given
T(10) 5 3(10) 5 30
Substitute 10 for x
So, applying the function rule T to the input value of 10 gives an output value of 30.
Common Error The expression f(x) does not mean “f times x.” It means the function f evaluated at x.
E X A M P L E
1
Using function notation with equations a. Given f(x) 5 2x2 1 3, evaluate f(5), f(0), and f(22) b. Evaluate g(0), g(2), and g(22) for the function gsxd 5
S O L U T I O N
1 x21
a. To evaluate f(5), we replace every x in the formula with 5. Given
f sxd 5 2x2 1 3
Substitute 5 for x
fs5 d 5 2s5d 2 1 3 5 2s5ds5d 1 3 5 53
Similarly, f(0) 5 2(0) 2 1 3 5 3 f (22) 5 2(22) 2 1 3 5 (2 # 4) 1 3 5 11 b. gs0d 5
1 5 21, 021
gs2d 5
1 5 1, 221
gs22d 5
1 1 1 5 52 22 2 1 23 3
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E X A M P L E
2
S O L U T I O N
Using function notation with data tables Use Table 1.13 to fill in the missing values: a. S(0) 5 ? b. S(21) 5 ? c. S(?) 5 4
The Language of Functions
Input, x Output, S(x)
31
21
–2
0
1
2
1
4
0
1
4
Table 1.13
a. S(0) means to evaluate S when the input x 0. The table says that the corresponding output is also 0, so S(0) 0. b. S(21) 5 1. c. S(?) 4 means to find the input when the output is 4. When the output is 4, the input is 22 or 2, so S(22) 5 4 and S(2) 5 4. 3
E X A M P L E
3
f (x) Using function notation with graphs Use the graph in Figure 1.20 to estimate the missing values: a. f(0) 5 ? 0 –15 15 b. f(25) 5 ? Figure 1.20 Graph of a c. ƒ(?) 5 0
function.
S O L U T I O N
Remember that by convention the horizontal axis represents the input (or independent variable) and the vertical axis represents the output (or dependent variable). a. f(0) 5 2 b. f(25) 5 1.5 c. f (10) 5 0 and f(210) 5 0 Rewriting equations using function notation In order to use function notation, an equation needs to be in the form output 5 some rule applied to input or equivalently dependent variable 5 some rule applied to independent variable Translating an equation into this format is called putting the equation in function form. Many graphing calculators and computer graphing programs accept only equations in function form as input. To put an equation into function form, we first need to identify the independent and the dependent variables. The choice is sometimes obvious, at other times arbitrary. If we use the mathematical convention that x represents the input or independent variable and y the output or dependent variable, when we put equations into function form, we want y 5 some rule applied to x
E X A M P L E
4
S O L U T I O N
Analyze the equation 4x 2 3y 5 6. Decide whether or not the equation represents a function. If it does, write the relationship using function notation. First, put the equation into function form. Assume y is the output. Given the equation
4x 2 3y 5 6
subtract 4x from both sides
23y 5 6 2 4x
divide both sides by 23
23y 6 2 4x 5 23 23
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6 24x 1 23 23
simplify
y5
simplify and rearrange terms
4 y5 x22 3
We now have an expression for y in terms of x. Using technology or by hand, we can generate a graph of the equation (see Figure 1.21). Since the graph passes the vertical line test, y is a function of x. 6
y
x –6
6
–6
Figure 1.21 Graph of y 5 43 x 2 2.
If we name our function f, then using function notation, we have y 5 f(x) E X A M P L E
5
S O L U T I O N
where f(x) 5 43x 2 2
Analyze the equation y2 2 x 5 0. Generate a graph of the equation. Decide whether or not the equation represents a function. If the equation represents a function, write the relationship using function notation. Assume y is the output. Put the equation in function form: Given the equation
y2 2 x 5 0
add x to both sides
y2 5 x
To solve this equation, we take the square root of both sides of the equation and we get y 5 6 "x This gives us two solutions for any value of x . 0 as shown in Table 1.14. For example, if x 5 4, then y can either be 2 or 22 since both 22 5 4 and s22d 2 5 4. 5
y
(4, 2)
x
y
0 1
0 1 or 21
2 4 9
"2 or 2 "2 2 or 22 3 or 23
Table 1.14
x –1
0
5
(4, –2)
–5
Figure 1.22 Graph of the
equation y2 5 x.
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The graph of the equation in Figure 1.22 does not pass the vertical line test. In particular, the solutions (4, 22) and (4, 2) lie on the same vertical line. So y is not a function of x and we cannot use function notation to represent this relationship.
Algebra Aerobics 1.4a 1. Given g(x) 5 3x, evaluate g(0), g(21), g(1), g(20), and g(100). 2. Consider the function f sxd 5 x2 2 5x 1 6. Find f(0), f (1), and f (23). 2 3. Given the function fsxd 5 , evaluate f (0), x21 f(21), f(1), and f(23). 4. Determine the value of t for which each of the functions has a value of 3. r(t) 5 5 2 2t
p(t) 5 3t 2 9
m(t) 5 5t 2 12
In Problems 5–7 solve for y in terms of x. Determine if y is a function of x. If it is, rewrite using f(x) notation. 5. 2(x 2 1) 2 3(y 1 5) 5 10 6. x2 1 2x 2 y 1 4 5 0 7. 7x 2 2y 5 5 8. From the graph in Figure 1.23, estimate f (24), f (21), f (0), and f (3). Find two approximate values for x such that f(x) 5 0.
y 2
–4
4
x
–4
Figure 1.23 Graph of f (x).
9. From Table 1.15 find f(0) and f (20). Find two values of x for which f(x) 5 10. Explain why f(x) is a function. x
f(x)
0 10 20 30 40
20 10 0 10 20
Table 1.15
Domain and Range A function is often defined only for certain values of the input (or independent variable). The set of all possible values for the input is called the domain of the function. The set of corresponding values of the output (or dependent variable) is called the range of the function. Domain and Range of a Function The domain of a function is the set of possible values of the input. The range is the set of corresponding values of the output.
E X A M P L E
6
Finding a reasonable domain and range In the sales tax example at the beginning of this section, we used the equation T 5 0.06P to represent the sales tax, T, as a function of the purchase price, P (where all units are in dollars). What are the domain and range of this function?
S O L U T I O N
Since negative values for P are meaningless, P is restricted to dollar amounts greater than or equal to zero. In theory there is no upper limit on prices, so we assume P has no maximum amount. In this example, the domain is all dollar values of P greater than or equal to 0
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We can express this more compactly with the symbol $, which means “greater than or equal to.” So, the domain is all dollar values of P such that P $ 0 the domain is P $ 0
or abbreviated to
What are the corresponding values for the tax T? The values for T in our model will also always be nonnegative. As long as there is no maximum value for P, there will be no maximum value for T. So, the range is all dollar values of T greater than or equal to 0 or we can shorten this to
the range is T $ 0
Representing the Domain and Range with Interval Notation Interval notation is often used to represent the domain and range of a function.
Interval Notation A closed interval [a, b] indicates all real numbers x for which a x b. Closed intervals include their endpoints. An open interval (a, b) indicates all real numbers x for which a x b. Open intervals exclude their endpoints. Half-open (or equivalently half-closed) intervals are represented by [a, b) which indicates all real numbers x for which a x b or (a, b] which indicates all real numbers x for which a x b.
For example, if the domain is values of n greater than or equal to 50 and less than or equal to 100, then domain 5 all n values with 50 # n # 100 5 [50, 100] If the domain is values of n greater than 50 and less than 100, then domain 5 all n values with 50 , n , 100 5 interval (50, 100) If we want to exclude 50 but include 100 as part of the domain, we would represent the interval as (50, 100]. The interval can be displayed on the real number line as: 50
100
In general, a hollow dot indicates exclusion and a solid dot inclusion. Note: Since the notation (a, b) can also mean the coordinates of a point, we will say the interval (a, b) when we want to refer to an interval. E X A M P L E
7
Finding the domain and range from a graph The graph in Figure 1.24 shows the water level of the tides in Pensacola, Florida, over a 24-hour period. Are the Pensacola tides a function of the time of day? If so, identify the independent and dependent variables. Use interval notation to describe the domain and range of this function.
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Water level (in centimeters)
1.4
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35
+10
0
–10
0
4
8 12 16 Time (in hours)
20
24
Figure 1.24 Diurnal tides in a 24-hour
period in Pensacola, Florida. Source: Adapted from Fig. 8.2 in Oceanography: An Introduction to the Planet Oceanus, by Paul R. Pinet. Copyright © 1992 by West Publishing Company, St. Paul, MN. All rights reserved.
S O L U T I O N
The Pensacola tides are a function of the time of day since the graph passes the vertical line test. The independent variable is time, and the dependent variable is water level. The domain is from 0 to 24 hours, and the range is from about 210 to 110 centimeters. Using interval notation: domain 5 [0, 24] range 5 [210, 10] When are there restrictions on the domain and range? When specifying the domain and range of a function, we need to consider whether the function is undefined for any values. For example, for the function y5
?
SOMETHING TO THINK ABOUT
What happens to the value of y 5 x1 as x S 0?
1 x
the expression 1/x is undefined when x 5 0. For any other value for x, the function is defined. Thus, the domain is all real numbers except 0 To find the range, we need to determine the possible output values for y. Sometimes it is easier to find the y values that are not possible. In this case, y can’t equal zero. Why? Our rule says to take 1 and divide by x, but it is impossible to divide 1 by a real number in our domain and get zero as a result. Thus, the range is all real numbers except 0 The interval (2infinity, 1infinity) or (2`, `) represents all real numbers. To represent all real numbers except 0 using interval notation, we use the union symbol, h, of two intervals: (2`, 0) h (0, `) Using the graph of the function in Figure 1.25 we can check to see if our domain and range are reasonable. The graph suggests that x comes very close to 0, but does not equal 0. The same is true for y.
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5
y
–5
x 5
–5
Figure 1.25 Graph of y 5
1 . x
Another expression that is undefined is "negative number , if we restrict ourselves to the real number system. E X A M P L E
8
Finding restrictions on the domain and range Specify the domain and range for each of the following functions: a. y 5 "x 2 4
S O L U T I O N
b. y 5
1 x22
a. First, we ask ourselves, “Is the domain restricted?” In this case, the answer is yes. The expression "x 2 4 is not defined for negative values, so we must have the expression x 2 4 $ 0 1 x $ 4. Corresponding y values must be $ 0. So, Domain: Range:
all real numbers $ 4 all real numbers $ 0.
Using interval notation, we get Domain: Range:
[4, 1 `) [0, 1 `)
1 b. The domain of y 5 x 2 2 is also restricted. In this case, the denominator cannot 1 equal 0 since 0 is not defined. This means the expression x 2 2 2 0 1 x 2 2. The range is also restricted. It is not possible for y 5 0, since 1 divided by any real number is never 0. So,
Domain: Range:
all real numbers except 2 all real numbers except 0
Using interval notation, we get Domain:
s2`, 2d d s2, 1 `d
Range:
s2`, 0d d s0, 1 `d
Two Cases Where the Domain and Range Are Restricted y5
1 x
y 5 "x
Domain: Range:
all real numbers except 0. all real numbers except 0
Domain: Range:
all real numbers $ 0 all real numbers $ 0
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37
When is a function undefined? Match each function graph in Figure 1.26 with the appropriate domain and range listed in parts (a) to (e). [Note: The dotted line in the graph of f (x) is not part of the function.]
y 10
y y 5
x –5
5
x –5
5
x –5
5
–2
–10
–2
Graph of g (x) = 12 x
Graph of f(x) = 1 x–1
Graph of h (x) = 冑 x
Figure 1.26 Graphs for three functions.
a. b. c. d. e. S O L U T I O N
Domain: [0, `) Domain: (0, `) Domain: (2`, 1) h (1, `) Domain: (2`, 1) h (1, `) Domain: (2`, 0) h (0, `)
Range: [0, `) Range: (0, `) Range: (2`, 1) h (1, `) Range: (2`, 0) h (0, `) Range: (0, `)
1 matches with (d) x21 1 g(x) 5 2 matches with (e) x f(x) 5
h(x) 5 "x matches with (a)
Algebra Aerobics 1.4b 1. Express each of the following using interval notation. a. x . 2 b. 4 # x , 20 c. t # 0 or t . 500 2. Express the given interval as an inequality. a. [23, 10) b. (22.5, 6.8] c. (2`, 5] h [12, `) 3. Express each of the following statements in interval notation. a. Harry’s GPA is at least 2.5 but at most 3.6. b. A good hitter has a batting average of at least 0.333.
c. Starting annual salary at a position is anything from $35,000 to $50,000 depending upon experience. In Problems 4–8 solve for y in terms of x. Determine if y is a function of x. If it is, rewrite using f (x) notation and determine the domain and range. 4. 2(x 1 1) 1 3y 5 5 5. x 1 2y 5 3x 2 4 6. y 5 "x 7. 2xy 5 6 y 8. 2x 1 3 5 1 9. Find values of x for which the function is undefined, and determine the domain and range. f (x) 5
x11 1 ,g(x) 5 , x15 x11
h(x) 5 "x 2 10
Exercises for Section 1.4 A graphing program is required for Exercise 6. 1. Given T(x) 5 x2 2 3x 1 2, evaluate T(0), T(21), T(1), and T(25).
2. Given f sxd 5 f(100).
x , evaluate f(0), f(21), f(1), f(20), and x21
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3. Assume that for persons who earn less than $20,000 a year, income tax is 16% of their income. a. Generate a formula that describes income tax in terms of income for people earning less than $20,000 a year. b. What are you treating as the independent variable? The dependent variable? c. Does your formula represent a function? Explain. d. If it is a function, what is the domain? The range? 4. Suppose that the price of gasoline is $3.09 per gallon. a. Generate a formula that describes the cost, C, of buying gas as a function of the number of gallons of gasoline, G, purchased. b. What is the independent variable? The dependent variable? c. Does your formula represent a function? Explain. d. If it is a function, what is the domain? The range? e. Generate a small table of values and a graph. 5. The cost of driving a car to work is estimated to be $2.00 in tolls plus 32 cents per mile. Write an equation for computing the total cost C of driving M miles to work. Does your equation represent a function? What is the independent variable? What is the dependent variable? Generate a table of values and then graph the equation. 6. (Graphing program required.) For each equation, write the equivalent equation that expresses y in terms of x. Use technology to graph each function and then estimate its domain and range. a. 3x 1 5x 2 y 5 3y c. x(x 2 1) 1 y 5 2x 2 5 b. 3x(5 2 x) 5 x 2 y d. 2(y 2 1) 5 y 1 5x(x 1 1) 7. If f sxd 5 x2 2 x 1 2 , find: a. f(2) b. f(21) c. f(0)
9. Look at the accompanying table. a. Find p(24), p(5), and p(1). b. For what value(s) of n does p(n) 5 2?
p(n)
24
23
22
21
0
1
2
3
0.063
0.125
0.25
0.5
1
2
4
8 16
4
11. From the accompanying graph of y 5 f(x):
5
5 32
12. Find f(3), if it exists, for each of the following functions: x11 a. f (x) 5 (x 2 3) 2 c. f (x) 5 x23 1 2x b. f (x) 5 d. f (x) 5 x x21 Determine the domain for each function. 13. If f sxd 5 s2x 2 1d 2, evaluate f(0), f(1), and f(22).
b. g(x) 5
5 2 2x 2
c. j(x) 5
1 x11
x
d. f(3), g(3)
16. Each of the following functions has a restricted domain and range. Find the domain and range for each function and explain why the restrictions occur. 3 x12
c. h(x) 5 2 d. k(x) 5
–5
e. f(x) 5 x2 1 3
15. Given f(x) 5 1 2 0.5x and gsxd 5 x2 1 1 , evaluate: a. f(0), g(0) c. f(2), f(1)
a. f (x) 5
3
x
a. Find f(22), f(21), f(0), and f(1). b. Find two values of x for which f(x) 5 23. c. Estimate the range of f. Assume that the arms of the graph extend upward indefinitely.
b. g(x) 5 "x 2 5 –3
5
–5
5
f (x)
f (x)
–5
b. f(–2), g(–3) 10. Consider the function y 5 f (x) graphed in the accompanying figure. y
y
14. Find the domain for each of the following functions: a. f(x) 5 300.4 1 3.2x d. k(x) 5 3
d. f(25)
8. If g(x) 5 2x 1 3, evaluate g(0), g(1), and g(21).
n
a. Find f(23), f(0), f(1), and f(2.5). b. Find two values of x such that f(x) 5 0.
1 2x 2 3
1 x224
e. l(x) 5 "x 1 3
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5
y
Visualizing Functions
39
17. For the functions ƒ(x) and g(x) shown on the accompanying graph, find the values of x that make the following true. a. ƒ(x) 5 0 b. g(x) 5 0 c. ƒ(x) 5 g(x)
f (x)
18. Determine the domain of each of the following functions. Explain your answers. x21 hsxd 5 f sxd 5 4 gsxd 5 3x 1 5 x22 x21 Hsxd 5 "x 2 2 Gsxd 5 2 Fsxd 5 x2 2 4 x 24
g(x) –5
5
x
–5
1.5 Visualizing Functions In this section we return to the question: How does change in one variable affect change in another variable? Graphs are one of the easiest ways to recognize change. We start with three basic questions:
Is There a Maximum or Minimum Value? If a function has a maximum (or minimum) value, then it appears as the highest point (or lowest point) on its graph. E X A M P L E
1
Determine if each function in Figure 1.27 has a maximum or minimum, then estimate its value.
y
y 100
200
h(x) g (x)
x
x 20
–20
–25
25
f (x)
–100
–200
Figure 1.27 Graphs of ƒ(x), g(x), and h(x).
S O L U T I O N
The function f(x) in Figure 1.27 appears to have a maximum value of 40 when x 5 25 but has no minimum value since both arms of the function extend indefinitely downward. The function g(x) appears to have a minimum value of 20 when x 5 5, but no maximum value since both arms of the function extend indefinitely upward. The function h(x) appears to have a maximum value of 200, which occurs when x 5 25, and a minimum value of 250, when x 5 5.
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Is the Function Increasing or Decreasing?
Increasing
E X A M P L E
2
Increasing and decreasing production Figure 1.28 shows a 75-year history of annual natural gas production in the United States. Create a 60-second summary about gas production from 1930 to 2005 in the U.S. U.S. Natural Gas Production 700
Production in billions of cubic meters
Decreasing
A function f is decreasing over a specified interval if the values of f (x) decrease as x increases over the interval. A function f is increasing over a specified interval if the values of f(x) increase as x increases over the interval. The graph of an increasing function climbs as we move from left to right. The graph of a decreasing function falls as we move from left to right.
600 500 400 300 200 100 0 1930 1935 1940 1945 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 Year
Figure 1.28 Source: Wikipedia. http://www.answers.com/topic/natural-gas-prices.
S O L U T I O N
Natural gas production in the United States increased from approximately 50 billion cubic meters in 1930 to a high of approximately 650 billion cubic meters in 1973. For the next 10 years production generally decreased to under 500 billion cubic meters in 1983. Between 1983 and 2005, annual production oscillated between 500 and 600 billion cubic meters, with production at approximately 550 billion in 2005.
Is the Graph Concave Up or Concave Down? What does the concavity of a graph mean? The graph of a function is concave up if it bends upward and it is concave down if it bends downward.
Concave up
Concave down
Concavity is independent of whether the function is increasing or decreasing. Increasing Concave down
Decreasing Concave up
Concave down
Concave up
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3
41
Graphs are not necessarily pictures of events12 The graph in Figure 1.29 shows the speed of a roller coaster car as a function of time.
Speed
E X A M P L E
Visualizing Functions
t0
t1
t2
t3
t4
Time
Figure 1.29
a. Describe how the speed of the roller coaster car changes over time. Describe the changes in the graph as the speed changes over time. b. Draw a picture of a possible track for this roller coaster. a. The speed of the roller coaster car increases from t0 to t2 , reaching a maximum for this part of the ride at t2 . The speed decreases from t2 to t4 . The graph of speed versus time is concave up and increasing from t0 to t1 and then concave down and increasing from t1 to t2 . From t2 to t3 the graph is concave down and decreasing, and from t3 to t4 it is concave up and decreasing. b. A picture for a possible track of the roller coaster is shown in Figure 1.30. Notice how the track is an upside-down picture of the graph of speed versus time. (When the roller coaster car goes down the speed increases, and when the roller coaster car goes up, the speed decreases.) Height of track
S O L U T I O N
t0
t1
t2
t3
t4
Time
Figure 1.30
E X A M P L E
4
Growth patterns Figure 1.31 shows the growth patterns for three areas in Virginia. Compare the differences in growth for these areas between 1900 and 2005. Richmond, Va., Population Growth Patterns, 1900–2005 300,000 Henrico, Va.
250,000
Chesterfield, Va. 200,000
Richmond City, Va.
150,000 100,000 50,000 0 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 2005
Figure 1.31 Source: www.savethebay.org/land/images. 12
Example 3 is adapted from Shell Centre for Mathematical Education, The Language of Functions and Graphs. Manchester, England: University of Nottingham, 1985.
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S O L U T I O N
From 1900 to about 1988, Richmond City consistently had the largest population; however, after about 1950 Henrico and Chesterfield counties were growing faster than Richmond City. By 1990 all three areas had populations of about 200,000. After 1990, Henrico’s and Chesterfield’s populations continued to grow and Richmond’s continued to decline.
Getting the Big Idea We now have the basic vocabulary for describing a function’s behavior. Think of a function as telling a story. We want to decipher not just the individual words, but the overall plot. In each situation we should ask, What is really happening here? What do the words tell us about the shape of the graph? What does the graph tell us about the underlying phenomenon?
S O L U T I O N
Generate a rough sketch of each of the following situations. a. A cup of hot coffee cooling. b. U.S. venture capital (money provided by investment companies to business start-ups) increased modestly but steadily in the mid-1990s, soared during the “dotcom bubble” (in the late 1990s), with a high in 2000, and then suffered a drastic decrease back to pre-dotcom levels. c. Using a simple predator-prey model: initially as the number of lions (the predators) increases, the number of gazelles (their prey) decreases. When there are not enough gazelles to feed all the lions, the number of lions decreases and the number of gazelles starts to increase. a.
b.
A Cup of Hot Coffee Cooling Over Time
Room temp.
1995
Time (minutes)
E X A M P L E
6
U.S. Venture Capital 1995–2005
c.
Predator-Prey Model of Lions vs. Gazelles Number of animals
5
Venture capital ($)
E X A M P L E
Temperature (°F)
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Lions
Gazelles
2005
Time
Year
You are a TV journalist. Summarize for your viewers the essence of each of the graphs in Figures 1.32 and 1.33. a. Note: The vertical scale is used in two different ways on this graph from the U.S. Bureau of the Census. Number in Poverty and Poverty Rate, 1959–2004 Numbers in millions, rates in percentage 45
Number in poverty
40
36.9 million
35 30 25 20
Poverty rate
15
12.7%
10 5 0 1959
1964
1969
1974
1979
1984
1989
1994
1999
2004
Figure 1.32 Source: U.S. Bureau of the Census www.census.gov/compendia/statab/incom_expenditures_wealth/household_incom.
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43
Data on the top 400 taxpayers in the United States The percentage of their income paid in taxes (in 2000 dollars)
Their mean income (in 2000 dollars) 30 In 2000 dollars 150
Percentage
Dollars (in millions)
200
100 50
20 10 0
0 '92 '94 '96 '98 '00
'92 '94 '96 '98 '00
Figure 1.33 Source: The New York Times, June 26, 2003.
S O L U T I O N
a. In 1959, according to the U.S. Bureau of the Census, there were about 40 million Americans in poverty, representing almost 23% of the population. Between 1959 and 1974 both the number and the percentage of Americans in poverty decreased to a low of about 23 million and 12%, respectively. From 1974 to 2004, the percentage in poverty continued to hover between 12% and 15%. During the same time period the number in poverty vacillated, but the overall trend was an increase to about 37 million in 2004. b. From 1994 to 2000 the mean income of the top 400 taxpayers grew steadily, but overall the percentage of their income that went toward taxes decreased.
We will spend the rest of this text examining patterns in functions and their graphs. We’ll study “families of functions”—linear, exponential, logarithmic, power, and polynomial—that will provide mathematical tools for describing the world around us.
Algebra Aerobics 1.5 1. Create a title for each of the following graphs. a.
b.
$4500
300
42-inch HD plasma TV
Average selling price
$3500 32-inch LCD TV $3000 $2500 $2000 $1500 50-inch digital rear-projection TV (DLP, LCD, LCoS) $1000 $500
iTunes Music, tracks sold (millions)
$4000
200
*Projected.
Source: Displaysearch.
Q1 2005
Q2 2005
Q3 2005
Q4 2005*
Q1 2006*
12/16/2004
150
10/14/2004
100
50
7/11/2004 3/15/2004 12/15/2003 9/8/2003
$0 Q4 2004
1/24/2005
250
0 Apr’03 Jul’03 Oct 03 Jan’04 Apr’04 Jul’04 Oct’04 Jan’05
Quarterly reports
Date reported
Source: Displaysearch.
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2. Use Figure 1.34 to answer the following questions. $47,000 $46,000 $45,000 $44,000 $43,000 $42,000 $41,000 $40,000 $39,000 1990
S
S
t
t
Graph A 1993
1996
1999
Graph C
2002 2004
Year
S
S
Figure 1.34 Median household income
adjusted for inflation (in constant 2004 dollars).
a. Estimate the maximum value for median household income during the time period represented on the graph. In what year does the maximum occur? What are the approximate coordinates at the maximum point? b. What is the minimum value for median household income? In what year does this occur? What are the coordinates of this point? c. Describe the changes in median household income from 1990 to 2004. 3. Choose the “best” graph in Figure 1.35 to describe the following situation. Speed (S) is on the vertical axis and time (t) is on the horizontal axis. A child in a playground tentatively climbs the steps of a large slide, first at a steady pace, then gradually slowing down until she reaches the top, where she stops to rest before sliding down.
t
t
Graph B
Graph D
Figure 1.35
4. Generate a rough sketch of the following situation. U.S. AIDS cases increased dramatically, reaching an all-time high for a relatively short period, and then consistently decreased, until a recent small increase. 5. Sketch a graph for each of the following characteristics, and then indicate with an arrow which arm of the graph is increasing and which is decreasing. a. Concave up with a minimum point at (22, 1). b. Concave down with a maximum point at (3, 22).
Exercises for Section 1.5 A graphing program is required for Exercises 12, 13, 20, 27, and 29.
y 6
1. Identify the graph (A or B) that a. Increases for 1, x,3 b. Increases for 2, x,5 c. Decreases for 22 ,x ,2 d. Decreases for 3, x ,5
4 2 –6
2
4
6
x
–4 –6
y
Graph B
4 2 –2
–2 –2
6
–4
–4
2 –2 –4 –6
Graph A
4
6
x
2. The Federal Reserve is the central bank of the United States that sets monetary policy. The Federal Reserve oversees money supply, interest rates, and credit with the goal of keeping the U.S. economy and currency stable. The federal funds rate is the interest rate that banks with excess reserves at a Federal Reserve district bank charge other banks that need overnight loans. Look at the accompanying graphs for the time period between 2000 and 2003.
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1.5
a. Describe the overall trends for this time period for the federal funds rate. b. Describe the overall trends for this time period for credit card rates. c. Describe the overall trends for this time period for the 30year mortgage rates. d. Estimate the maximum federal funds rate for this time period. When did it occur? e. Approximate the minimum federal funds rate for this time period. When did it occur? f. Write a topic sentence that compares the federal funds rate with the consumer loan rates for credit cards and mortgages for this time period.
Federal Funds Rate
Visualizing Functions
45
b. Over which interval is the function increasing? c. Does the function appear to have a minimum? If so, where? d. Does the function appear to have a maximum? If so, where? e. Describe the concavity.
4
y
y
12
x –12
4
Average Loan Rates
Overnight loan rate between banks
x
10
–4
16%
8%
14 12
6
–4
–12
Graph B
Graph A
Credit cards*
10 8
4
5. For the following function,
6 4
2
30-Year mortgage
2 0
y 12
0 '01
'00
'02
'03
'00
'01
'02
'03
* Average annual percentage rate for all credit card accounts in banks in the United States that respond to the Federal Reserve's Survey. National average for a 30-year fixed rate on a single-family home.
Source: www.federalreserve.gov
–6
3. The accompanying graphs show the price of shares of stock of two companies over the one-week period January 9 to 16, 2004. Describe the changes in price over this time period.
$17
$34
$16 $32 $15 $30 $14 $28 F
M
T
W
Rockwell Collins NYSE: COL
T
F
$13 F
M
T
W
T
Transkaryotic Therapies NNM: TKTX
Source: The New York Times, Jan. 17, 2004.
4. For each of the following functions, a. Over which interval is the function decreasing?
F
6
x
–12
a. b. c. d. e.
Over which interval(s) is the function positive? Over which interval(s) is the function negative? Over which interval(s) is the function decreasing? Over which interval(s) is the function increasing? Does the function appear to have a minimum? If so, where? f. Does the function appear to have a maximum? If so, where? 6. Choose which graph(s) at the top of the next page match the description: As x increases, the graph is: a. Increasing and concave up b. Increasing and concave down c. Concave up and appears to have a minimum value at (23, 2) d. Concave down and appears to have a maximum value at (23, 2)
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y 5
4
d. Over which interval(s) is f(x) decreasing? e. How would you describe the concavity of f (x) over the interval (0, 5) for x? Over (5, 8) for x? f. Find a value for x when f(x) 5 4. g. f (28) 5 ?
y
x 2
–8
x –3
3 –1
–10
Graph A
Graph C
y 6 10
y
10. Match each graph with the best description of the function. Assume that the horizontal axis represents time, t. i. The height of a ball thrown straight up is a function of time. ii. The distance a truck travels at a constant speed is a function of time. iii. The number of daylight hours is a function of the day of the year. iv. The temperature of a pie baking in an oven is a function of time. Q
Q x 6
–2
x –8
–2
2
–2
Graph B
Graph D
t
7. Examine each of the graphs in Exercise 6. Assume each graph describes a function f (x). The arrows indicate that the graph extends indefinitely in the direction shown. a. For each function estimate the domain and range. b. For each function estimate the x interval(s) where f(x) . 0. c. For each function estimate the x interval(s) where f(x) , 0.
t
Graph A
Graph C Q
Q
8. Look at the graph of y 5 f(x) in the accompanying figure.
t
t
y 8
f (x) x –8
Graph D
Graph B
12
11. Choose the best graph to describe the situation. a. A student in a large urban area takes a local bus whose route ends at the college. Time, t, is on the horizontal axis and speed, s, is on the vertical axis. s
s
s
–6
a. b. c. d. e.
Find f(26), f(2), and f(12). Find f(0). For what values of x is f(x) 5 0? Is f(8) . 0 or is f(8) , 0? How many times would the line y 51 intersect the graph of f(x)? f. What are the domain and range of f(x)? g. What is the maximum? The minimum?
9. Use the graph of Exercise 8 to answer the following questions about ƒ(x). a. Over which interval(s) is f(x) , 0? b. Over which interval(s) is f(x) . 0? c. Over which interval(s) is f(x) increasing?
t
t
t
Graph A
Graph B
Graph C
b. What graph depicts the total distance the student traveled in the bus? Time, t, is on the horizontal axis and distance, d, is on the vertical axis. d
d
d
t Graph D
t
t Graph E
Graph F
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1.5
12. (Graphing program required.) Use technology to graph each function. Then approximate the x intervals where the function is concave up, and then where it is concave down a. ƒsxd 5 x3 b. gsxd 5 x3 2 4x 13. (Graphing program required.) Use technology to graph each function. Then approximate the x intervals where the function is concave up, and then where it is concave down. a. hsxd 5 x4 b. ksxd 5 x4 2 24x 1 50 (Hint: Use an interval of [25, 5] for x and [0, 200] for y.)
17. Examine the accompanying graph, which shows the populations of two towns. Comparison of Populations, 1900–1990
5
–5
5
x
6 Population in 100,000’s
y
y
–4
4
x
–5
Graph A
Graph B
Johnsonville Palm City
5 4 3 2 1 0
–5
47
b. The temperature during a 24-hour period in your home town during one day in July. c. The amount of water inside your fishing boat if your boat leaks a little and your fishing partner bails out water every once in a while. d. The total hours of daylight each day of the year. e. The temperature of an ice-cold drink left to stand.
14. a. In the accompanying graphs, estimate the coordinates of the maximum and minimum points (if any) of the function. b. Specify the interval(s) over which each function is increasing. 5
Visualizing Functions
0 10 20 30 40 50 60 70 80 90 Years after 1900
a. What is the range of population size for Johnsonville? For Palm City? b. During what years did the population of Palm City increase? c. During what years did the population of Palm City decrease? d. When were the populations equal?
15. Consider the accompanying graph of U.S. military sales to foreign governments from 1995 to 2003. U.S. Military Sales to Foreign Goverments 18 16
18. In Section 1.2 we examined the annual federal budget surplus or deficit. The federal debt takes D A T A into account the cumulative effect of all the deficits and surpluses for each year together with any interest or payback of principal. The accompanying graph shows the accumulated gross federal debt from 1950 to 2005. (See related Excel or graph link file FEDDEBT.) Create a topic sentence for this graph for a newspaper article.
Billions of dollars
14 12 10 8 6 4 2 0 1995
1996
1997
1998
1999
2000
2001
2002
2003
U.S. Accumulated Gross Federal Debt (in billions)
Year
8,000
Source: U.S. Department of Defense, Defense Security Cooperation Agency, DSCA Data and Statistics.
7,000
a. b. c. d.
Between what years did sales increase? Between what years did sales decrease? Estimate the maximum value for sales. Estimate the minimum value for sales.
16. Sketch a plausible graph for each of the following and label the axes. a. The amount of snow in your backyard each day from December 1 to March 1.
Billions of dollars
6,000 5,000 4,000 3,000 2,000 1,000 0 1950
1960
1970
1980 Year
1990
Source: U.S. Dept. of Treasury, www.ustreas.gov
2000
2010
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19. Consider the accompanying table. Y 1990 1991 1992 1993 1994 1995
23. Examine the graphs of military reserve enlisted personnel over the years 1990 to 2004. P $1.4 $2.3 $0 2$0.5 $1.4 $1.2
a. Is P a function of Y? b. What is the domain? What is the range? c. What is the maximum value of P? In what year did this occur? d. During what intervals was P increasing? Decreasing? e. Now consider P as the independent variable and Y as the dependent variable. Is Y a function of P?
Male Enlisted Reserve Personnel 1,500 1,400 Number (thousands)
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1,300 1,200 1,100 1,000
800 700 1990
For each function, a. Estimate the maximum value of y on each interval. b. Estimate the minimum value of y on each interval. 21. The accompanying graph shows the world population over time and future predictions.
1994
1996
1998
2000
2002
2004
2002
2004
Female Enlisted Reserve Personnel 220 210 Number (thousands)
y2 5 0.5x3 2 2x 2 1
1992
Year
20. (Graphing program required.) Using technology, graph each function over the intervals [26, 6] for x and [220, 20] for y. y1 5 x2 2 3x 1 2
900
200 190 180 170 160 150 1990
1992
1994
1996
1998
2000
Year
Historical World Population with Predictions 12,000
Source: U.S. Dept. of Defense, Official Guard and Reserve Manpower Strengths and Statistics, annual.
Population (millions)
10,000
a. In what year(s) did female and male enlisted personnel reach a maximum? b. What was the maximum and minimum for both men and women over the time interval [1990, 2004]? c. Describe the trends in number of female and male reserve enlisted personnel. d. Given the current state of world affairs, what would you expect would happen to the enlistment numbers for 2004 to the present?
8,000 6,000 4,000 2,000 0 0
250
500
750 1000 1250 1500 1750 2000 2250 Year
Source: www.worldonline.n1/invd/world/whist2.html
a. Over what interval does the world population show dramatic growth? b. Does the dramatic growth slow down? c. Write a topic sentence for a report for the United Nations. 22. Make a graph showing what you expect would be the relative ups and downs throughout the year of sales (in dollars) of a. Turkeys c. Bathing suits in your state b. Candy d. Textbooks at your school bookstore
24. Generate a rough sketch of a graph of internal pressure vs. time for the following situation: When a soda is removed from the fridge, the internal pressure is slightly above the surrounding air pressure. With the can unopened, the internal pressure soon more than doubles, stabilizing at a level three times the surrounding air pressure. 25. A student breaks her ankle and is taken to a doctor, who puts a cast on her leg and tells her to keep the foot off the ground altogether. After 2 weeks she is given crutches and can begin to walk around more freely, but at the beginning of the third week she falls and is resigned to keeping stationary again for a while. After 6 weeks from her first fall, she is given a walking cast in which she can begin to put her foot on the
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Chapter Summary
ground again. She is now able to limp around using crutches. Her walking speed slowly progresses. At 12 weeks she hits a plateau and, seeing no increase in her mobility, starts physical therapy. She rapidly improves. At 16 weeks the cast is removed and she can walk freely. Make a graph of the student’s mobility level during her recovery. 26. Every January 1 a hardy group called the L Street Brownies celebrates the New Year by going for a swim at the L Street Beach in South Boston. The water is always very cold, and swimmers adopt a variety of strategies for getting into it. The graph shows the progress of three different friends who join in the event, with percentage of body submerged on the vertical axis and time on the horizontal axis. Match the graphs to the descriptions below of how each of the friends manages to get completely submerged in the icy ocean.
49
27. (Graphing program required.) Use technology to graph the following functions and then complete both sentences for each function. 1 1 , y1 5 x3 , y2 5 x2 , y3 5 y4 5 1 2 x13 x a. As x approaches positive infinity, y approaches __________. b. As x approaches negative infinity, y approaches __________. 28. Describe the behavior of ƒ(x) in the accompanying figure over the interval (2`, 1`) for x, using such words as “increases,” “decreases,” “concavity,” “maximum/minimum,” and “approaches infinity.” y 4
Swimmers in South Boston
Percent submerged
100
f(x) 75 50
x 3
–3 25 0
0
1
2
3
4
5
Time
Minutes
a. Ali has done this before and confidently walks in until his head is underwater; then he puts his head out and swims around a few minutes; then he walks out. b. Ben dashes in until the water is up to his knee, trips over a hidden rock, and falls in completely; he stands up and, since he is now totally wet, runs back out of the water. c. Cat puts one foot in, takes it out again, and shivers. She makes up her mind to get it over with, runs until she is up to her waist, dives in, swims back as close to the water’s edge as she can get, stands up, and steps out of the water.
29. (Graphing program required.) This exercise is to be done with a partner. Name the partners person #1 and person #2. a. Person #1, using technology, graphs the function fsxd 5 0.5sx 2 3dsx 1 2d 2, but does not show the graph to person #2. b. Person #1 describes to person #2 the behavior of the graph of ƒ(x) so that he/she can sketch it on a piece of paper. c. Switch roles; now person #2, using technology, graphs gsxd 5 20.5sx 2 3dsx 1 2d 2, but does not show the graph to person #1. d. Person #2 describes to person #1 the behavior of the graph of g(x) so that he/she can sketch it on a piece of paper. e. Compare the accuracy of the graphs and compare the shapes of the two graphs. What do ƒ(x) and g(x) have in common? How do they differ?
Single-Variable Data Single-variable data are often represented with bar charts, pie charts, and histograms.
Two-Variable Data Graphs of two-variable data show how change in one variable affects change in the other. The accompanying graph is called a time series because it shows changes over time.
Water level (in centimeters)
CHAPTER SUMMARY +10 0 –10
0
4
8 12 16 Time (in hours)
20
24
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Equations in Two Variables
Function Notation
The solutions of an equation in two variables x and y are the ordered pairs (x, y) that make the equation a true statement. For example, (3, 25) is one solution of the equation 2x 1 y 5 1. The graph of an equation in two variables displays the set of points that are solutions to the equation.
The expression f (x) means the rule f is applied to the input value x to give the output value, f(x):
Functions
f(input) 5 output For example, f(x) 2x 5 tells us f is the name of a function where the input is x, the output is f(x), and the rule is to multiply the input by 2 and add 5. To evaluate this function when the input is 4, we write f(4) (2 • 4) 5 1 f(4) 13
A variable y is a function of a variable x if each value of x determines a unique (one and only one) value of y. Functions can be represented with words, graphs, equations, and tables.
Visualizing Functions
When a set of ordered pairs represents a function, then each ordered pair is written in the form
Graphs of functions may demonstrate the following properties: Maximum
(independent variable, dependent variable) (input, output) (x, y) By convention, on the graph of a function, the independent variable is represented on the horizontal axis and the dependent variable on the vertical axis. A graph does not represent a function if it fails the vertical line test. If you can draw a vertical line that crosses the graph two or more times, the graph does not represent a function. The domain of a function is the set of all possible values of the independent variable. The range is the set of corresponding values of the dependent variable.
Minimum
Increasing
Decreasing
Concave up
Concave down
C H E C K Y O U R U N D E R S TA N D I N G I. Is each of the statements in Problems 1–27 true or false?
4. The graph shows the changes in mean snowfall for each month of the year over the years 1948–2005.
1. Histograms, bar charts, and pie charts are used to graph single-variable data.
5. The maximum snowfall occurs in March, with mean snowfall of about 52 inches.
2. Means and medians, both measures of central tendency, can be used interchangeably.
6. The accompanying figure, of the wolf population in Yellowstone National Park (YNP) and the Northern
3. A scatter plot is a plot of data points (x, y) for twovariable data.
Yellowstone National Park Wolf Population, 1995–2005
Problems 4 and 5 refer to the accompanying figure. 200 Mean Snowfall (inches), Blue Canyon California, 1948–2005
YNP
NR
174
Number of wolves
Mean snowfall (in)
60 50 40 30 20
150
J
F
M
A
M
J
J
A
S
O
Month
Source: Western Regional Climate Center, www.wrcc.dri.edu.
N
D
132 119
118 106
100
50
10 0
169
148
83
80
48 37 24 21 21
32
72
72
77
84
85 54
44 48
0 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005
Source: Yellowstone Wolf Project, Annual Report 2005, U.S. Department of the Interior.
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Check Your Understanding
Region (NR) of Yellowstone, shows that the wolf population in the Greater Yellowstone area is increasing.
Problems 20–22 refer to the accompanying graph of f(x). y 3
7. The independent variable of a function is also called the output of the function. 8. The graph of the equation describing the sales revenue R ($ million) as a function of the amount spent on advertising A ($ thousand) is the set of ordered pairs (R, A) that satisfy the equation. 9. If sales revenue R is a function of advertising A, then R is the dependent variable and A is the independent variable.
x 5
–3
f (x) –3
20. The function f (x) has both a maximum and a minimum over the interval [21, 3). 21. The function f(x) decreases over the interval (0, 2).
10. If M 5 F(q), then F is a function of q. Problems 11 and 12 refer to the function Fsqd 5 11. Fs21d 5
51
2 . 3q
22. The function f(x) is concave up for x . 1. Problems 23 and 24 refer to the accompanying figure.
22 . 3
y 9 (–1, 7)
12. The domain of F(q) is the set of all real numbers.
6
13. In the accompanying graph y is a function of x.
(2, 4) 3
y
g(x) x
–3
x
0
3
23. The domain of the function g(x) is the interval [21, 2]. 24. The range of the function g(x) is the interval [4, 7]. Problems 25–27 refer to the accompanying figure. U.S. Bicycle Use by Age in Year 2000
Problems 14–16 refer to the accompanying graph of h(x). Number (in millions)
25
y (1, 3)
x
20 15 10 5 0
h(x)
18 to 44 years 45 years and older
14. The function h (x) has a maximum value at (1, 3). 15. The function h (x) is concave down. 16. The function h (x) increases to the right of x 5 1 and increases to the left of x 5 1. Problems 17 and 18 refer to the following table.
Source: National Survey on Recreation and the Environment, www.bts.gov
25. The histogram describes the number of bicycle riders by age. 26. There are about twice as many bicycle riders who are 18 to 44 years old as there are riders who are 45 years or older. 27. The total number of bicycle riders is over 30 million.
R
21
2
0
3
S
10
8
6
4
17. The variable S is a function of R. 18. The variable S is decreasing over the domain for R. 19. A set of ordered pairs of the form (M, C) implies that M is a function of C.
II. In Problems 28–34, give an example of a graph, relationship, function, or functions with the specified properties. 28. A relationship between two variables w and z described with an equation where z is not a function of w. 29. A relationship between two variables w and z described as a table where w is a function of z but z is not a function of w. 30. A graph of a function that is increasing and concave down.
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31. A graph of a function that is decreasing and concave up. 32. A graph of a function that is concave up and has a minimum value at the point (22, 0).
38. The graph in the accompanying figure is decreasing and concave down. y
33. A graph of a function where the domain is the set of real numbers and that has no maximum or minimum values. 34. A topic sentence describing the wolf pups that were born and survived in Yellowstone Park as shown in the accompanying graph. Yellowstone National Park, Pups Born and Survived 80
x
39. The graph in the accompanying figure is increasing and concave up. y
70
Number of pups
60 50 40
x
30 20 number born number survived
10 0
1995
1996
1997
1998
1999
2000
2001
2002
Source: Yellowstone Wolf Project, Annual Report 2002, U.S. Department of the Interior.
40. All functions have at least one minimum value. 41. Sometimes the choice as to which variable will be the independent variable for a function is arbitrary. 42. The graph in the accompanying figure represents P as a function of Q. P
III. Is each of the statements in Problems 35–42 true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample or explain why it is false. 35. A function can have either a maximum or a minimum but not both. 36. Neither horizontal nor vertical lines are functions.
Q
0
37. A function is any relationship between two quantities.
CHAPTER 1 REVIEW: PUTTING IT ALL TOGETHER 1. A man weighed 160 pounds at age 20. Now, at age 60, he weighs 200 pounds. a. What percent of his age 20 weight is his weight gain? b. If he went on a diet and got back to 160 pounds, what percent of his age 60 weight would be his weight loss? c. Explain why the weight gain is a different percentage than the weight loss even though it is the same number of pounds. 2. Data from the World Health Organization Report 2005 shows vast differences in average spending per person on health for different countries. Data on health spending from selected countries is shown here.
Country China India Iraq North Korea South Korea United States
As % of Gross Domestic Product (GDP)
$ per Capita
5.8 6.1 1.5 4.6 5.0 14.6
63 30 11 0.3 532 5274
a. North and South Korea spent close to the same percentage of their gross domestic product on health, but a very
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different amount per person (per capita) in 2005. How can this be? b. In 2005, how many Iraqis could receive health care (at the Iraqi level of spending) for the amount the United States spent on one person? If Iraq spent the same percent of its GDP on health care as the United States, how much would they be spending per person? c. China and India had the two largest populations in 2005, with China at 1.31 billion and India at 1.08 billion. Estimate how much each of these countries is spending on health care. 3. In April 2005 there was considerable debate in the media about whether “average” incomes had gone up or down in the United Kingdom (UK). The Institute for Fiscal Studies produced a report in which they stated that the mean household income in the UK in 2004 fell by 0.2% over the previous year and that median household income in the UK rose for the same period by 0.5%. Explain how the mean income could go down while the median income rose. 4. In 2001 there were 130,651 thousand metric tons of commercial fish caught worldwide. The chart below lists the world’s 10 leading commercial fishing nations.
The 10 Leading Commercial Fishing Nations in 2001 Fish caught (in thousands of metric tons) China Peru India Japan United States Indonesia Chile Russia Thailand Norway
42,579 7,996 5,897 5,515 5,424 5,137 4,363 3,718 3,657 3,198
53
5. The following table shows the four leading causes of death around the world. The Four Leading Causes of Death Worldwide Cause
% of All Deaths
Heart disease Stroke Lower respiratory infections (e.g., pneumonia, emphysema, bronchitis) HIV/AIDS
12.6 9.7 6.8
4.9
Source: World Health Organization (WHO), The World Health Report, 2003.
a. What is the leading cause of death worldwide? b. Create a bar chart using these data. c. What percentage of worldwide deaths is not accounted for in the table? List at least two other diseases or conditions that can cause death. Could either of them account for more that 5% of all deaths worldwide? 6. Before doctors transfuse blood they must know the patient’s blood type and which types of blood are compatible with that type. a. On the following graph, types O1 and O2 make up approximately 37% and 6%, respectively, of all U.S. blood types. Estimate the percentage of the population in the United States for each of the other blood types. b. How could you determine the number of people with each blood type? % U.S. Population by Blood Type in 2006 B–
AB+ AB–
B+ 0+ A–
Source: U.S. Bureau of the Census, Statistical Abstract of the United States: 2006.
A+
a. What are the mean and median for this set of data? b. China caught what percent of the commercial fish among the leading 10 nations? Of the world? c. Will the mean and median for the 10 leading nations change if China substantially increases the amount of commercial fish it catches while other nations remain at the same level?
0–
Source: http://bloodcenter.stanford.edu/ about_blood/blood_types.html.
7. In a certain state, car buyers pay an excise tax of 2.5%. This means that someone who buys a car must pay the state a onetime fee of 2.5% of the car’s value. Use P to represent the price of the car and E to represent excise tax, then express E as a function of P.
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8. A waitress makes $2.50 an hour in wages and receives tips totaling 18% of the price of each meal she serves. Let H represent the hours she works, P represent the total cost of the meals she served, and E represent what she earns in a week before taxes. Write an equation expressing E in terms of H and P. 9. The following graph shows the number of babies per million babies that were named Emma in the United States between 1880 and 2004. Create a title for the graph and explain in a few sentences what is happening over time to the name Emma.
b. Create a table of values for the variables A and B, where B is a function of A but A is not a function of B. c. Create two graphs, one that represents a function and one that does not represent a function. 12. Find an equation that represents the relationship between x and y in each of the following tables. Specify in each case if y is a function of x. a. x b. x c. x y y y
8000 7000
Per million babies
11. a. Determine whether each set of points represents a function. i. (6, 21), (5, 21), (8, 3), (23, 2) ii. (5, 3), (7, 21), (5, 22), (4, 23) iii. (7, 23), (8, 23), (22, 23), (9, 23) iv. (4, 0), (4, 23), (4, 7), (4 2 1)
6000
0 1 2 3 4
5000 4000 3000
Emma 2000 1000
1880 1890 1900 1910 1920 1930 1940 19501960 19701980 1990 2004
Source: thebabywizard.ivillage.com.
150
1 2 9 28 65
60
50
40
30 O N D J F M A M J 2004 2003
$ (billions)
0 1 2 3 4
Favor bringing home troops in the next year Favor waiting until stable government established
% Financial costs of natural disasters during the 1990s
21 0 3 8 15
0 1 2 3 4
13. The following graph shows changes in U.S. opinion regarding the war in Iraq over the period between October 2003 and June 2005.
10. During the 1990s natural disasters worldwide took more than 666,000 lives and cost over $683 billion (US $). Using the accompanying chart, write a 60-second summary describing the worldwide financial costs of natural disasters during this period. 200
2 4 6 8 10
J
A S O N D J F M A MJ 2005
Source: Harris Interactive. 100
50
Source: Australian government, Department of Meteorology.
Tsunami
Drought/famine
Heat-wave
Bushfire
Severe storm
Tropical cyclone
Other/geophysical
Flood
0
a. In February 2004, what was the approximate percentage of Americans who favored bringing troops home in the next year? b. In the period covered by this graph, when did more people favor waiting for the establishment of a stable government than favored a withdrawal in the next year? c. During which interval(s) did the amount of people in favor of waiting for a stable government increase? d. On which date was the difference in the popularity of the two opinions the greatest? Estimate this difference. 14. The following graph shows fuel economy in relation to speed. a. Does the graph represent a function? If not, why not? If so, what are the domain and range?
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b. Describe in words how speed relates to fuel economy.
c. For the equations in part (a) that represent functions, what is the independent variable? The dependent variable? What is a reasonable domain? What is a reasonable range? d. Generate small tables of values and graphs for C1 and C2. How do the two graphs compare? e. How do you think these graphs would compare to the corresponding graphs for a 200 lb. person? Explain.
35 Fuel economy (mpg)
30 25 20
17. Sketch the graph of a function f (t) with domain 5 [0, 10] and range 5 [0, 100] that satisfies all of the given conditions:
15 10
f(0) 5 100, f(5) 5 60, f(10) 5 0 f(t) is decreasing and concave down for interval (0, 5) f(t) is decreasing and concave up for interval (5, 10)
5 0
55
5
15
25
35 45 Speed (mph)
55
65
75
18. Below is the graph of f(x).
(mpg 5 miles per gallon of gas, mph 5 miles per hour) Source: www.fueleconomy.gov.
y 5
15. In 2006 the exchange rate for the Chinese yuan and U.S. dollar was about 8 to 1; that is, you could exchange 8 yuan for 1 U.S. dollar. a. Let y 5 number of Chinese yuan and d 5 number of U.S. dollars. Complete the following table. 1
2
3
4
10
20
–5
b. Find an equation that expresses y as a function of d. What is the independent variable? The dependent variable? c. Complete the following table: y
8
12
16
20
50
80
100
d d. Find an equation that expresses d as a function of y. What is the independent variable? The dependent variable? 16. The average number of calories burned with exercising varies by the weight of the person. The accompanying table gives the approximate calories burned per hour from disco dancing. Number of Calories Burned/hr from Disco Dancing Weight calories burned per hour
x
100
100 lb. person 264
125 lb. person 330
150 lb. person 396
175 lb. person 462
200 lb. person 528
Source: http://www.fitresource.com/Fitness.
Let t be the number of hours spent exercising and C the total number of calories burned. a. Write an equation expressing C1 in terms of t for a 125 lb. person disco dancing and an equation for C2 for a 175 lb. person disco dancing. b. Do your equations in part (a) represent functions? If not, why not?
a. Determine the values for f(21), f(0), f(2). b. For what value(s) of x does f(x) 5 22 ? c. Specify the interval (s) over which the function is i. Increasing ii. Decreasing iii. Concave up iv. Concave down 19. The following graph shows the life expectancy in Botswana compared with world life expectancy. The dotted lines on the graph represent projections. Life Expectancy, World and Botswana, 1970–2010 80 Life expectancy (years)
d y
4
–4
75 70 65 60
World
55
Botswana
50 45 40 1970
1980
1990 Year
2000
2010
Source: UN Population Division, 1999.
a. Does the graph for Botswana represent a function? If not, why not? If so, what are the domain and range? b. Over the given time period, when did life expectancy in Botswana reach its maximum? Estimate this maximum value.
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c. When was life expectancy in Botswana projected to reach its minimum? Estimate this minimum value. What might have caused the sudden drop in life expectancy? d. Over what interval(s) was life expectancy (actual and projected) in Botswana increasing? Decreasing? e. Over what interval(s) is Botswana’s graph concave down? Concave up? f. Write a short description of life expectancy in Botswana compared with world life expectancy.
e. Write a 60-second summary about where you would suggest UNICEF focus its efforts in childhood education.
Where Children Are Out of School Middle East/ North Africa 9%
CEE/CIS and Baltic states 3%
East Asia/ Pacific 10%
20. Below is a graph of data collected by the FBI.
Juvenile Arrests for Murder
Latin America/ Caribbean 3%
South Asia 36%
4000
Industrialized countries 2%
Number of arrests
3500 3000 2500 2000 1500 1000 500 0 1980
1985
1995
1990
2000
Sub-Saharan Africa 37%
2005
Year
Source: U.S. Bureau of The Census, Statistical Abstract of the United States: 2006.
If we let N(x) 5 number of juvenile arrests for murder in year x, a. Estimate N(1993). What are the coordinates of this point? What does this point represent? b. Estimate the coordinates of the minimum point. What does this point represent? c. Over what interval is the function increasing? Over what interval is the function decreasing? d. What is the domain of N(x)? What is the range? e. Write a brief summary about juvenile arrests between 1980 and 2003. 21. UNICEF has made children’s education worldwide a top priority. According to UNICEF, 25 years ago only half of the world’s children received a primary school education. Today, 86% receive a primary education. The following pie chart shows for each region the percentage of the total number of worldwide children out of school. a. Explain what the label 3% means next to the region Latin America/Caribbean. b. Does this graph tell us what percentage of children in Sub-Saharan Africa are out of school? Explain why or why not. c. What does this graph not show? Why might this be confusing? d. What is the region with the largest percentage of the world’s children who are out of school? What is the total percentage of the two regions with the most children out of school? What is the region with the smallest percentage?
22. a. Given g(x) 5 2x2 1 3x 2 1, evaluate g(0) and g(21). b. Find the domain of g(x). 1 , evaluate f(0), f(21), and f(2). x22 b. Find the domain of f(x).
23. a. Given f(x) 5
24. Find the domain and range of the following functions: a. hsxd 5 "x b. f sxd 5 "x 2 6 25. The following table shows the cost each month in a small condominium in Maine for natural gas, which is used for cooking, heating, and hot water. Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec $ gas
220 234 172
83
51
42
30
32
51
89
102 150
a. What is the month with the maximum cost for gas? The month with the minimum? Do these numbers seem reasonable? Why or why not? b. What is the mean monthly cost of gas for the year? What is the median monthly cost of gas? Describe in words what these descriptors tell you about the cost of gas in this condominium. c. Create a table with intervals $0–49, $50–99, $100–149, and so on, and next to each interval show the number of months that fall within that cost interval. Use the table to construct a histogram of these data (with the cost intervals on the horizontal axis). d. Describe two patterns in the cost of gas for this condominium.
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26. The following graphs are population pyramids for Yemen (a developing country) and Japan (a highly developed country).
Write a 60-second summary comparing and contrasting the populations in the year 2000 in these two countries.
2000
Japan 2000 Female Male
Age
Age
Yemen 2000 100+ 95–99 90–94 85–89 80–84 75–79 70–74 65–69 60–64 55–59 50–54 45–49 40–44 35–39 30–34 25–29 20–24 15–19 10–14 5–9 0–4 1500 1000
500 0 500 1000 1500 Population, in thousands
Source. WHO World Population Prospects, 2002.
57
2000
100+ 95–99 90–94 85–89 80–84 75–79 70–74 65–69 60–64 55–59 50–54 45–49 40–44 35–39 30–34 25–29 20–24 15–19 10–14 5–9 0–4 2000
Female Male
4000
2000 0 2000 Population, in thousands
4000
6000
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E X P L O R AT I O N 1 . 1 Collecting, Representing, and Analyzing Data Objectives • explore issues related to collecting data. • learn techniques for organizing and graphing data using a computer (with a spreadsheet program) or a graphing calculator. • describe and analyze the overall shape of single-variable data using frequency and relative frequency histograms. • use the mean and median to represent single-variable data. Material/Equipment • • • • • • •
class questionnaire measuring tapes in centimeters and inches optional measuring devices: eye chart, flexibility tester, measuring device for blood pressure computer with spreadsheet program or graphing calculator with statistical plotting features data from class questionnaire or other small data set formatted either as spreadsheet or graph link file overhead projector and projection panel for computer or graphing calculator transparencies for printing or drawing graphs for overhead projector (optional)
Related Readings (On the web at www.wiley.com/college/kimeclark) “U.S. Government Definitions of Census Terms” “Health Measurements” Related Software “F1: Histograms,” in FAM 1000 Census Graphs Procedure This exploration may take two class periods. Day One In a Small Group or with a Partner 1. Pick (or your instructor will assign you) one of the undefined variables on the questionnaire. Spend about 15 minutes coming up with a workable definition and a strategy for measuring that variable. Be sure there is a way in which a number or single letter can be used to record each individual’s response on the questionnaire. 2. Consult the reading “Health Measurements” if you decide to collect health data. Class Discussion After all of the definitions are recorded on the board, discuss your definition with the class. Is it clear? Does everyone in the class fall into one of the categories of your definition? Can anyone think of someone who might not fit into any of the categories? Modify the definition until all can agree on some wording. As a class, decide on the final version of the questionnaire and record it in your class notebook.
58
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Exporation 1.1
59
In a Small Group or with a Partner Help each other when necessary to take measurements and fill out the entire questionnaire. Questionnaires remain anonymous, and you can leave blank any question you can’t or don’t want to answer. Hand in your questionnaire to your instructor by the end of class. Exploration-Linked Homework Read “U.S. Government Definitions of Census Terms” for a glimpse into the federal government’s definitions of the variables you defined in class. How do the “class” definitions and the “official” ones differ? Day Two Class Demonstration 1. If you haven’t used a spreadsheet or graphing calculator before, you’ll need a basic technical introduction. (Note: If you are using a TI-83 or TI-84 graphing calculator, there are basic instructions in the Graphing Calculator Manual on www.wiley.com/college/kimeclark.) Then you’ll need an electronic version of the data set from which you will choose one variable for the whole class to study (e.g., age from the class data). a. If you’re using a spreadsheet: • Copy the column with the data onto a new spreadsheet and graph the data. What does this graph tell you about the data? • Sort and replot the data. Is this graph any better at conveying information about the data? b. If you’re using a graphing calculator: • Discuss window sizes, changing interval sizes, and statistical plot procedures. 2. Select an interval size and then construct a frequency histogram and a relative frequency histogram. If possible, label one of these carefully and print it out. If you have access to a laser printer, you can print onto an overhead transparency. 3. Calculate the mean and median using spread sheet or graphing calculator functions. In a Small Group or with a Partner Choose another variable from your data. Pick an interval size, and then generate both a frequency histogram and a relative frequency histogram. If possible, make copies of the histograms for both your partner and yourself. Calculate the mean and median. Discussion/Analysis With your partner(s), analyze and jot down patterns that emerge from the data. How could you describe your results? What are some limitations of the data? What other questions are raised and how might they be resolved? Record your ideas for a 60-second summary. Exploration-Linked Homework Prepare a verbal 60-second summary to give to the class. If possible, use an overhead projector with a transparency of your histogram or a projector linked to your graphing calculator. If not, bring in a paper copy of your histogram. Construct a written 60-second summary. (See Section 1.2 for some writing suggestions.)
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ALGEBRA CLASS QUESTIONNAIRE (You may leave any category blank) 1.
Age (in years)
2.
Sex (female 5 1, male 5 2)
3.
Your height (in inches)
4.
Distance from your navel to the floor (in centimeters)
5.
Estimate your average travel time to school (in minutes)
6.
What is your most frequent mode of transportation to school? (F 5 by foot, C 5 car, P 5 public transportation, B 5 bike, O 5 other)
The following variables will be defined in class. We will discuss ways of coding possible responses and then use the results to record our personal data. 7.
The number of people in your household
8.
Your employment status
9.
Your ethnic classification
10.
Your attitude toward mathematics
Health Data 11. Your pulse rate before jumping (beats per minute) 12.
Your pulse rate after jumping for 1 minute (beats per minute)
13.
Blood pressure: systolic (mm Hg)
14.
Blood pressure: diastolic (mm Hg)
15.
Flexibility (in inches)
16.
Vision, left eye
17.
Vision, right eye
Other Data
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E X P L O R AT I O N 1 . 2 Picturing Functions
Objective • develop an intuitive understanding of functions. Material/Equipment None required Procedure Part I Class Discussion
Water level
Bridget, the 6-year-old daughter of a professor at the University of Pittsburgh, loves playing with her rubber duckie in the bath at night. Her mother drew the accompanying graph for her math class. It shows the water level (measured directly over the drain) in Bridget’s tub as a function of time.
0
t1
t2
t3
t4
t5
t6
t7
t8
Time
Pick out the time period during which: • • • • •
The tub is being filled Bridget is entering the tub She is playing with her rubber duckie She leaves the tub The tub is being drained
With a Partner Create your own graph of a function that tells a story. Be as inventive as possible. Some students have drawn functions that showed the decibel levels during a phone conversation of a boyfriend and girlfriend, number of hours spent doing homework during one week, and amount of money in one’s pocket during the week.
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Class Discussion Draw your graph on the blackboard and tell its story to the class. Part II With a Partner Generate a plausible graph for each of the following: 1. Time spent driving to work as a function of the amount of snow on the road. (Note: The first inch or so may not make any difference; the domain may be only up to about a foot of snow since after that you may not be able to get to work.) 2. The hours of nighttime as a function of the time of year. 3. The temperature of ice cream taken out of the freezer and left to stand. 4. The distance that a cannonball (or javelin or baseball) travels as a function of the angle of elevation at which is it is launched. (The maximum distance is attained for angles of around 458 from the horizontal.) 5. Assume that you leave your home walking at a normal pace, realize you have forgotten your homework and run back home, and then run even faster to school. You sit for a while in a classroom and then walk leisurely home. Now plot your distance from home as a function of time. Bonus Question Assume that water is pouring at a constant rate into each of the containers shown. The height of water in the container is a function of the volume of liquid. Sketch a graph of this function for each container.
(a)
(b)
(c)
Discussion/Analysis Are your graphs similar to those generated by the rest of the class? Can you agree as a class as to the basic shape of each of the graphs? Are there instances in which the graphs could look quite different?
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E X P L O R AT I O N 1 . 3 Deducing Formulas to Describe Data
Objective • find and describe patterns in data. • deduce functional formulas from data tables. • extend patterns using functional formulas.
Material/Equipment None required Procedure Class Discussion 1. Examine data tables (a) and (b). In each table, look for a pattern in terms of how y changes when x changes. Explain in your own words how to find y in terms of x. a. x b. x y y 0 1 2 3 4
0.0 0.5 1.0 1.5 2.0
0 1 2 3 4
5 8 11 14 17
2. Assuming that the pattern continues indefinitely, use the rule you have found to extend the data table to include negative numbers for x. 3. Check your extended data tables. Did you find only one value for y given a particular value for x? 4. Use a formula to describe the pattern that you have found. Do you think this formula describes a function? Explain. On Your Own 1. For each of the data tables (c) to (h), explain in your own words how to find y in terms of x. Then extend each table using the rules you have found. c. x d. x e. x y y y 0 1 2 3 4
0 1 4 9 16
0 1 2 3 4
0 1 8 27 64
0 1 2 3 4
0 2 12 36 80
[Hint: For table (e) think about some combination of data in tables (c) and (d).]
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f.
x
y
22 0 5 10 100
0 10 35 60 510
g.
x
y
0 1 2 3 4
21 0 3 8 15
h.
x
y
0 10 20 30 100
3 8 13 18 53
2. For each table, construct a formula to describe the pattern you have found. Discussion/Analysis With a Partner Compare your results. Do the formulas that you have found describe functions? Explain. Class Discussion Does the rest of the class agree with your results? Remember that formulas that look different may give the same results. Exploration-Linked Homework 1. a. For data tables (i) and (j), explain in your own words how to find y in terms of x. Using the rules you have found, extend the data tables to include negative numbers. i. x j. x y y 210 0 3 8 10
10.0 0.0 0.9 6.4 10.0
0 1 2 3 4
23 1 5 9 13
b. For each table, find a formula to describe the pattern you have found. Does your formula describe a function? Explain. 2. Make up a functional formula, generate a data table, and bring the data table on a separate piece of paper to class. The class will be asked to find your rule and express it as a formula. k.
x
y
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CHAPTER 2
RATES OF CHANGE AND LINEAR FUNCTIONS OVERVIEW How does the U.S. population change over time? How do children’s heights change as they age? Average rates of change provide a tool for measuring how change in one variable affects a second variable. When average rates of change are constant, the relationship is linear. After reading this chapter you should be able to • calculate and interpret average rates of change • understand how representations of data can be biased • recognize that a constant rate of change denotes a linear relationship • construct a linear equation given a table, graph, or description • derive by hand a linear model for a set of data
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2.1 Average Rates of Change In Chapter 1 we looked at how change in one variable could affect change in a second variable. In this section we’ll examine how to measure that change.
Describing Change in the U.S. Population over Time We can think of the U.S. population as a function of time. Table 2.1 and Figure 2.1 are two representations of that function. They show the changes in the size of the U.S. population since 1790, the year the U.S. government conducted its first decennial census. Time, as usual, is the independent variable and population size is the dependent variable.
DATA
USPOP
Change in population Figure 2.1 clearly shows that the size of the U.S. population has been growing over the last two centuries, and growing at what looks like an increasingly rapid rate. How can the change in population over time be described quantitatively? One way is to pick two points on the graph of the data and calculate how much the population has changed during the time period between them. Suppose we look at the change in the population between 1900 and 1990. In 1900 the population was 76.2 million; by 1990 the population had grown to 248.7 million. How much did the population increase? change in population 5 (248.7 2 76.2) million people 5 172.5 million people This difference is portrayed graphically in Figure 2.2, at the top of the next page.
Year
Population in Millions
1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
3.9 5.3 7.2 9.6 12.9 17.1 23.2 31.4 39.8 50.2 63.0 76.2 92.2 106.0 123.2 132.2 151.3 179.3 203.3 226.5 248.7 281.4
Table 2.1
change in years 5 (1990 2 1900) years 5 90 years 300
250
Population in millions
Population of the United States: 1790–2000
Change in time Knowing that the population increased by 172.5 million tells us nothing about how rapid the change was; this change clearly represents much more dramatic growth if it happened over 20 years than if it happened over 200 years. In this case, the length of time over which the change in population occurred is
200
150
100
50
0 1800
1840
1880
1920
1960
Year
Figure 2.1 Population of the United States. Source: U.S. Bureau of the Census, Statistical Abstract of the United States: 2002.
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2.1 Average Rates of Change
67
300 Population in millions
300 Population in millions
250 200 172.5 million people
150 100 50
250 200 150 100 90 years
50 0
0 1800
1840
1880 1920 Year
1960
1800
2000
1840
1880 1920 Year
1960
2000
Figure 2.3 Time change: 90 years.
Figure 2.2 Population change: 172.5 million people.
This interval is indicated in Figure 2.3 above. Average rate of change To find the average rate of change in population per year from 1900 to 1990, divide the change in the population by the change in years: average rate of change 5 5
change in population change in years 172.5 million people 90 years
< 1.92 million people/year In the phrase “million people/year” the slash represents division and is read as “per.” So our calculation shows that “on average,” the population grew at a rate of 1.92 million people per year from 1900 to 1990. Figure 2.4 depicts the relationship between time and population increase.
300 Population in millions
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250 Average rate of change 1.92 million per year
200
172.5 million people
150 100 90 years
50 0 1800
1840
1880 1920 Year
1960
2000
Figure 2.4 Average rate of change: 1900–1990.
Defining the Average Rate of Change The notion of average rate of change can be used to describe the change in any variable with respect to another. If you have a graph that represents a plot of data points of the form (x, y), then the average rate of change between any two points is the change in the y value divided by the change in the x value.
The average rate of change of y with respect to x 5
change in y change in x
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If the variables represent real-world quantities that have units of measure (e.g., millions of people or years), then the average rate of change should be represented in terms of the appropriate units: units of the average rate of change 5
units of y units of x
For example, the units might be dollars/year (read as “dollars per year”) or pounds/ person (read as “pounds per person”).
EXAMPLE
1
SOLUTION
Between 1850 and 1950 the median age in the United States rose from 18.9 to 30.2, but by 1970 it had dropped to 28.0. a. Calculate the average rate of change in the median age between 1850 and 1950. b. Compare your answer in part (a) to the average rate of change between 1950 and 1970. a. Between 1850 and 1950, average rate of change 5
change in median age change in years
11.3 years s30.2 2 18.9d years 5 s1950 2 1850d years 100 years 5 0.113 years/year 5
The units are a little confusing. But the results mean that between 1850 and 1950 the median age increased an average of 0.113 years each calendar year. b. Between 1950 and 1970, average rate of change 5 5
change in median age change in years s28.0 2 30.2d years 22.2 years 5 s1970 2 1950d years 20 years
5 20.110 years/year Note that since the median age dropped in value between 1950 and 1970, the average rate is negative. The median age decreased by 0.110 years/year between 1950 and 1970, whereas the median age increased by 0.113 years/year beween 1850 and 1950.
Limitations of the Average Rate of Change The average rate of change is an average. Average rates of change have the same limitations as any average. Although the average rate of change of the U.S. population from 1900 to 1990 was 1.92 million people/year, it is highly unlikely that in each year the population grew by exactly 1.92 million. The number 1.92 million people/year is, as the name states, an average. Similarly, if the arithmetic average, or mean, height of students in your class is 67 inches, you wouldn’t expect every student to be 67 inches tall. In fact, it may be the case that not even one student is 67 inches tall. The average rate of change depends on the end points. If the data points do not all lie on a straight line, the average rate of change varies for different intervals. For instance, the average rate of change in population for the time interval 1840 to 1940 is 1.15 million people/year and from 1880 to 1980 is 1.76 million people/year. (See Table 2.2. Note: Here we abbreviate “million people” as “million.”) You can see on the graphs that the line segment is much steeper from 1880 to 1980 than from 1840 to 1940 (Figures 2.5 and 2.6). Different intervals give different impressions of the rate of change in the U.S. population, so it is important to state which end points are used.
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2.1 Average Rates of Change
Time Interval
Change in Time
Change in Population
Average Rate of Change
69
1840–1940
100 yr
132.2 2 17.1 5 115.1 million
115.1 million < 1.15 million/yr 100 yr
1880–1980
100 yr
226.5 2 50.2 5 176.3 million
176.3 million < 1.76 million/yr 100 yr
Table 2.2 300 Population in millions
Population in millions
300 250 200 150 100
Average rate of change 1.15 million per year
50 0
1800
1840 1880 1920 Year
1960 2000
250 200 Average rate of change 1.76 million per year
150 100 50 0
1800
1840 1880 1920 Year
1960 2000
Figure 2.5 Average rate of change:
Figure 2.6 Average rate of change:
1840–1940.
1880–1980.
The average rate of change does not reflect all the fluctuations in population size that may occur between the end points. For more specific information, the average rate of change can be calculated for smaller intervals.
Algebra Aerobics 2.1 1. Suppose your weight five years ago was 135 pounds and your weight today is 143 pounds. Find the average rate of change in your weight with respect to time. 2. Table 2.3 shows data on U.S. international trade as reported by the U.S. Bureau of the Census.
Year
U.S. Exports U.S. Imports (billions of $) (billions of $)
1990 2006
537.2 820.2
618.4 1273.2
U.S. Trade Balance 5 Exports 2 Imports (billions of $) 281.2 2453.0
Table 2.3
a. What is the average rate of change between 1990 and 2006 for: i. Exports? ii. Imports? iii. The trade balance, the difference between what we sell abroad (exports) and buy from abroad (imports)? b. What do these numbers tell us? 3. Table 2.4 indicates the number of deaths in motor vehicle accidents in the United States as listed by the U.S. Bureau of the Census.
Annual Deaths in Motor Vehicle Accidents (thousands) 1980 52.1
1990 44.6
2000 41.8
2004 42.6
Table 2.4
4.
5.
6.
7.
Find the average rate of change: a. From 1980 to 2000 b. From 2000 to 2004 Be sure to include units. A car is advertised to go from 0 to 60 mph in 5 seconds. Find the average rate of change (i.e., the average acceleration) over that time. According to the National Association of Insurance Commissioners, the average cost for automobile insurance has gone from $689 in 2000 to $867 in 2006. What is the average rate of change? A football player runs for 1056 yards in 2002 and for 978 yards in 2006. Find the average rate of change in his performance. The African elephant is an endangered species, largely because poachers (people who illegally hunt elephants) kill elephants to sell the ivory from their tusks. In the African country of Kenya in the last 10 years, the elephant population has dropped from 150,000 to 30,000. Calculate the average rate of change and describe what it means.
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Exercises for Section 2.1 1. If r is measured in inches, s in pounds, and t in minutes, identify the units for the following average rates of change: change in r change in t change in s a. b. c. change in s change in r change in r 2. Assume that R is measured in dollars, S in ounces, T in dollars per ounce, and V in ounces per dollar. Write a product of two of these terms whose resulting units will be: a. Dollars b. Ounces 3. Your car’s gas tank is full and you take a trip. You travel 212 miles, then you fill your gas tank up again and it takes 10.8 gallons. If you designate your change in distance as 212 miles and your change in gallons as 10.8, what is the average rate of change of gasoline used, measured in miles per gallon? 4. The gas gauge on your car is broken, but you know that the car averages 22 miles per gallon. You fill your 15.5-gallon gas tank and tell your friend, “I can travel 300 miles before I need to fill up the tank again.” Explain why this is true. 5. The consumption of margarine (in pounds per person) decreased from 9.4 in 1960 to 7.5 in 2000. What was the annual average rate of change? (Source: www.census.gov) 6. The percentage of people who own homes in the United States has gone from 65.5% in 1980 to 69.0% in 2005. What is the average rate of change in percentage points per year? 7. The accompanying table shows females’ SAT scores in 2000 and 2005.
Year
Average Female Verbal SAT
Average Female Math SAT
2000 2005
504 505
498 504
Source: www.collegeboard.com.
Find the average rate of change: a. In the math scores from 2000 to 2005 b. In the verbal scores from 2000 to 2005 8. a. In 1992 the aerospace industry showed a net loss (negative profit) of $1.84 billion. In 2002 the industry had a net profit of $8.97 billion. Find the average annual rate of change in net profits from 1992 to 2002. b. In 2005, aerospace industry net profits were $2.20 billion. Find the average rate of change in net profits: i. From 1992 to 2005 ii. From 2002 to 2005
9. According to the U.S. Bureau of the Census, in elementary and secondary schools, in the academic year ending in 1985 there were about 630,000 computers being used for student instruction, or about 84.1 students per computer. In the academic year ending in 2005, there were about 13,600,000 computers being used, or about 4.0 students per computer. Find the average rate of change from 1985 to 2005 in: a. The number of computers being used b. The number of students per computer 10. According to the U.S. Bureau of the Census, the percentage of persons 25 years old and over completing 4 or more years of college was 4.6 in 1940 and 27.6 in 2005. a. Plot the data, labeling both axes and the coordinates of the points. b. Calculate the average rate of change in percentage points per year. c. Write a topic sentence summarizing what you think is the central idea to be drawn from these data. 11. Though reliable data about the number of African elephants are hard to come by, it is estimated that there were about 4,000,000 in 1930 and only 500,000 in 2000. a. What is the average annual rate of change in elephants over time? Interpret your result. b. During the 1980s it was estimated that 100,000 elephants were being killed each year due to systematic poaching for ivory. How does this compare with your answer in part (a)? What does this tell you about what was happening before or after the 1980s? (Source: www.panda.org) 12. a. According to the U.S. Bureau of the Census, between 1980 and 2004 domestic new car sales declined from 6581 thousand cars to 5357 thousand. (Note: This does not include trucks, vans, or SUVs.) Calculate the annual average rate of change. b. During the same period Japanese car sales in the United States dropped from 1906 thousand to 798 thousand. Calculate the average rate of change. c. What do the two rates suggest about car sales in the United States? 13. Use the information in the accompanying table to answer the following questions. Percentage of Persons 25 Years Old and Over Who Have Completed 4 Years of High School or More
All White Black Asian/Pacific Islander
1940
2004
24.5 26.1 7.3 22.6
85.2 85.8 80.6 85.0
Source: U.S. Bureau of the Census, Statistical Abstract of the United States: 2006.
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a. What was the average rate of change (in percentage points per year) of completion of 4 years of high school from 1940 to 2004 for whites? For blacks? For Asian/Pacific Islanders? For all? b. If these rates continue, what percentages of whites, of blacks, of Asian/Pacific Islanders, and of all will have finished 4 years of high school in the year 2007? Check the Internet to see if your predictions are accurate.
15. Use the accompanying table on life expectancy to answer the following questions. Average Number of Years of Life Expectancy in the United States by Race and Sex Since 1900 Life Expectancy at Birth by Year 1900 1950 2000 2005
c. Write a 60-second summary describing the key elements in the high school completion data. Include rates of change and possible projections for the near future. d. If these rates continue, in what year will 100% of whites have completed 4 years of high school or more? In what year 100% of blacks? In what year 100% of Asian/Pacific Islanders? Do these projections make sense? 14. The accompanying data show U.S. consumption and exports of cigarettes.
Year 1960 1980 2000 2005
U.S. Consumption (billions) 484 631 430 378
Exports (billions) 20 82 148 113
Source: U.S. Department of Agriculture.
a. Calculate the average rates of change in U.S. cigarette consumption from 1960 to 1980, from 1980 to 2005, and from 1960 to 2005. b. Compute the average rate of change for cigarette exports from 1960 to 2005. Does this give an accurate image of cigarette exports? c. The total number of cigarettes consumed in the United States in 1960 was 484 billion, very close to the number consumed in 1995, 487 billion. Does that mean smoking was as popular in 1995 as it was in 1960? Explain your answer. d. Write a paragraph summarizing what the data tell you about the consumption and exports of cigarettes since 1960, including average rates of change.
71
White Males
White Females
Black Males
Black Females
46.6 66.5 74.8 75.4
48.7 72.2 80.0 81.1
32.5 58.9 68.2 69.9
33.5 62.7 74.9 76.8
Source: U.S. National Center for Health Statistics, Statistical Abstract of the United States, 2007.
a. What group had the highest life expectancy in 1900? In 2005? What group had the lowest life expectancy in 1900? In 2005? b. Which group had the largest average rate of change in life expectancy between 1900 and 2005? c. Write a short summary of the patterns in U.S. life expectancy from 1900 to 2005 using average rates of change to support your points. 16. The accompanying table gives the number of unmarried males and females over age 15 in the United States. Marital Status of Population 15 Years Old and Older Year
Number of unmarried males (in thousands)
Number of unmarried females (in thousands)
1950 1960 1970 1980 1990 2000 2004
17,735 18,492 23,450 30,134 36,121 43,429 41,214
19,525 22,024 29,618 36,950 43,040 50,133 47,616
Source: U.S. Bureau of the Census, www.census.gov.
a. Calculate the average rate of change in the number of unmarried males between 1950 and 2004. Interpret your results. b. Calculate the average rate of change in the number of unmarried females between 1950 and 2004. Interpret your results. c. Compare the two results. d. What does this tell you, if anything, about the percentages of unmarried males and females?
2.2 Change in the Average Rate of Change We can obtain an even better sense of patterns in the U.S. population if we look at how the average rate of change varies over time. One way to do this is to pick a fixed interval period for time and then calculate the average rate of change for each successive time period. Since we have the U.S. population data in 10-year intervals, we can calculate the average rate of change for each successive decade. The third column in Table 2.5 shows the results of these calculations. Each entry represents the average population growth per year (the average annual rate of change) during the previous decade. A few of these calculations are worked out in the last column of the table.
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SOMETHING TO THINK ABOUT
What might be some reasons for the slowdown in population growth from the 1960s through the 1980s?
Average Annual Rates of Change of U.S. Population: 1790–2000
Year
Population (millions)
Average Annual Rate for Prior Decade (millions/yr)
1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
3.9 5.3 7.2 9.6 12.9 17.1 23.2 31.4 39.8 50.2 63.0 76.2 92.2 106.0 123.2 132.2 151.3 179.3 203.3 226.5 248.7 281.4
Data not available 0.14 0.19 0.24 0.33 0.42 0.61 0.82 0.84 1.04 1.28 1.32 1.60 1.38 1.72 0.90 1.91 2.80 2.40 2.32 2.22 3.27
Sample Calculations 0.14 5 (5.3 2 3.9)/(1800 2 1790)
0.42 5 (17.1 2 12.9)/(1840 2 1830)
0.90 5 (132.2 2 123.2)/(1940 2 1930)
Table 2.5 Source: U.S. Bureau of the Census, Statistical Abstract of the United States: 2002.
What is happening to the average rate of change over time? Start at the top of the third column and scan down the numbers. Notice that until 1910 the average rate of change increases every year. Not only is the population growing every decade until 1910, but it is growing at an increasing rate. It’s like a car that is not only moving forward but also accelerating. A feature that was not so obvious in the original data is now evident: In the intervals 1910 to 1920, 1930 to 1940, and 1960 to 1990 we see an increasing population but a decreasing rate of growth. It’s like a car decelerating—it is still moving forward but it is slowing down. The graph in Figure 2.7, with years on the horizontal axis and average rates of change on the vertical axis, shows more clearly how the average rate of change 3.50 Rate of change (in millions per year) over prior decade
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3.00 2.50 2.00 1.50 1.00 0.50 0.00 1800 1820 1840 1860 1880 1900 1920 1940 1960 1980 2000 Year
Figure 2.7 Average rates of change in the U.S. population by decade.
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73
fluctuates over time. The first point, corresponding to the year 1800, shows an average rate of change of 0.14 million people/year for the decade 1790 to 1800. The rate 1.72, corresponding to the year 1930, means that from 1920 to 1930 the population was increasing at a rate of 1.72 million people/year. What does this tell us about the U.S. population? The pattern of growth was fairly steady up until about 1910. Why did it change? A possible explanation for the slowdown in the decade prior to 1920 might be World War I and the 1918 flu epidemic, which by 1920 had killed nearly 20,000,000 people, including about 500,000 Americans. In Figure 2.7, the steepest decline in the average rate of change is between 1930 and 1940. One obvious suspect for the big slowdown in population growth in the 1930s is the Great Depression. Look back at Figure 2.1, the original graph that shows the overall growth in the U.S. population. The decrease in the average rate of change in the 1930s is large enough to show up in our original graph as a visible slowdown in population growth. The average rate of change increases again between 1940 and 1960, then drops off from the 1960s through the 1980s. The rate increases once more in the 1990s. This latest surge in the growth rate is attributed partially to the “baby boom echo” (the result of baby boomers having children) and to a rise in birth rates and immigration.
Algebra Aerobics 2.2 1. Table 2.6 and Figure 2.8 show estimates for world population between 1800 and 2050. a. Fill in the third column of the table by calculating the average annual rate of change. b. Graph the annual average rate of change versus time. World Population
Year
Total Population (millions)
Annual Average Rate of Change (over prior 50 years)
1800 1850 1900 1950 2000 2050
980 1260 1650 2520 6090 9076 (est.)
n.a.
Table 2.6 Source: Population Division of the United Nations, www.un.org/popin.
High School Completers
10000 Population (in millions)
c. During what 50-year period was the average annual rate of change the largest? d. Describe in general terms what happened (and is predicted to happen) to the world population and its average rate of change between 1800 and 2050. 2. A graph illustrating a corporation’s profits indicates a positive average rate of change between 2003 and 2004, another positive rate of change between 2004 and 2005, a zero rate of change between 2005 and 2006, and a negative rate of change between 2006 and 2007. Describe the graph and the company’s financial situation over the years 2003–2007. 3. Table 2.7 shows educational data collected on 18- to 24-year-olds between 1960 and 2004 by the National Center for Educational Statistics. The table shows the number of students who graduated from high school or completed a GED (a high school equivalency exam) during the indicated year.
8000 6000 4000 2000
1800
1850
1900
1950 Year
Figure 2.8 World population.
2000
2050
Year
Number (thousands)
1960 1970 1980 1990 2000 2004
1679 2757 3089 2355 2756 2752
Table 2.7
Average Rate of Change (thousands per year) n.a.
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a. Fill in the blank cells with the appropriate average rates of change for high school completers. b. Describe the pattern in the number of high school completers between 1960 and 2004. c. What does it mean here when the average rate of change is positive? Give a specific example from your data.
d. What does it mean here when the average rate of change is negative? Give another specific example. e. What does it mean when two adjacent average rates of change are positive, but the second one is smaller than the first?
Exercises for Section 2.2 Technology is optional for Exercises 4 and 11. 1. Calculate the average rate of change between adjacent points for the following function. (The first few are done for you.)
x
ƒ(x)
0 1 2 3 4 5
0 1 8 27 64 125
Average Rate of Change n.a. 1 7
a. Is the function ƒ(x) increasing, decreasing, or constant throughout? b. Is the average rate of change increasing, decreasing, or constant throughout? 2. Calculate the average rate of change between adjacent points for the following function. The first one is done for you.
x
ƒ(x)
0 1 2 3 4 5
0 1 16 81 256 625
3. The accompanying table shows the number of registered motor vehicles in the United States.
Year
Registered Motor Vehicles (millions)
1960 1970 1980 1990 2000
74 108 156 189 218
a. Is the function ƒ(x) increasing, decreasing, or constant throughout? b. Is the average rate of change increasing, decreasing, or constant throughout?
n.a.
a. Fill in the third column in the table. b. During which decade was the average rate of change the smallest? c. During which decade was the average rate of change the largest? d. Write a paragraph describing the change in registered motor vehicles between 1960 and 2000. 4. (Graphing program optional.) The accompanying table indicates the number of juvenile arrests (in thousands) in the United States for aggravated assault.
Average Rate of Change n.a. 1
Annual Average Rate of Change (over prior decade)
Year
Juvenile Arrests (thousands)
Annual Average Rate of Change over Prior 5 Years
1985 1990 1995 2000 2005
36.8 54.5 68.5 49.8 36.9
n.a.
a. Fill in the third column in the table by calculating the annual average rate of change. b. Graph the annual average rate of change versus time.
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c. During what 5-year period was the annual average rate of change the largest? d. Describe the change in aggravated assault cases during these years by referring both to the number and to the annual average rate of change. 5. Calculate the average rate of change between adjacent points for the following functions and place the values in a third column in each table. (The first entry is “n.a.”)
7. Each of the following functions has a graph that is increasing. If you calculated the average rate of change between sequential equal-size intervals, which function can be said to have an average rate of change that is: a. Constant? b. Increasing? c. Decreasing? 40
y
15
x
0
x
ƒ(x)
x
g(x)
0 10 20 30 40 50
5 25 45 65 85 105
0 10 20 30 40 50
270 240 210 180 150 120
a. Are the functions ƒ(x) and g(x) increasing, decreasing, or constant throughout? b. Is the average rate of change of each function increasing, decreasing, or constant throughout? 6. Calculate the average rate of change between adjacent points for each of the functions in Tables A–D and place the values in a third column in each table. (The first entry is “n.a.”) Then for each function decide which statement best describes it.
x
ƒ(x)
x
h(x)
0 1 2 3 4 5
1 3 9 27 81 243
0 10 20 30 40 50
50 55 60 65 70 75
Table B
a. b. c. d.
Table D
x
g(x)
x
k(x)
0 15 30 45 60 75
200 155 110 65 20 25
0 1 2 3 4 5
40 31 24 19 16 15
As x increases, the function increases at a constant rate. As x increases, the function increases at an increasing rate. As x increases, the function decreases at a constant rate. As x increases, the function decreases at an increasing rate.
6
Graph A
y
0
6
6
Graph B
x
y
0
Graph C
6
x
8. Match the data table with its graph. Table A
15
Table C
Table B
x
y
x
y
x
y
0 1 2 3 4 5 6
2 5 8 11 14 17 20
0 1 2 3 4 5 6
0 0.5 2 4.5 8 12.5 18
0 1 4 9 16 25 36
0 2 4 6 8 10 12
y
0
Table C
Table A
75
20
x
Graph D
y
y 20
x
0
40
8
x
0
Graph E
6
Graph F
9. Refer to the first two data tables (A and B) in Exercise 8. Insert a third column in each table and label the column “average rate of change.” a. Calculate the average rate of change over adjacent data points. b. Identify whether the table represents an average rate of change that is constant, increasing, or decreasing. c. Explain how you could tell this by looking at the corresponding graph in Exercise 8. 10. Following are data on the U.S. population over the time period 1830–1930 (extracted from Table 2.1). U.S. Population Year 1830 1850 1870 1890 1910 1930
Population (in millions)
Average Rate of Change (millions/yr)
12.9 23.2
n.a.
63.0 92.2
0.83 1.16 1.55
Source: U.S. Bureau of the Census, www.census.gov.
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a. Fill in the missing parts of the chart. b. Which 20-year interval experienced the largest average rate of change in population? c. Which 20-year interval experienced the smallest average rate of change in population?
Number of U.S. Newspapers
11. (Technology recommended.) The accompanying data give a picture of the two major methods of D A T A news communication in the United States. (See also Excel or graph link files NEWPRINT and ONAIRTV.) a. Use the U.S. population numbers from Table 2.1 (at the beginning of this chapter) to calculate and compare the number of copies of newspapers per person in 1920 and in 2000. b. Create a table that displays the annual average rate of change in TV stations for each decade since 1950. Create a similar table that displays the annual average rate of change in newspapers published for the same period. Graph the results. c. If new TV stations continue to come into existence at the same rate as from 1990 to 2000, how many will there be by the year 2010? Do you think this is likely to be a reasonable projection, or is it overly large or small judging from past rates of growth? Explain. d. What trends do you see in the dissemination of news as reflected in these data?
Year
Newspapers (thousands of copies printed)
Number of Newspapers Published
1920 1930 1940 1950 1960 1970 1980 1990 2000
27,791 39,589 41,132 53,829 58,882 62,108 62,202 62,324 55,800
2042 1942 1878 1772 1763 1748 1745 1611 1480
Source: U.S. Bureau of the Census, www.census.gov.
Number of U.S. Commercial TV Stations Year
Number of Commercial TV Stations
1950 1960 1970 1980 1990 2000
98 515 677 734 1092 1248
Source: U.S. Bureau of the Census, www.census.gov.
2.3 The Average Rate of Change Is a Slope Calculating Slopes
The reading “Slopes” describes many of the practical applications of slopes, from cowboy boots to handicap ramps.
On a graph, the average rate of change is the slope of the line connecting two points. The slope is an indicator of the steepness of the line. If sx1, y1 d and sx2, y2 d are two points, then the change in y equals y2 2 y1 (see Figure 2.9). This difference is often denoted by y, read as “delta y,” where is the Greek letter capital D (think of D as representing difference): y 5 y2 2 y1. Similarly, the change in x (delta x) can be represented by x 5 x2 2 x1. Then slope 5
y2 2 y1 y change in y 5 5 x2 2 x1 x change in x
(x2,y2)
y = y2 – y1
(x1,y1)
x = x2 – x1
(x2,y1)
Figure 2.9 Slope 5 y/x.
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77
The average rate of change represents a slope. Given two points sx1, y1 d and sx2, y2 d, average rate of change 5
change in y y y 2 y1 5 slope 5 5 2 change in x x x2 2 x1
When calculating a slope, it doesn’t matter which point is first Given two points, sx1, y1 d and sx2, y2 d, it doesn’t matter which one we use as the first point when we calculate the slope. In other words, we can calculate the slope between sx1, y1 d and sx2, y2 d as y2 2 y1 y 2 y2 or as 1 x2 2 x1 x1 2 x2 The two calculations result in the same value. We can show that the two forms are equivalent. slope 5
Given
5
21 y2 2 y1 ? 21 x2 2 x1
simplify
5
2y2 1 y1 2x2 1 x1
rearrange terms
5
y1 2 y2 x1 2 x2
multiply by
21 21
y2 2 y1 x2 2 x1
In calculating the slope, we do need to be consistent in the order in which the coordinates appear in the numerator and the denominator. If y1 is the first term in the numerator, then x1 must be the first term in the denominator.
EXAMPLE
1
SOLUTION 12
Plot the two points (22, 26) and (7, 12) and calculate the slope of the line passing through them. Treating (22, 26) as sx1, y1 d and (7, 12) as sx2, y2 d (Figure 2.10), then
y (7, 12)
slope 5
y2 2 y1 12 2 s26d 18 5 5 52 x2 2 x1 7 2 s22d 9
12 – (–6)
x –8
8
(–2, –6)
7 – (–2)
We could also have used 26 and 22 as the first terms in the numerator and denominator, respectively:
slope 5
26 2 12 218 y1 2 y2 5 5 52 x1 2 x2 22 2 7 29
–12
Figure 2.10
Either way we obtain the same answer.
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EXAMPLE
2
The percentage of the U.S. population living in rural areas decreased from 84.7% in 1850 to 21.0% in 2000. Plot the data, then calculate and interpret the average rate of change in the rural population over time.
SOLUTION
If we treat year as the independent and percentage as the dependent variable, our given data can be represented by the points (1850, 84.7) and (2000, 21.0). See Figure 2.11. 100 (1850, 84.7) Percentage
80 60 40
(2000, 21.0)
20
1800
1850
1900
1950
2000
2050
Year
Figure 2.11 Percentage of the U.S. population
living in rural areas. Source: U.S. Bureau of the Census, www.census.gov.
The average rate of change 5
change in the percentage of rural population change in time
5
s21.0 2 84.7d% s2000 2 1850d years
5
263.7% 150 years
< 20.42% per year
?
SOMETHING TO THINK ABOUT
What kind of social and economic implications does a population shift of this magnitude have on society?
The sign of the average rate of change is negative since the percentage of people living in rural areas was decreasing. (The negative slope of the graph in Figure 2.11 confirms this.) The value tells us that, on average, the percentage living in rural areas decreased by 0.42 percentage points (or about one-half of 1%) each year between 1850 and 2000. The change per year may seem small, but in a century and a half the rural population went from being the overwhelming majority (84.7%) to about one-fifth (21%) of the population. If the slope between any two points (the average rate of change of y with respect to x) is positive, then the graph of the relationship rises when read from left to right. This means that as x increases in value, y also increases in value. If the slope is negative, the graph falls when read from left to right. As x increases, y decreases. If the slope is zero, the graph is flat. As x increases, there is no change in y.
EXAMPLE
3
Given Table 2.8, a listing of civil disturbances over time, plot and then connect the points, and (without doing any calculations) indicate on the graph when the average rate of change between adjacent data points is positive (1), negative (2), and zero (0).
SOLUTION
The data are plotted in Figure 2.12. Each line segment is labeled 1, 2, or 0, indicating whether the average rate of change between adjacent points is positive, negative, or zero. The largest positive average rate of change, or steepest upward slope, seems to be
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Oct.–Dec.
0 July–Sept.
Jan.–Mar.
3 8 5 5
1971
–
+ Apr.–June
Jan.–Mar. Apr.–June July–Sept. Oct.–Dec.
Oct.–Dec.
1972
1970
July–Sept.
12 21 5 1
Jan.–Mar.
Jan.–Mar. Apr.–June July–Sept. Oct.–Dec.
1969
– + Apr.–June
1971
1968
Oct.–Dec.
26 24 20 6
Apr.–June
Jan.–Mar. Apr.–June July–Sept. Oct.–Dec.
– +
July–Sept.
1970
+
Jan.–Mar.
5 27 19 6
+ –
– Oct.–Dec.
Jan.–Mar. Apr.–June July–Sept. Oct.–Dec.
–
+
+
–
Apr.–June
1969
–
–
July–Sept.
6 46 25 3
Jan.–Mar.
Jan.–Mar. Apr.–June July–Sept. Oct.–Dec.
Oct.–Dec.
1968
–
Apr.–June
Number of Disturbances
July–Sept.
Period
50 45 40 35 + 30 25 20 15 10 5 0 Jan.–Mar.
Year
Number of disturbances
Civil Disturbances in U.S. Cities
79
1972
Figure 2.12 Civil disturbances: 1968–1972. Source: D. S. Moore and G. P. McCabe, Introduction to the Practice of Statistics. Copyright © 1989 by W.H. Freeman and Company. Used with permission.
between the January-to-March and April-to-June counts in 1968. The largest negative average rate of change, or steepest downward slope, appears later in the same year (1968) between the July-to-September and October-to-December counts. Civil disturbances between 1968 and 1972 occurred in cycles: The largest numbers generally occurred in the spring months and the smallest in the winter months. The peaks decrease over time. What was happening in America that might correlate with the peaks? This was a tumultuous period in our history. Many previously silent factions of society were finding their voices. Recall that in April 1968 Martin Luther King was assassinated and in January 1973 the last American troops were withdrawn from Vietnam.
Table 2.8
Algebra Aerobics 2.3
Fatalities in the United States Due to Tornadoes 140 120 100 80 60 40 20
2,000
0 1998 1,500
2000
2002 Year
2004
2006
Figure 2.14 Deaths due to tornadoes: 1998–2005.
1,000
Year
Figure 2.13 Violent crimes in the United States.
2003
2002
2001
2000
1999
1998
1997
1996
1995
1994
0
1993
500
1992
Violent crimes (in millions)
2,500
3. Specify the intervals on the graph in Figure 2.14 for which the average rate of change between adjacent data points appears positive, negative, or zero.
Number of deaths
1. a. Plot each pair of points and then calculate the slope of the line that passes through them. i. (4, 1) and (8, 11) ii. (23, 6) and (2, 6) iii. (0, 23) and (25, 21) b. Recalculate the slopes in part (a), reversing the order of the points. Check that your answers are the same. 2. Specify the intervals on the graph in Figure 2.13 for which the average rate of change between adjacent data points is approximately zero.
4. What is the missing y-coordinate that would produce a slope of 4, if a line were drawn through the points (3, 22) and (5, y)? 5. Find the slope of the line through the points (2, 9) and (2 1 h, 9 1 2h).
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6. Consider points P1 5 (0, 0), P2 5 (1, 1), P3 5 (2, 4), and P4 5 (3, 9). a. Verify that these four points lie on the graph of y 5 x2.
b. Find the slope of the line segments connecting P1 and P2, P2 and P3, and P3 and P4. c. What do these slopes suggest about the graph of the function within those intervals?
Exercises for Section 2.3 1. Find the slope of a straight line that goes through: a. (25, 26) and (2, 3) b. (25, 6) and (2, 23)
y 8
E –8
2. Find the slope of each line using the points where the graph intersects the x and y axes. 12
y
8
A
x 12
B C
D
F
–16
y
x
6. Plot each pair of points and calculate the slope of the line that passes through them. a. (3, 5) and (8, 15) d. s22, 6d and s2, 26d b. s21, 4d and (7, 0) e. s24, 23d and s2, 23d c. (5, 9) and s25, 9d
3. Find the slope of each line using the points where the line crosses the x- or y-axis.
7. The following problems represent calculations of the slopes of different lines. Solve for the variable in each equation. 150 2 75 182 2 150 a. c. 5m 54 20 2 10 28 2 x
x –6
–6
6
6 –4
–8 Graph A
2
Graph B
y
6
y
x –4
b.
4
x –1
1
2
3
70 2 y 5 0.5 028
d.
620 5 0.6 x 2 10
4
8. Find the slope m of the line through the points (0, b) and (x, y), then solve the equation for y. –6
–10
Graph A
Graph B
4. Examine the line segments A, B, and C. a. Which line segment has a slope that is positive? That is negative? That is zero? b. Calculate the exact slope for each line segment A, B, and C. 12
y
10
A
8
C
6 4
B
2 0
x 0
2
4
6
5. Given the graph at the top of the next column: a. Estimate the slope for each line segment A–F. b. Which line segment is the steepest? c. Which line segment has a slope of zero?
9. Find the value of t if m is the slope of the line that passes through the given points. a. (3, t) and s22, 1d , m 5 24 b. (5, 6) and (t, 9), m 5 23 10. a. Find the value of x so that the slope of the line through (x, 5) and (4, 2) is 13 . b. Find the value of y so that the slope of the line through s1, 23d and s24, yd is 22. c. Find the value of y so that the slope of the line through s22, 3d and (5, y) is 0. d. Find the value of x so that the slope of the line through (22, 2) and (x, 10) is 2. e. Find the value of y so that the slope of the line through 1 (2100, 10) and (0, y) is 210 . f. Find at least one set of values for x and y so that the slope of line through (5, 8) and (x, y) is 0. 11. Points that lie on the same straight line are said to be collinear. Determine if the following points are collinear. a. (2, 3), (4, 7), and (8, 15) b. s23, 1d , (2, 4), and (7, 8)
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12. a. Find the slope of the line through each of the following pairs of points. i. s21, 4d and s22, 4d ii. s7, 23d and s27, 23d iii. s22, 26d and s5, 26d b. Summarize your findings.
Unemployment Rate
Percentage of population unemployed
7
13. Graph a line through each pair of points and then calculate its slope. a. The origin and s6, 22d b. The origin and s24, 7d 14. Find some possible values of the y-coordinates for the points s23, y1 d and s6, y2 d such that the slope m 5 0. 15. Calculate the slope of the line passing through each of the following pairs of points. a. s0, "2d and s "2, 0d b. Q0, 232R and Q232, 0R c. (0, b) and (b, 0) d. What do these pairs of points and slopes all have in common?
19. Handicapped Vietnam veterans successfully lobbied for improvements in the architectural standards for wheelchair access to public buildings. a. The old standard was a 1-foot rise for every 10 horizontal feet. What would the slope be for a ramp built under this standard? b. The new standard is a 1-foot rise for every 12 horizontal feet. What would the slope of a ramp be under this standard? c. If the front door is 3 feet above the ground, how long would the handicapped ramp be using the old standard? Using the new? 20. a. The accompanying graph gives information about the unemployment rate. Specify the intervals on the graph for which the slope of the line segment between adjacent data points appears positive. For which does it appear negative? For which zero? b. What might have caused the increase in the unemployment rate just after 2000?
5 4 3 2 1 1998
2000
2002
2004
2006
Year
Source: U.S. Department of Labor, Department of Labor Statistics, www.bls.gov
21. Read the Anthology article “Slopes” on the course website and then describe two practical applications of slopes, one of which is from your own experience. 22. The following graph shows the relationship between education and health in the United States. U.S. Average 90
Perc en
tage w ho did not r teste have d for a flu H.I.V . shot in la O ve r st ye weig ar ht o r ob ese Ha ve at le a st on ea Have suffe lco red from pa in in last ye ho ar lic dri Sm nk oke pe rm on th Have a heart condition Neve r had a ma mmo In po gram o r or (wom fair h en 4 ealth 0+)
Neve
80 70 Percentage of group
18. A study on numerous streams examined the effects of a warming climate. It found an increase in water temperature of about 0.78C for every 18C increase in air temperature. a. Find the rate of change in water temperature with respect to air temperature. What are the units? b. If the air temperature increased by 58C, by how much would you expect the stream temperature to increase?
6
0 1996
16. Which pairs of points produce a line with a negative slope? a. s25, 25d and s23, 23d d. (4, 3) and (12, 0) b. s22, 6d and s21, 4d e. (0, 3) and s4, 210d c. (3, 7) and s23, 27d f. (4, 2) and (6, 2) 17. In the previous exercise, which pairs of points produce a line with a positive slope?
81
60 50 40 30 20
Have an ulcer
10 Ha ve diabetes
0
2
4
6
8
10
12
14
16
Years spent in school
Source: “A Surprising Secret to a Long Life, According to Studies: Stay in School,” New York Times, January 3, 2007.
a. Estimate the percentages for those with 8 years of education and for those with 16 years of education (college graduates) who: i. Are overweight or obese ii. Have at least one alcoholic drink per month iii. Smoke b. For each category in part (a), calculate the average rate of change with respect to years of education. What do they all have in common? c. What does the graph suggest about the link between education and health? Could there be other factors at play?
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2.4 Putting a Slant on Data Whenever anyone summarizes a set of data, choices are being made. One choice may not be more “correct” than another. But these choices can convey, either accidentally or on purpose, very different impressions.
Slanting the Slope: Choosing Different End Points Within the same data set, one choice of end points may paint a rosy picture, while another choice may portray a more pessimistic outcome.
1
The data in Table 2.9 and the scatter plot in Figure 2.15 show the number of people below the poverty level in the United States from 1960 to 2005. How could we use the information to make the case that the poverty level has decreased? Has increased?
People in Poverty in the United States Year
Number of People in Poverty (in thousands)
1960 1970 1980 1990 2000 2005
39,851 25,420 29,272 33,585 31,581 36,950
Table 2.9 Source: U.S. Bureau of the Census, www. census.gov.
Number of people in poverty (thousands)
EXAMPLE
45,000 Slope = –64 thousand people/year
40,000
35,000
30,000 Slope = 329 thousand people/year 25,000
20,000 1960
1970
1980
1990
2000
2010
Year
Figure 2.15 Number of people in poverty in the United
States between 1960 and 2005. SOLUTION
Optimistic case: To make an upbeat case that poverty numbers have decreased, we could choose as end points (1960, 39851) and (2005, 36950). Then average rate of change 5
change in no. of people in poverty s000sd change in years
36,950 2 39,851 2005 2 1960 22901 5 45 < 264 thousand people/year 5
So between 1960 and 2005 the number of impoverished individuals decreased on average by 64 thousand (or 64,000) each year. We can see this reflected in Figure 2.15 in the negative slope of the line connecting (1960, 39851) and (2005, 36950). Pessimistic case: To make a depressing case that poverty numbers have risen, we could choose (1970, 25420) and (2005, 36950) as end points. Then average rate change of change 5 5
change in no. of people in poverty s000sd change in years 36,950 2 25,420 2005 2 1970
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83
11,530 35 < 329 thousand people/year 5
So between 1970 and 2005 the number of impoverished individuals increased on average by 329 thousand (or 329,000) per year. This is reflected in Figure 2.15 in the positive slope of the line connecting (1970, 25420) and (2005, 36950). Both average rates of change are correct, but they give very different impressions of the changing number of people living in poverty in America.
Slanting the Data with Words and Graphs See “C4: Distortion by Clipping and Scaling” in Rates of Change.
If we wrap data in suggestive vocabulary and shape graphs to support a particular viewpoint, we can influence the interpretation of information. In Washington, D.C., this would be referred to as “putting a spin on the data.” Take a close look at the following three examples. Each contains exactly the same underlying facts: the same average rate of change calculation and a graph with a plot of the same two data points (1990, 248.7) and (2000, 281.4), representing the U.S. population (in millions) in 1990 and in 2000. The U.S. population increased by only 3.27 million/year between 1990 and 2000 Stretching the scale of the horizontal axis relative to the vertical axis makes the slope of the line look almost flat and hence minimizes the impression of change (Figure 2.16).
Population (in millions)
300
290
Population (in millions)
280
250 200 150 100 50 0 1990
270
1995 Year
2000
Figure 2.16 “Modest” growth in the U.S. population. 260
250
240 1990
1995 Year
2000
Figure 2.17
“Explosive” growth in the U.S. population.
The U.S. population had an explosive growth of over 3.27 million/year between 1990 and 2000 Cropping the vertical axis (which now starts at 240 instead of 0) and stretching the scale of the vertical axis relative to the horizontal axis makes the slope of the line look steeper and strengthens the impression of dramatic change (Figure 2.17). The U.S. population grew at a reasonable rate of 3.27 million/year during the 1990s Visually, the steepness of the line in Figure 2.18 seems to lie roughly halfway between the previous two graphs. In fact, the slope of 3.27 million/year is precisely the same for all three graphs. How could you decide upon a “fair” interpretation of the data? You might try to put the data in context by asking, How does the growth between 1990 and 2000 compare with other decades in the history of the United States? How does it compare with growth in other countries at the same time? Was this rate of growth easily accommodated, or did it strain national resources and overload the infrastructure?
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Population (in millions)
300
275
250
225
200 1990
1995 Year
2000
Figure 2.18 “Reasonable” growth in the U.S.
population.
Exploration 2.1 gives you a chance to put your own “spin” on data.
A statistical claim is never completely free of bias. For every statistic that is quoted, others have been left out. This does not mean that you should discount all statistics. However, you will be best served by a thoughtful approach when interpreting the statistics to which you are exposed on a daily basis. By getting in the habit of asking questions and then coming to your own conclusions, you will develop good sense about the data you encounter.
Algebra Aerobics 2.4 1. Table 2.10 shows the percentage of the U.S. population in poverty between 1960 and 2005. In each case identify two end points you could use to make the case that poverty:
Year
% of Population in Poverty
1960 1970 1980 1990 2000 2005
22.2 12.6 13.0 13.5 11.3 12.6
Table 2.10 Source: U.S. Bureau of the Census, www.census.gov.
a. You are an antiwar journalist reporting on American casualties during a war. In week 1 there were 17, in week 2 there were 29, and in week 3 there were 26. b. You are the president’s press secretary in charge of reporting war casualties [listed in part (a)] to the public. 4. The graph in Figure 2.19 appeared as part of an advertisement in the Boston Globe on June 27, 2003. Identify at least three strategies used to persuade you to buy gold. Gold Explodes Experts Predict $1,500.00 an Ounce $380 $370 $360
Up $80 in 11 Months
$350 $340 $330 $320 $310
Figure 2.19 Prices for an ounce of gold.
Feb '03
Jan '03
Dec '02
Nov '02
Oct '02
Sept '02
Aug '02
July '02
June '02
May '02
$300 Apr '02
a. Has declined dramatically b. Has remained stable c. Has increased substantially 2. Assume you are the financial officer of a corporation whose stock earnings were $1.02 per share in 2005, and $1.12 per share in 2006, and $1.08 per share in 2007. How could you make a case for: a. Dramatic growth? b. Dramatic decline? 3. Sketch a graph and compose a few sentences to forcefully convey the views of the following persons.
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85
Exercises for Section 2.4 Graphing program optional for Exercise 10. Course software is required for Exercise 12. 1. Examine the three graphs A, B, and C. All report the same data on 30-year mortgage interest rates for the first six months in 2006. Create a different title for each graph exaggerating their differences.
3. Compare the accompanying graphs. a. Which line appears to have the steeper slope? b. Use the intercepts to calculate the slope. Which graph has the more negative (and hence steeper) slope? 24 18 12 6
Interest rate (%)
6.8 6.7 6.6
5 4 3 2 1
y
x 2
x
6.5
–2
6.4
–1 –6
1
–1
–1
2
Graph A
6.3 6.2 6.1
y
1
2
4
3
5
6
Month Graph A 6.7
Interest rate (%)
6.6
1
Graph B
4. The following graphs show the same function graphed on different scales. a. In which graph does Q appear to be growing at a faster rate? b. In which graph does Q appear to be growing at a near zero rate? c. Explain why the graphs give different impressions. 100
6.5
Q
10
Q
t
6.4
5
–5
t 50
–50
–100
6.3
–10
Graph A
Graph B
6.2
0
2
4
5. The following graph shows both federal spending on K–12 education under the Education and Secondary Education Act (ESEA) and National Assessment of Educational Progress (NAEP) reading scores (for age 9).
6
Month
8 7.5 7 6.5 6 5.5 5
0
1
2
3 Month
4
5
6
Graph C
Graph B
2004
Graph A
2002
4 –12
2000
–4
1998
–8
0
1996
2
5 1994
–2
235
10
1992
–2
Reading score
285
15
1990
x
x –4
20
1988
2 –6
ESEA funding (constant dollars)
185
a. Summarize the message that this chart conveys concerning NAEP reading scores and federal spending on ESEA. b. What strategy is used to make the reading scores seem lower?
y 24
4
–8
25
1986
y 6
30
1984
2. Compare the accompanying graphs. a. Which line appears to have the steeper slope? b. Use the intercepts to calculate the slope. Which graph actually does have the steeper slope?
NAEP Reading Scores (Age 9) and ESEA Funding (in 2004 dollars) NAEP scale score
Interest rate (%)
Graph B
Funding (in billions of dollars)
6.1
8
6. Generate two graphs and on each draw a line through the points (0, 3) and (4, 6), choosing x and y scales such that: a. The first line appears to have a slope of almost zero. b. The second line appears to have a very large positive slope.
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7. a. Generate a graph of a line through the points s0, 22d and s5, 22d . b. On a new grid, choose different scales so that the line through the same points appears to have a large positive slope. c. What have you discovered? 8. What are three adjectives (like “explosive”) that would imply rapid growth? 9. What are three adjectives (like “severe”) that would imply rapid decline? 10. (Graphing program optional.) Examine the data given on women in the U.S. military forces.
12. (Course software required.) Open “L6: Changing Axis Scales” in Linear Functions. Generate a line in the upper left-hand box. The same line will appear graphed in the three other boxes but with the axes scaled differently. Describe how the axes are rescaled in order to create such different impressions. 13. The accompanying graphs show the same data on the median income of a household headed by a black person as a percentage of the median income of a household headed by a white person. Describe the impression each graph gives and how that was achieved. (Data from the U.S. Bureau of the Census.) Black Household Median Income as a Percentage of White
Women in Uniform: Female Active-Duty Military Personnel Year
Total
Army
Navy
Marine Corps
Air Force
1970 1980 1990 2000 2005
41,479 171,418 227,018 204,498 189,465
16,724 69,338 83,621 72,021 71,400
8,683 34,980 59,907 53,920 54,800
2,418 6,706 9,356 9,742 8,498
13,654 60,394 74,134 68,815 54,767
66% 65% 64% 63% 62%
Source: U.S. Defense Department. 61%
a. Make the case with graphs and numbers that women are a growing presence in the U.S. military. b. Make the case with graphs and numbers that women are a declining presence in the U.S. military. c. Write a paragraph that gives a balanced picture of the changing presence of women in the military using appropriate statistics to make your points. What additional data would be helpful?
60% 2000
Black Household Median Income as a Percentage of White 100% 80% 60%
90,000 400,000 60,000
300,000 200,000
30,000 100,000
40% 20% 0 2000
New AIDS cases
1995
2000 Deaths
2004 Living
a. Find something encouraging to say about these data by using numerical evidence, including estimated average rates of change. b. Find numerical support for something discouraging to say about the data. c. How might we explain the enormous increase in new AIDS cases reported from 1985 to a peak in 1992–1993, and the drop-off the following year?
2002 2003 Graph B
2004
2005
Nobel Prizes Awarded in Science, for Selected Countries, 1901–1974 30
0 1990
2001
14. The accompanying graph shows the number of Nobel Prizes awarded in science for various countries between 1901 and 1974. It contains accurate information but gives the impression that the number of prize winners declined drastically in the 1970s, which was not the case. What flaw in the construction of the graph leads to this impression?
Number of Nobel Prizes
0 1985
Living with AIDS
Deaths and new AIDS diagnoses
11. The first case in the United States of what later came to be called AIDS appeared in June 1981. The accompanying graph shows the progress of AIDS cases in the United States as reported by the Centers for Disease Control and Prevention (CDC). Estimated New AIDS Cases, Deaths Among Persons with AIDS and People Living with AIDS, 1985–2004
2002 2004 Graph A
25 United States 20 15 10
United Kingdom
Germany
5 0 1901– 1910
U.S.S.R
France 1911– 1920
1921– 1930
1931– 1940
1941– 1950
1951– 1960
1961– 1971– 1970 1974
Source: E. R. Tufte, The Visual Display of Quantitative Information (Cheshire, Conn.: Graphics Press, 1983).
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2.5 Linear Functions: When Rates of Change Are Constant In many of our examples so far, the average rate of change has varied depending on the choice of end points. Now we will examine the special case when the average rate of change remains constant.
What If the U.S. Population Had Grown at a Constant Rate? A Hypothetical Example
Experiment with varying the average velocities and then setting them all constant in “C3: Average Velocity and Distance” in Rates of Change.
In Section 2.2 we calculated the average rate of change in the population between 1790 and 1800 as 0.14 million people/year. We saw that the average rate of change was different for different decades. What if the average rate of change had remained constant? What if in every decade after 1790 the U.S. population had continued to grow at the same rate of 0.14 million people/year? That would mean that starting with a population estimated at 3.9 million in 1790, the population would have grown by 0.14 million each year. The slopes of all the little line segments connecting adjacent population data points would be identical, namely 0.14 million people/year. The graph would be a straight line, indicating a constant average rate of change. On the graph of actual U.S. population data, the slopes of the line segments connecting adjacent points are increasing, so the graph curves upward. Figure 2.20 compares the actual and hypothetical results.
300
Population (in millions)
250
Actual growth
200
150
100 Hypothetical growth at a constant rate
50
0 1790
1830
1870
1910
1950
1990
Year
Figure 2.20 U.S. population: A hypothetical example.
Any function that has the same average rate of change on every interval has a graph that is a straight line. The function is called linear. This hypothetical example represents a linear function. When the average rate of change is constant, we can drop the word “average” and just say “rate of change.” A linear function has a constant rate of change. Its graph is a straight line.
Real Examples of a Constant Rate of Change EXAMPLE
1
According to the standardized growth and development charts used by many American pediatricians, the median weight for girls during their first six months of life increases at an almost constant rate. Starting at 7.0 pounds at birth, female median weight
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increases by approximately 1.5 pounds per month. If we assume that the median weight for females, W, is increasing at a constant rate of 1.5 pounds per month, then W is a linear function of age in months, A. a. Generate a table that gives the median weight for females, W, for the first six months of life and create a graph of W as a function of A. b. Find an equation for W as a function of A. What is an appropriate domain for this function? c. Express the equation for part (b) using only units of measure. SOLUTION
a. For female infants at birth sA 5 0 monthsd , the median weight is 7.0 lb sW 5 7.0 lbd . The rate of change, 1.5 pounds/month, means that as age increases by 1 month, weight increases by 1.5 pounds. See Table 2.11. The dotted line in Figure 2.21 shows the trend in the data. Median Weight for Girls A, Age (months)
7.0 8.5 10.0 11.5 13.0 14.5 16.0
Weight (in pounds)
0 1 2 3 4 5 6
W 20
W, Weight (lb)
15
10
5
0
Table 2.11 Source: Data derived from the Ross Growth and Development Program, Ross Laboratories, Columbus, OH.
2 4 Age (in months)
6
A
Figure 2.21 Median weight for girls.
b. To find a linear equation for W (median weight in pounds) as a function of A (age in months), we can study the table of values in Table 2.11. W 5 initial weight 1 weight gained 5 initial weight 1 rate of growth ? number of months 5 7.0 lb 1 1.5 lb/month ? number of months The equation would be W 5 7.0 1 1.5A An appropriate domain for this function would be 0 # A # 6. c. Since our equation represents quantities in the real world, each term in the equation has units attached to it. The median weight W is in pounds (lb), rate of change is in lb/month, and age, A, is in months. If we display the equation
?
W 5 7.0 1 1.5A
SOMETHING TO THINK ABOUT
If the median birth weight for baby boys is the same as for baby girls, but boys put on weight at a faster rate, which numbers in the model would change and which would stay the same? What would you expect to be different about the graph?
showing only the units, we have lb 5 lb 1 a
lb bmonth month
The rules for canceling units are the same as the rules for canceling numbers in fractions. So, lb 5 lb 1 a lb 5 lb 1 lb
lb bmonth month
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89
This equation makes sense in terms of the original problem since pounds (lb) added to pounds (lb) should give us pounds (lb).
2
You spend $1200 on a computer and for tax purposes choose to depreciate it (or assume it decreases in value) to $0 at a constant rate over a 5-year period. a. Calculate the rate of change of the assumed value of the equipment over 5 years. What are the units? b. Create a table and graph showing the value of the equipment over 5 years. c. Find an equation for the value of the computer as a function of time in years. Why is this a linear function? d. What is an appropriate domain for this function? What is the range?
SOLUTION
a. After 5 years, your computer is worth $0. If V is the value of your computer in dollars and t is the number of years you own the computer, then the rate of change of V from t 5 0 to t 5 5 is rate of change 5
change in value V 5 change in time t
52
$1200 5 2$240/year 5 years
Thus, the worth of your computer drops at a constant rate of $240 per year. The rate of change in V is negative because the worth of the computer decreases over time. The units for the rate of change are dollars per year.
Value of Computer Depreciated over 5 Years Numbers of Years
Value of Computer ($)
0 1 2 3 4 5
1200 960 720 480 240 0
Table 2.12
$1400 Value of computer
EXAMPLE
$1200 $1000 $800 $600 $400 $200 $0
0
1
2 3 Number of years
4
5
Figure 2.22 Value of computer over 5 years.
b. Table 2.12 and Figure 2.22 show the depreciated value of the computer. c. To find a linear equation for V as a function of t, think about how we found the table of values. value of computer 5 initial value 1 srate of declined ? snumber of yearsd V 5 $1200 1 s2$240/yeard ? t V 5 1200 2 240t This equation describes V as a function of t because for every value of t, there is one and only one value of V. It is a linear function because the rate of change is constant. d. The domain is 0 # t # 5 and the range is 0 # V # 1200.
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The General Equation for a Linear Function The equations in Examples 1 and 2 can be rewritten in terms of the output (dependent variable) and the input (independent variable).
Explorations 2.2A and 2.2B (along with “L1: m & b Sliders” in Linear Functions) allow you to examine the effects of m and b on the graph of a linear function.
weight 5 initial weight 1 rate of growth ? number of months value of computer 5 initial value
1 rate of decline ? number of years
output 5 initial value 1 5 rate of change ? 3 input 3 5 x y m b Thus, the general linear equation can be written in the form y 5 b 1 mx where we use the traditional mathematical choices of y for the output (dependent variable) and x for the input (independent variable). We let m stand for the rate of change; thus y change in y 5 change in x x 5 slope of the graph of the line
y m (slope) = b (vertical intercept)
m5
y x y
In our general equation, we let b stand for the initial value. Why is b called the initial value? The number b is the value of y when x 5 0. If we let x 5 0, then
x x
given
y 5 b 1 mx y5b1m?0 y5b
The point (0, b) satisfies the equation and lies on the y-axis. The point (0, b) is technically the vertical intercept. However, since the coordinate b tells us where the line crosses the y-axis, we often just refer to b as the vertical or y-intercept (see Figure 2.23).
Figure 2.23 Graph of
y 5 b 1 mx.
A Linear Function A function y 5 ƒ(x) is called linear if it can be represented by an equation of the form y 5 b 1 mx Its graph is a straight line where m is the slope, the rate of change of y with respect to x, so y x b is the vertical or y-intercept and is the value of y when x 5 0. m5
The equation y 5 b 1 mx could, of course, be written in the equivalent form y 5 mx 1 b. In linear mathematical models, b is often the initial or starting value of the output, so it is useful to place it first in the equation. EXAMPLE
3
SOLUTION
For each of the following equations, identify the value of b and the value of m. a. y 5 24 1 3.25x c. y 5 24x 1 3.25 b. y 5 3.25x 2 4 d. y 5 3.25 2 4x a. b. c. d.
b 5 24 and m 5 3.25 b 5 24 and m 5 3.25 b 5 3.25 and m 5 24 b 5 3.25 and m 5 24
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EXAMPLE
4
91
In the following equations, L represents the legal fees (in dollars) charged by four different law firms and h represents the number of hours of legal advice. L1 5 500 1 200h
L3 5 800 1 350h
L2 5 1000 1 150h
L4 5 500h
a. Which initial fee is the highest? b. Which rate per hour is the highest? c. If you need 5 hours of legal advice, which legal fee will be the highest? SOLUTION
a. L2 has the highest initial fee of $1000. b. L4 has the highest rate of $500 per hour. c. Evaluate each equation for h 5 5 hours. L1 5 500 1 200h 5 500 1 200(5) 5 $1500
L3 5 800 1 350h 5 800 1 350(5) 5 $2550
L2 5 1000 1 150h 5 1000 1 150(5) 5 $1750
L4 5 500h 5 500(5) 5 $2500
For 5 hours of legal advice, L3 has the highest legal fee.
Algebra Aerobics 2.5 1. From Figure 2.21, estimate the weight W of a baby girl who is 4.5 months old. Then use the equation W 5 7.0 1 1.5A to calculate the corresponding value for W. How close is your estimate? 2. From the same graph, estimate the age of a baby girl who weighs 11 pounds. Then use the equation to calculate the value for A. 3. a. If C 5 15P 1 10 describes the relationship between the number of persons (P) in a dining party and the total cost in dollars (C) of their meals, what is the unit of measure for 15? For 10? b. The equation W 5 7.0 1 1.5A (modeling weight as a function of age) expressed in units of measure only is lb 5 lb 1 a
lb b months month
Express C 5 15P 1 10 from part (a) in units of measure only. 4. (True story.) A teenager travels to Alaska with his parents and wins $1200 in a rubber ducky race. (The race releases 5000 yellow rubber ducks marked with successive integers from one bridge over a river and collects them at the next bridge.) Upon returning home he opens up a “Rubber Ducky Savings Account,” deposits his winnings, and continues to deposit $50 each month. If D 5 amount in the account and M 5 months since the creation of the account,
then D 5 1200 1 50 M describes the amount in the account after M months. a. What are the units for 1200? For 50? b. Express the equation using only units of measure. 5. Assume S 5 0.8Y 1 19 describes the projected relationship between S, sales of a company (in millions of dollars), for Y years from today. a. What are the units of 0.8 and what does it represent? b. What are the units of 19 and what does it represent? c. What would be the projected company sales in three years? d. Express the equation using only units of measure. 6. Assume C 5 0.45N represents the total cost C (in dollars) of operating a car for N miles. a. What does 0.45 represent and what are its units? b. Find the total cost to operate a car that has been driven 25,000 miles. c. Express the equation using only units of measure. 7. The relationship between the balance B (in dollars) left on a mortgage loan and N, the number of monthly payments, is given by B 5 302,400 2 840N. a. What is the monthly mortgage payment? b. What does 302,400 represent? c. What is the balance on the mortgage after 10 years? 20 years? 30 years? (Hint: Remember there are 12 months in a year.)
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8. Identify the slope, m, and the vertical intercept, b, of the line with the given equation. a. y 5 5x 1 3 e. ƒ(x) 5 7.0 2 x b. y 5 5 1 3x f. h(x) 5 211x 1 10 c. y 5 5x g. y 5 1 2 23x d. y 5 3 h. 2y 1 6 5 10x 9. If ƒ(x) 5 50 2 25x: a. Why does ƒ(x) describe a linear function? b. Evaluate ƒ(0) and ƒ(2). c. Use your answers in part (b) to verify that the slope is 225.
8
y
2
y
m4
–1
1
–2
4
m–2 –1
–2
10. Identify the functions that are linear. For each linear function, identify the slope and the vertical intercept. a. ƒ(x) 5 3x 1 5 c. ƒsxd 5 3x2 1 2 b. ƒ(x) 5 x d. ƒsxd 5 4 2 23x 11. Write an equation for the line in the form y 5 mx 1 b for the indicated values. a. m 5 3 and b 5 4 b. m 5 21 and passes through the origin c. m 5 0 and b 5 23 12. Write the equation of the graph of each line in Figure 2.24 in y 5 mx 1 b form. Use the y-intercept and the slope indicated on each graph.
2
1
y
4
x 2
x x
Graph A
–2
Graph B
m1
m0
x –3
y
–1 Graph C
2
4
–2
–1
1
2
–3 Graph D
Figure 2.24 Four linear graphs.
Exercises for Section 2.5 You might wish to hone your algebraic mechanical skills with three programs in the course software; Linear Functions: “L1: Finding m & b”, “L3: Finding a Line through 2 Points” and “L4: Finding 2 Points on a Line.” They offer practice in predicting values for m and b, generating linear equations, and finding corresponding solutions. 1. Consider the equation E 5 5000 1 100n. a. Find the value of E for n 5 0, 1, 20. b. Express your answers to part (a) as points with coordinates (n, E). 2. Consider the equation G 5 12,000 1 800n. a. Find the value of G for n 5 0, 1, 20. b. Express your answers to part (a) as points with coordinates (n, G). 3. Determine if any of the following points satisfy one or both of the equations in Exercises 1 and 2. a. (5000, 0) b. (15, 24000) c. (35, 40000) 4. Suppose during a 5-year period the profit P (in billions of dollars) for a large corporation was given by P 5 7 1 2Y , where Y represents the year. a. Fill in the chart. Y P
0
1
2
3
4
b. What are the units of P? c. What does the 2 in the equation represent, and what are its units? d. What was the initial profit? 5. Consider the equation D 5 3.40 1 0.11n a. Find the values of D for n 5 0, 1, 2, 3, 4. b. If D represents the average consumer debt, in thousands of dollars, over n years, what does 0.11 represent? What are its units? c. What does 3.40 represent? What are its units? 6. Suppose the equations E 5 5000 1 1000n and G 5 12,000 1 800n give the total cost of operating an electrical (E) versus a gas (G) heating/cooling system in a home for n years. a. Find the cost of heating a home using electricity for 10 years. b. Find the cost of heating a home using gas for 10 years. c. Find the initial (or installation) cost for each system. d. Determine how many years it will take before $40,000 has been spent in heating/cooling a home that uses: i. Electricity ii. Gas 7. If the equation E 5 5000 1 1000n gives the total cost of heating/cooling a home after n years, rewrite the equation using only units of measure.
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8. Over a 5-month period at Acadia National Park in Maine, the average night temperature increased on average 5 degrees Fahrenheit per month. If the initial temperature is 25 degrees, create a formula for the night temperature N for month t, where 0 # t # 4. 9. A residential customer in the Midwest purchases gas from a utility company that charges according to the formula C(g) 5 11 1 10.50(g), where C(g) is the cost, in dollars, for g thousand cubic feet of gas. a. Find C(0), C(5), and C(10). b. What is the cost if the customer uses no gas? c. What is the rate per thousand cubic feet charged for using the gas? d. How much would it cost if the customer uses 96 thousand cubic feet of gas (the amount an average Midwest household consumes during the winter months)? 10. Create the formula for converting degrees centigrade, C, to degrees Fahrenheit, F, if for every increase of 5 degrees centigrade the Fahrenheit temperature increases by 9 degrees, with an initial point of (C, F) 5 (0, 32). 11. Determine the vertical intercept and the rate of change for each of these formulas: a. P 5 4s c. C 5 2pr b. C 5 pd d. C 5 59 F 2 17.78 12. A hiker can walk 2 miles in 45 minutes. a. What is his average speed in miles per hour? b. What formula can be used to find the distance traveled, d, in t hours? 13. The cost C(x) of producing x items is determined by adding fixed costs to the product of the cost per item and the number of items produced, x. Below are three possible cost functions C(x), measured in dollars. Match each description [in parts (e) 2 (g)] with the most likely cost function and explain why. a. C(x) 5 125,000 1 42.50x b. C(x) 5 400,000 1 0.30x c. C(x) 5 250,000 1 800x e. The cost of producing a computer f. The cost of producing a college algebra text g. The cost of producing a CD
b. Rewrite S(x) as an equation using only units of measure. c. Evaluate S(x) for x values of 0, 5, and 10 years. d. How many years will it take for a person to earn an annual salary of $43,000? 17. The following represent linear equations written using only units of measure. In each case supply the missing unit. a. inches 5 inches 1 sinches/hourd ? s?d b. miles 5 miles 1 s?d ? sgallonsd c. calories 5 calories 1 s?d ? sgrams of fatd 18. Identify the slopes and the vertical intercepts of the lines with the given equations. a. y 5 3 1 5x d. Q 5 35t 2 10 b. ƒ(t) 5 2t e. ƒ(E) 5 10,000 1 3000E c. y 5 4 19. For each of the following, find the slope and the vertical intercept, then sketch the graph. (Hint: Find two points on the line.) a. y 5 0.4x 2 20 b. P 5 4000 2 200C 20. Construct an equation and sketch the graph of its line with the given slope, m, and vertical intercept, b. (Hint: Find two points on the line.) a. m 5 2, b 5 23 c. m 5 0, b 5 50 b. m 5 234, b 5 1 In Exercises 21 to 23 find an equation, generate a small table of solutions, and sketch the graph of a line with the indicated attributes. 21. A line that has a vertical intercept of 22 and a slope of 3. 22. A line that crosses the vertical axis at 3.0 and has a rate of change of 22.5. 23. A line that has a vertical intercept of 1.5 and a slope of 0. 24. Estimate b (the y-intercept) and m (the slope) for each of the accompanying graphs. Then, for each graph, write the corresponding linear function. 8
y
16. a. If S(x) 5 20,000 1 1000x describes the annual salary in dollars for a person who has worked for x years for the Acme Corporation, what is the unit of measure for 20,000? For 1000?
8
x
–8
14. A new $25,000 car depreciates in value by $5000 per year. Construct a linear function for the value V of the car after t years. 15. The state of Pennsylvania has a 6% sales tax on taxable items. (Note: Clothes, food, and certain pharmaceuticals are not taxed in Pennsylvania.) a. Create a formula for the total cost (in dollars) of an item C(p) with a price tag p. (Be sure to include the sales tax.) b. Find C(9.50), C(115.25), and C(1899). What are the units?
93
8
x
–8
–8
8
–8
Graph A
Graph C
8
y
8
x
–8
8
8
–8 Graph D
y
x
–8
–8 Graph B
y
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2.6 Visualizing Linear Functions The values for b and m in the general form of the linear equation, y 5 b 1 mx, tell us about the graph of the function. The value for b tells us where to anchor the line on the y-axis. The value for m tells us whether the line climbs or falls and how steep it is.
The Effect of b In the equation y 5 b 1 mx, the value b is the vertical intercept, so it anchors the line at the point (0, b) (see Figure 2.25). 8y y = 3 + 0.5x
(b = 3)
y = 0.5x
(b = 0)
y = –2 + 0.5x
(b = –2)
(0, 3) (0, 0) –8
x 8
(0, –2)
–8
Figure 2.25 The effect of b, the vertical intercept.
1
Explain how the graph of y 5 4 1 5x differs from the graphs of the following functions: a. y 5 8 1 5x b. y 5 2 1 5x c. y 5 5x 1 4
SOLUTION
Although all of the graphs are straight lines with a slope of 5, they each have a different vertical intercept. The graph of y 5 4 1 5x has a vertical intercept at 4. a. The graph of y 5 8 1 5x intersects the y-axis at 8, four units above the graph of y 5 4 1 5x. b. The graph of y 5 2 1 5x intersects the y-axis at 2, two units below the graph of y 5 4 1 5x. c. Since y 5 5x 1 4 and y 5 4 1 5x are equivalent equations, they have the same graph.
EXAMPLE
The Effect of m The sign of m The sign of m in the equation y 5 b 1 mx determines whether the line climbs (slopes up) or falls (slopes down) as we move left to right on the graph. If m is positive, the line climbs from left to right (as x increases, y increases). If m is negative, the line falls from left to right (as x increases, y decreases) (see Figure 2.26).
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25
y
95
y
y 5x
y mx (m 0)
x –5
x
5
y mx (m 0)
y –5x
–25
Figure 2.26 The effect of the sign of m.
EXAMPLE
2
Match the following functions to the lines in Figure 2.27. ƒ(x) 5 3 2 2x g(x) 5 3 1 2x h(x) 5 25 2 2x j(x) 5 25 1 2x
10
y
A
10
y
C
x –10
10
–10
B
x –10
10
–10
D
Figure 2.27 Matching graphs.
SOLUTION
?
SOMETHING TO THINK ABOUT
Describe the graph of a line where m 5 0.
A is the graph of g(x). B is the graph of ƒ(x). C is the graph of j(x). D is the graph of h(x). The magnitude of m The magnitude (absolute value) of m determines the steepness of the line. Recall that the absolute value of m is the value of m stripped of its sign; for example, k 23 k 5 3. The greater the magnitude k m k , the steeper the line. This makes sense since m is the slope or the rate of change of y with respect to x. Notice how the steepness of each line in Figure 2.28 (next page) increases as the magnitude of m increases. For example, the slope (m 5 5) of y 5 7 1 5x is steeper than the slope (m 5 3) of y 5 7 1 3x. In Figure 2.29 we can see that the slope (m 5 25) of y 5 7 2 5x is steeper than the slope (m 5 23) of y 5 7 2 3x since k 25 k 5 5 . k 23 k 5 3. The lines y 5 7 2 5x and y 5 7 1 5x have the same steepness of 5 since k 25 k 5 k 5 k 5 5.
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16
y 75x y 73x y 7x
y
16
y
y 7x x
x –8
–8
8
8
EXAMPLE
3
y 73x
–16
–16
y 75x
Figure 2.28 Graphs with
Figure 2.29 Graphs with negative
positive values for m.
values for m.
Pair each graph in Figure 2.30 with a matching equation. ƒ(x) 5 9 2 0.4x g(x) 5 24 1 x h(x) 5 9 2 0.2x i(x) 5 24 1 0.25x j(x) 5 9 1 0.125x k(x) 5 4 2 0.25x
10
y
A B C D x
E –10
SOLUTION
EXAMPLE
4
SOLUTION
EXAMPLE
5
A is the graph of j(x). B is the graph of h(x). C is the graph of ƒ(x). D is the graph of k(x). E is the graph of g(x). F is the graph of i(x).
10
F
–10
Figure 2.30 Graphs of multiple
linear functions.
Without graphing the following functions, how can you tell which graph will have the steepest slope? a. ƒ(x) 5 5 2 2x b. g(x) 5 5 1 4x c. h(x) 5 3 2 6x The graph of the function h will be steeper than the graphs of the functions f and g since the magnitude of m is greater for h than for f or g. The greater the magnitude of m, the steeper the graph of the line. For ƒ(x), the magnitude of the slope is k 22 k 5 2. For g(x), the magnitude of the slope is k 4 k 5 4. For h(x), the magnitude of the slope is k 26 k 5 6. Which of the graphs in Figure 2.31 has the steeper slope? 50
y
25
50
y
25
A
B
x –10
–5
5
10
x –50
–25
25
–25
–25
–50
–50
Graph A
Figure 2.31 Comparing slopes.
Graph B
50
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2.6 Visualizing Linear Functions
SOLUTION
97
Remember from Section 2.4 that the steepness of a linear graph is not related to its visual impression, but to the numerical magnitude of the slope. The scales of the horizontal axes are different for the two graphs in Figure 2.31, so the impression of relative steepness is deceiving. Line A passes through (0, 0) and (10, 25), so its slope is (25 2 0)/(10 2 0) 5 2.5. Line B passes through (0, 0) and (50, 50) so its slope is (50 2 0)/(50 2 0) 5 1. So line A has a steeper slope than line B.
The Graph of a Linear Equation Given the general linear equation, y 5 b 1 mx, whose graph is a straight line: The y-intercept, b, tells us where the line crosses the y-axis. The slope, m, tells us how fast the line is climbing or falling. The larger the magnitude (or absolute value) of m, the steeper the graph. If the slope, m, is positive, the line climbs from left to right. If m is negative, the line falls from left to right.
Algebra Aerobics 2.6 1. Place these numbers in order from smallest to largest. k 212 k , k 27 k , k 23 k , k 21 k , 0, 4, 9 2. Without graphing the function, explain how the graph y 5 6x 2 2 differs from the graph of a. y 5 6x c. y 5 22 1 3x b. y 5 2 1 6x d. y 5 22 2 2x 3. Without graphing, order the graphs of the functions from least steep to steepest. a. y 5 100 2 2x c. y 5 23x 2 5 b. y 5 1 2 x d. y 5 3 2 5x
7. Match the four graphs in Figure 2.32 with the given functions. a. ƒ(x) 5 2 1 3x c. hsxd 5 12x 2 2 b. g(x) 5 22 2 3x d. k(x) 5 2 2 3x y
–10
4. On an x-y coordinate system, draw a line with a positive slope and label it ƒ(x). a. Draw a line g(x) with the same slope but a y-intercept three units above ƒ(x). b. Draw a line h(x) with the same slope but a y-intercept four units below ƒ(x). c. Draw a line k(x) with the same steepness as ƒ(x) but with a negative slope. 5. Which function has the steepest slope? Create a table of values for each function and graph the function to show that this is true. a. ƒ(x) 5 3x 2 5 b. g(x) 5 7 2 8x 6. Create three functions with a y-intercept of 4 and three different negative slopes. Indicate which function has the steepest slope.
10
10
10
x
–10
10
x
–10
–10
Graph A 10
y
Graph C y
y
–10
10
10
–10
Graph B
x
–10
10
–10
Graph D
Figure 2.32 Four graphs of linear functions.
x
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Exercises for Section 2.6 1. Assuming m is the slope, identify the graph(s) where: a. m 5 3 b. k m k 5 3 c. m 5 0 d. m 5 23 y
6
6
5. Match the graph with the correct equation. a. y 5 x b. y 5 2x c. y 5 2x d. y 5 4x
y
15
y y 6
x
x
x –6
–6
6
–3
3
x –10
10
9 –6 –15
–6
Graph C
Graph A 6
Graph A
–6
y 6
3
Graph C 15
y
y
y x –3
x –6
6
–6
6
x
3
x 6
–6
–3 –15 Graph D
Graph B
–6
–6
Graph B
Graph D
6. Match the function with its graph. (Note: There is one graph that has no match.) a. y 5 24 1 3x b. y 5 23x 1 4 c. y 5 4 1 3x 6
2. Which line(s), if any, have a slope m such that: a. k m k 5 2? b. m 5 12 ?
10
5
y
–5 –5
A
5
5
B C –5
–4 Graph A
5
Graph C
4
–5
0
5
x
3. For each set of conditions, construct a linear equation and draw its graph. a. A slope of zero and a y-intercept of 23 b. A positive slope and a vertical intercept of 23 c. A slope of 23 and a vertical intercept that is positive 4. Construct a linear equation for each of the following conditions. a. A negative slope and a positive y-intercept b. A positive slope and a vertical intercept of 210.3 c. A constant rate of change of $1300/year
7
–5
5
–5 –6 Graph B
5 –3 Graph D
7. In each part construct three different linear equations that all have: a. The same slope b. The same vertical intercept 8. Which equation has the steepest slope? a. y 5 2 2 7x b. y 5 2x 1 7 c. y 5 22 1 7x
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9. Given the function Q(t) 5 13 2 5t, construct a related function whose graph: a. Lies five units above the graph of Q(t) b. Lies three units below the graph of Q(t) c. Has the same vertical intercept d. Has the same slope e. Has the same steepness, but the slope is positive
99
13. Given the following graphs of three straight lines: a. Which has the steepest slope? b. Which has the flattest slope? c. If the slope of the lines A, B, and C are m1, m2, and m3, respectively, list them in increasing numerical order. d. In this example, why does the line with the steepest slope have the smallest numerical value?
10. Given the equation C(n) 5 30 1 15n, construct a related equation whose graph: a. Is steeper b. Is flatter (less steep) c. Has the same steepness, but the slope is negative
y
A
B
C
11. On the same axes, graph (and label with the correct equation) three lines that go through the point (0, 2) and have the following slopes: a. m 5 12 b. m 5 2 c. m 5 56
x
12. On the same axes, graph (and label with the correct equation) three lines that go through the point (0, 2) and have the following slopes: a. m 5 212 b. m 5 22 c. m 5 256
2.7 Finding Graphs and Equations of Linear Functions Finding the Graph 1
Given the equation A forestry study measured the diameter of the trunk of a red oak tree over 5 years. The scientists created a linear model D 5 1 1 0.13Y, where D 5 diameter in inches and Y 5 number of years from the beginning of the study. a. What do the numbers 1 and 0.13 represent in this context? b. Sketch a graph of the function model.
SOLUTION
a. The number 1 represents a starting diameter of 1 inch. The number 0.13 represents the annual growth rate of the oak’s diameter (change in diameter/change in time), 0.13 inches per year. 1.8 b. The linear equation D 5 1 1 0.13Y tells us 1.6 that 1 is the vertical intercept, so the point 1.4 (1, 1.13) (0, 1) lies on the graph. The graph 1.2 represents solutions to the equation. So to 1.0 find a second point, we can evaluate D for 0.8 any other value of Y. If we set Y 5 1, then 0.6 D 5 1 1 s0.13 ? 1d 5 1 1 0.13 5 1.13. 0.4 So (1, 1.13) is another point on the line. 0.2 Since two points determine a line, we can 0 0 1 2 3 4 5 sketch our line through (0, 1) and (1, 1.13) Year of study (see Figure 2.33). Diameter (inches)
EXAMPLE
The program “L4: Finding 2 Points on a Line” in Linear Functions will give you practice in finding solutions to equations.
Figure 2.33 The diameter of a red
EXAMPLE
2
oak over time.
Given a point off the y-axis and the slope Given a point (2, 3) and a slope of m 5 25/4, describe at least two ways you could find a second point to plot a line with these characteristics without constructing the equation.
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EXAMPLE
3
SOLUTION
y Plot the point (2, 3). 6 a. If we write the slope as (25)/4, then a change of 4 in x corresponds to a change of 25 in y. So 4 starting at (2, 3), moving horizontally four units (2, 3) to the right (adding 4 to the x-coordinate), and –5 then moving vertically five units down x (subtracting 5 from the y-coordinate) gives us a 8 second point on the line at (2 1 4, 3 2 5) 5 –2 (6, 22). Now we can plot our second point (6, –2) (6, 22) and draw the line through it and our –4 original point, (2, 3) (see Figure 2.34). b. If we are modeling real data, we are more likely Figure 2.34 Graph of the line to convert the slope to decimal form. In this case through (2, 3) with slope 25/4. m 5 (25)/4 5 21.25. We can generate a new point on the line by starting at (2, 3) and moving 1 unit to the right and down (since m is negative) 21.25 units to get the point (2 1 1, 3 2 1.25) 5 (3, 1.75).
Given a general description A recent study reporting on the number of smokers showed: a. A linear increase in Georgia b. A linear decrease in Utah c. A nonlinear decrease in Hawaii d. A nonlinear increase in Oklahoma Generate four rough sketches that could represent these situations. See Figure 2.35.
Utah No. of smokers
No. of smokers
Georgia
Year
Hawaii
Year
Oklahoma No. of smokers
SOLUTION
No. of smokers
100
10/11/07
Year
Year
Figure 2.35 The change in the number of smokers over time in four states.
Finding the Equation The program “L.3: Finding a Line Through 2 Points” in Linear Functions will give you practice in this skill.
EXAMPLE
4
SOLUTION
To determine the equation of any particular linear function y 5 b 1 mx, we only need to find the specific values for m and b. Given two points Find the equation of the line through the two points (3, 5) and (4, 9). If we think of these points as being in the form (x, y), then the slope m of the line connecting them is change in y 9 2 s25d 14 5 5 52 change in x 4 2 s23d 7
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So the equation is of the form y b 2x. To find b, the y-intercept, we can substitute either point into the equation. Substituting in (4, 9) we get 9 b 1 (2 ? 4) ➪ 9 b 8 ➪ b 1. So the equation is y 1 2x. EXAMPLE
5
SOLUTION y
From a graph Find the equation of the linear function graphed in Figure 2.36. We can use any two points on the graph to calculate m, the slope. If, for example, we take (23, 0) and (3, 22), then
4
slope 5 x 4
–4
change in y 22 2 0 22 21 5 5 5 change in x 3 2 s23d 6 3
From the graph we can estimate the y-intercept as 21. So b 5 21. Hence the equation is y 5 21 2 13x.
–4
Figure 2.36 Graph of a linear function.
EXAMPLE
6
From a verbal description In 2006 AT&T’s One Rate plan charged a monthly base fee of $3.95 plus $0.07 per minute for long-distance calls. Construct an equation to model a monthly phone bill.
SOLUTION
In making the transition from words to an equation, it’s important to first identify which is the independent and which the dependent variable. We usually think of the phone bill, B, as a function of the number of minutes you talk, N. If you haven’t used any phone minutes, then N 5 0 and your bill B 5 $3.95. So $3.95 is the vertical intercept. The number $0.07 is the rate of change of the phone bill with respect to number of minutes talked. The rate of change is constant, making the relationship linear. So the slope is $0.07/minute and the equation is B 5 3.95 1 0.07N
Snowplow speed (mph)
EXAMPLE
7
The top speed a snowplow can travel on dry pavement is 40 miles per hour, which decreases by 0.8 miles per hour with each inch of snow on the highway. a. Construct an equation describing the relationship between snowplow speed and snow depth. b. Determine a reasonable domain and then graph the function.
SOLUTION
a. If we think of the snow depth, D, as determining the snowplow speed, P, then we need an equation of the form P 5 b 1 mD. If there is no snow, then the snowplow can travel at its maximum speed of 40 mph; that is, when D 5 0, then P 5 40. So the point (0, 40) lies on the line, making the vertical intercept b 5 40. The rate of change, m, (change in snowplow speed)/(change in snow depth) 5 20.8 mph per inch of snow. So the desired equation is
50
P 5 40 2 0.8D b. Consider the snowplow as only going forward (i.e., not backing up). Then the snowplow speed does not go below 0 mph. So if we let P 5 0 and solve for D, we have
40 30 20 10 0
10 20 30 40 50 Snow depth (inches)
60
Figure 2.37 Snowplow speed
versus snow depth.
0 5 40 2 0.8D 0.8D 5 40 D 5 50 So when the snow depth reaches 50 inches, the plow is no longer able to move. A reasonable domain then would be 0 # D # 50. (See Figure 2.37.)
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EXAMPLE
8
SOLUTION
Pediatric growth charts suggest a linear relationship between age (in years) and median height (in inches) for children between 2 and 12 years. Two-year-olds have a median height of 35 inches, and 12-year-olds have a median height of 60 inches. a. Generate the average rate of change of height with respect to age. (Be sure to include units.) Interpret your result in context. b. Generate an equation to describe height as a function of age. What is an appropriate domain? c. What would this model predict as the median height of 8-year-olds? change in height 60 2 35 25 5 5 5 2.5 inches/year change in age 12 2 2 10 The chart suggests that, on average, children between the ages of 2 and 12 grow 2.5 inches each year. b. If we think of height, H (in inches), depending on age, A (in years), then we want an equation of the form H 5 b 1 m ? A. From part (a) we know m 5 2.5, so our equation is H 5 b 1 2.5A. To find b, we can substitute the values for any known point into the equation. When A 5 2, then H 5 35. Substituting in, we get a. Average rate of change 5
H 5 b 1 2.5A 35 5 b 1 s2.5 ? 2d 35 5 b 1 5 30 5 b So the final form of our equation is H 5 30 1 2.5A where the domain is 2 # A # 12. c. When A 5 8 years, our model predicts that the median height is H 5 30 1 s2.5 ? 8d 5 30 1 20 5 50 inches.
EXAMPLE
9
From a table a. Determine if the data in Table 2.13 represent a linear relationship between values of blood alcohol concentration and number of drinks consumed for a 160-pound person. (One drink is defined as 5 oz of wine, 1.25 oz of 80-proof liquor, or 12 oz of beer.) D, Number of Drinks
A, Blood Alcohol Concentration
2 4 6 10
0.047 0.094 0.141 0.235
Table 2.13
Note that federal law requires states to have 0.08 as the legal BAC limit for driving drunk. b. If the relationship is linear, determine the corresponding equation. SOLUTION
a. We can generate a third column in the table that represents the average rate of change between consecutive points (see Table 2.14). Since the average rate
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of change of A with respect to D remains constant at 0.0235, these data represent a linear relationship. D
A
Average Rate of Change
2
0.047
4
0.094
6
0.141
10
0.235
n.a. 0.047 0.094 2 0.047 5 5 0.0235 422 2 0.047 0.141 2 0.094 5 5 0.0235 624 2 0.094 0.235 2 0.141 5 5 0.0235 10 2 6 4
Table 2.14
b. The rate of change is the slope, so the corresponding linear equation will be of the form A 5 b 1 0.0235D
(1)
To find b, we can substitute any of the original paired values for D and A, for example, (4, 0.094), into Equation (1) to get 0.094 5 b 1 s0.0235 ? 4d 0.094 5 b 1 0.094 0.094 2 0.094 5 b 05b So the final equation is or just
A 5 0 1 0.0235D A 5 0.0235D
So when D, the number of drinks, is 0, A, the blood alcohol concentration, is 0, which makes sense.
Algebra Aerobics 2.7 For Problems 1 and 2, find an equation, make an appropriate table, and sketch the graph of: 1. A line with slope 1.2 and vertical intercept 24. 2. A line with slope 2400 and vertical intercept 300 (be sure to think about scales on both axes). 3. Write an equation for the line graphed in Figure 2.38. 5
y
4. a. Find an equation to represent the current salary after x years of employment if the starting salary is $12,000 with annual increases of $3000. b. Create a small table of values and sketch a graph. 5. a. Plot the data in Table 2.15. Years of Education
Hourly Wage
8 10 13
$5.30 $8.50 $13.30
Table 2.15 x –5
5
–5
Figure 2.38 Graph of a
linear function.
b. Is the relationship between hourly wage and years of education linear? Why or why not? c. If it is linear, construct a linear equation to model it. 6. Complete this statement regarding the graph of the line with equation y 5 6.2 1 3x: Beginning with any point on the graph of the line, we could find another point by
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moving up ___ units for each unit that we move horizontally to the right. 7. Given the equation y 5 8 2 4x, complete the following statements. a. Beginning with the vertical intercept, if we move one unit horizontally to the right, then we need to move down ___ units vertically to stay on the line and arrive at point (__, __). b. Beginning with the vertical intercept, if we move one unit horizontally to the left, then we need to move up ___ units vertically to stay on the line and arrive at point (__, __). 8. The relationship between the number of payments P made and the balance B (in dollars) of a $10,800 car loan can be represented by Table 2.16.
b. What is the cost per ticket? c. Find the revenue generated by 120 ticket purchases. 10. For each graph in Figure 2.39, identify two points on each line, determine the slope, then write an equation for the line.
5
y
10
B, Amount of Loan Balance ($)
0 1 2 3 4 5 6
10,800 10,500 10,200 9,900 9,600 9,300 9,000
x
x –5
5
–10
–5
P, Number of Monthly Payments
y
–10
Graph A
5
10
Graph C
Q
–5
5
t
Table 2.16 –5
a. Based on the table, develop a linear equation for the amount of the car loan balance B as a function of the number of monthly payments P. b. What is the monthly car payment? c. What is the balance after 24 payments? d. How many months are needed to produce a balance of zero? 9. The relationship between the number of tickets purchased for a movie and the revenue generated from that movie is indicated in Table 2.17. Number of Tickets Purchased, T
Revenue, R ($)
0 10 20 30 40
0 75 150 225 300
Table 2.17
a. Based on this table, construct a linear equation for the relationship between revenue, R, and the number of tickets purchased, T.
Graph B
Figure 2.39
11. The revenue from one season of a college baseball team is the sum of allotted funds from alumni gifts and ticket sales from home games. In 2007 a college baseball team began the season with $15,000 in allotted funds. Tickets are sold at an average price of $12 each. a. Write an equation for the relationship between tickets sold at home games, T, and revenue, R (in dollars). b. Find the revenue for the team if 40,000 home game tickets are sold over the entire season. 12. Write the equations of three lines each with vertical intercept of 6. 13. Write the equations of three lines with slope 3. 14. Write an equation for the line that passes through: a. The point (22, 1) and has a slope of 4 b. The point (3, 5) and has a slope of 2/3 c. The point (1, 3) and has a slope of 10 d. The point (1.2, 4.5) and has a slope of 2
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e. y 2 3(x 1) f. y 4 5(x 2)
15. Write an equation for the line: a. Through (2, 5) and (4, 11) b. Through (3, 2) and (6, 1) c. Through (4, 1) and (2, 7) 16. Write each of the following in the form y 5 mx 1 b. a. 3x 4y 12 b. 7x y 5 c. 2x 8y 1 d. x 2y 0
17. Sketch the graph of the line: a. With slope 2 and vertical intercept 5 b. With slope 1/2 and vertical intercept 6 c. With slope 3/4 and passing through the point (4, 2) d. With slope 3 and passing through the point (5, 6)
Exercises for Section 2.7 Graphing program optional in Exercises 8, 10 and 23. 1. Write an equation for the line through (22, 3) that has slope: a. 5 b. 234 c. 0
6. Find the equation for each of the lines A–C on the accompanying graph. 4
y
A
2. Write an equation for the line through (0, 50) that has slope: a. 220 b. 5.1 c. 0
B 5 –5
x
C
3. Calculate the slope and write an equation for the linear function represented by each of the given tables. a. b. x y A W 2 4
7.6 5.1
5 7
–6
12 16
4. Determine which of the following tables represents a linear function. If it is linear, write the equation for the linear function. a. c. e. x y x g(x) x h(x) 0 1 2 3 4 b.
3 8 13 18 23
q
R
0 1 2 3 4
0.0 2.5 5.0 7.5 10.0
0 1 2 3 4 d.
0 1 4 9 16
t
r
10 20 30 40 50
5.00 2.50 1.67 1.25 1.00
20 40 60 80 100 f.
20 260 2140 2220 2300
p
T
5 10 15 20 25
0.25 0.50 0.75 1.00 1.25
7. Put the following equations in y 5 mx 1 b form, then identify the slope and the vertical intercept. a. 2x 2 3y 5 6 d. 2y 2 3x 5 0 b. 3x 1 2y 5 6 e. 6y 2 9x 5 0 c. 13 x 1 12 y 5 6 f. 12 x 2 23 y 5 216 8. (Graphing program optional.) Solve each equation for y in terms of x, then identify the slope and the y-intercept. Graph each line by hand. Verify your answers with a graphing utility if available. a. 24y 2 x 2 8 5 0 c. 24x 2 3y 5 9 1 1 b. 2 x 2 4 y 5 3 d. 6x 2 5y 5 15 9. Complete the table for each of the linear functions, and then sketch a graph of each function. Make sure to choose an appropriate scale and label the axes. a. x ƒ(x) 5 0.10x 1 10 2100 0 100 b.
5. Plot each pair of points, then determine the equation of the line that goes through the points. a. (2, 3), (4, 0) c. (2, 0), (0, 2) b. (22, 3), (2, 1) d. (4, 2), (25, 2)
x 20.5 0 0.5
h(x) 5 50x 100
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10. (Graphing program optional.) The equation K 5 4F 2 160 models the relationship between F, the temperature in degrees Fahrenheit, and K, the number of chirps per minute for the snow tree cricket. a. Assuming F is the independent variable and K is the dependent variable, identify the slope and vertical intercept in the given equation. b. Identify the units for K, 4, F, and 2160. c. What is a reasonable domain for this model? d. Generate a small table of points that satisfy the equation. Be sure to choose realistic values for F from the domain of your model. e. Calculate the slope directly from two data points. Is this value what you expected? Why? f. Graph the equation, indicating the domain. 11. Find an equation to represent the cost of attending college classes if application and registration fees are $150 and classes cost $120 per credit. 12. a. Write an equation that describes the total cost to produce x items if the startup cost is $200,000 and the production cost per item is $15. b. Why is the total average cost per item less if the item is produced in large quantities? 13. Your bank charges you a $2.50 monthly maintenance fee on your checking account and an additional $0.10 for each check you cash. Write an equation to describe your monthly checking account costs. 14. If a town starts with a population of 63,500 that declines by 700 people each year, construct an equation to model its population size over time. How long would it take for the population to drop to 53,000? 15. A teacher’s union has negotiated a uniform salary increase for each year of service up to 20 years. If a teacher started at $26,000 and 4 years later had a salary of $32,000: a. What was the annual increase? b. What function would describe the teacher’s salary over time? c. What would be the domain for the function? 16. Your favorite aunt put money in a savings account for you. The account earns simple interest; that is, it increases by a fixed amount each year. After 2 years your account has $8250 in it and after 5 years it has $9375. a. Construct an equation to model the amount of money in your account. b. How much did your aunt put in initially? c. How much will your account have after 10 years? 17. You read in the newspaper that the river is polluted with 285 parts per million (ppm) of a toxic substance, and local officials estimate they can reduce the pollution by 15 ppm each year. a. Derive an equation that represents the amount of pollution, P, as a function of time, t.
b. The article states the river will not be safe for swimming until pollution is reduced to 40 ppm. If the cleanup proceeds as estimated, in how many years will it be safe to swim in the river? 18. The women’s recommended weight formula from Harvard Pilgrim Healthcare says, “Give yourself 100 lb for the first 5 ft plus 5 lb for every inch over 5 ft tall.” a. Find a mathematical model for this relationship. Be sure you clearly identify your variables. b. Specify a reasonable domain for the function and then graph the function. c. Use your model to calculate the recommended weight for a woman 5 feet, 4 inches tall; and for one 5 feet, 8 inches tall. 19. In 1977 a math professor bought her condominium in Cambridge, Massachusetts, for $70,000. The value of the condo has risen steadily so that in 2007 real estate agents tell her the condo is now worth $850,000. a. Find a formula to represent these facts about the value of the condo V(t), as a function of time, t. b. If she retires in 2010, what does your formula predict her condo will be worth then? 20. The y-axis, the x-axis, the line x 5 6, and the line y 5 12 determine the four sides of a 6-by-12 rectangle in the first quadrant (where x . 0 and y . 0) of the xy plane. Imagine that this rectangle is a pool table. There are pockets at the four corners and at the points (0, 6) and (6, 6) in the middle of each of the longer sides. When a ball bounces off one of the sides of the table, it obeys the “pool rule”: The slope of the path after the bounce is the negative of the slope before the bounce. (Hint: It helps to sketch the pool table on a piece of graph paper first.) a. Your pool ball is at (3, 8). You hit it toward the y-axis, along the line with slope 2. i. Where does it hit the y-axis? ii. If the ball is hit hard enough, where does it hit the side of the table next? And after that? And after that? iii. Show that the ball ultimately returns to (3, 8). Would it do this if the slope had been different from 2? What is special about the slope 2 for this table? b. A ball at (3, 8) is hit toward the y-axis and bounces off it at Q0, 16 3 R. Does it end up in one of the pockets? If so, what are the coordinates of that pocket? c. Your pool ball is at (2, 9). You want to shoot it into the pocket at (6, 0). Unfortunately, there is another ball at (4, 4.5) that may be in the way. i. Can you shoot directly into the pocket at (6, 0)? ii. You want to get around the other ball by bouncing yours off the y-axis. If you hit the y-axis at (0, 7), do you end up in the pocket? Where do you hit the line x 5 6? iii. If bouncing off the y-axis at (0, 7) didn’t work, perhaps there is some point (0, b) on the y-axis from which the ball would bounce into the pocket at (6, 0). Try to find that point.
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21. Find the equation of the line shown on the accompanying graph. Use this equation to create two new graphs, taking care to label the scales on your new axes. For one of your graphs, choose scales that make the line appear steeper than in the original graph. For your second graph, choose scales that make the line appear less steep than in the original graph. 4
y
x
–4
4
107
b. In either table, is d a linear function of t? If so, construct a linear equation relating d and t for that table. 25. Adding minerals or organic compounds to water lowers its freezing point. Antifreeze for car radiators contains glycol (an organic compound) for this purpose. The accompanying table shows the effect of salinity (dissolved salts) on the freezing point of water. Salinity is measured in the number of grams of salts dissolved in 1000 grams of water. So our units for salinity are in parts per thousand, abbreviated ppt. Is the relationship between the freezing point and salinity linear? If so, construct an equation that models the relationship. If not, explain why. Relationship between Salinity and Freezing Point
–4
22. The exchange rate that a bank gave for euros in October 2006 was 0.79 euros for $1 U.S. They also charged a constant fee of $5 per transaction. The bank’s exchange rate from euros to British pounds was 0.66 pounds for 1 euro, with a transaction fee of 4.1 euros. a. Write a general equation for how many euros you got when changing dollars. Use E for euros and D for dollars being exchanged. Draw a graph of E versus D. b. Would it have made any sense to exchange $10 for euros? c. Find a general expression for the percentage of the total euros converted from dollars that the bank kept for the transaction fee. d. Write a general equation for how many pounds you would get when changing euros. Use P for British pounds and E for the euros being exchanged. Draw a graph of P versus E. 23. (Graphing program optional.) Suppose that: • For 8 years of education, the mean annual earnings for women working full-time are approximately $19,190. • For 12 years of education, the mean annual earnings for women working full-time are approximately $31,190. • For 16 years of education, the mean annual earnings for women working full-time are approximately $43,190. a. Plot this information on a graph. b. What sort of relationship does this information suggest between earnings and education for women? Justify your answer. c. Generate an equation that could be used to model the data from the limited information given (letting E 5 years of education and M 5 mean earnings). Show your work. 24. a. Create a third column in Tables A and B, and insert values for the average rate of change. (The first entry will be “n.a.”.) t
d
t
d
0 1 2 3 4 5
400 370 340 310 280 250
0 1 2 3 4 5
1.2 2.1 3.2 4.1 5.2 6.1
Table A
Table B
Salinity (ppt)
Freezing Point (8C)
0 5 10 15 20 25
0.00 20.27 20.54 20.81 21.08 21.35
Source: Data adapted from P.R. Pinel, Oceanography: An Introduction to the Planet Oceanus (St. Paul, MN: West, 1992), p. 522.
26. The accompanying data show rounded average values for blood alcohol concentration (BAC) for people of different weights, according to how many drinks (5 oz wine, 1.25 oz 80-proof liquor, or 12 oz beer) they have consumed. Blood Alcohol Concentration for Selected Weights Number of Drinks
100 lb
140 lb
180 lb
2 4 6 8 10
0.075 0.150 0.225 0.300 0.375
0.054 0.107 0.161 0.214 0.268
0.042 0.083 0.125 0.167 0.208
a. Examine the data on BAC for a 100-pound person. Are the data linear? If so, find a formula to express blood alcohol concentration, A, as a function of the number of drinks, D, for a 100-pound person. b. Examine the data on BAC for a 140-pound person. Are the data linear? If they’re not precisely linear, what might be a reasonable estimate for the average rate of change of blood alcohol concentration, A, with respect to number of drinks, D? Find a formula to estimate blood alcohol concentration, A, as a function of number of drinks, D, for a 140-pound person. Can you make any general conclusions about BAC as a function of number of drinks for all of the weight categories? c. Examine the data on BAC for people who consume two drinks. Are the data linear? If so, find a formula to express blood alcohol concentration, A, as a function of weight, W, for people who consume two drinks. Can you make any general conclusions about BAC as a function of weight for any particular number of drinks?
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2.8 Special Cases Direct Proportionality The simplest relationship between two variables is when one variable is equal to a constant multiple of the other. For instance, in the previous example A 5 0.0235D; blood alcohol concentration A equals a constant, 0.0235, times D, the number of drinks. We say that A is directly proportional to D. How to recognize direct proportionality Linear functions of the form y 5 mx
sm 2 0d
describe a relationship where y is directly proportional to x. If two variables are directly proportional to each other, the graph will be a straight line that passes through the point (0, 0), the origin. Figure 2.40 shows the graphs of two relationships in which y is directly proportional to x, namely y 5 2x and y 5 2x.
y –x
4
y y 2x
x –4
4
(0, 0) origin
–4
Figure 2.40 Graphs of two
relationships in which y is directly proportional to x. Note that both graphs are lines that go through the origin.
?
Direct Proportionality In a linear equation of the form
SOMETHING TO THINK ABOUT
y 5 mx
If y is directly proportional to x, is x directly proportional to y?
(m ? 0)
we say that y is directly proportional to x. Its graph will go through the origin.
EXAMPLE
1
Braille is a code, based on six-dot cells, that allows blind people to read. One page of regular print translates into 2.5 Braille pages. Construct a function describing this relationship. Does it represent direct proportionality? What happens if the number of print pages doubles? Triples?
SOLUTION
If P 5 number of regular pages and B 5 number of Braille pages, then B 5 2.5P. So B is directly proportional to P. If the number of print pages doubles, the number of Braille pages doubles. If the number of print pages triples, the number of Braille pages will triple.
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EXAMPLE
2
SOLUTION
109
You are traveling to Canada and need to exchange American dollars for Canadian dollars. On that day the exchange rate is approximately 1 American dollar for 1.13 Canadian dollars. a. Construct an equation for converting American to Canadian dollars. Does it represent direct proportionality? b. Suppose the Exchange Bureau charges a $2 flat fee to change money. Alter your equation from part (a) to include the service fee. Does the new equation represent direct proportionality? a. If we let A 5 American dollars and C 5 Canadian dollars, then the equation C 5 1.13A describes the conversion from American (the input) to Canadian (the output). The amount of Canadian money you receive is directly proportional to the amount of American money you exchange. b. If there is a $2 service fee, you would have to subtract $2 from the American money you have before converting to Canadian. The new equation is C 5 1.13(A 2 2) or equivalently, C 5 1.13A 2 2.26 where 2.26 is the service fee in Canadian dollars. Then C is no longer directly proportional to A.
EXAMPLE
3
A prominent midwestern university decided to change its tuition cost. Previously there was a ceiling on tuition (which included fees). Currently the university uses what it calls a linear model, charging $106 per credit hour for in-state students and $369 per credit hour for out-of-state students. a. Is it correct to call this pricing scheme a linear model for in-state students? For outof-state students? Why? b. Generate equations and graphs for the cost of tuition for both in-state and out-ofstate students. If we limit costs to one semester during which the usual maximum credit hours is 15, what would be a reasonable domain? c. In each case is the tuition directly proportional to the number of credit hours?
SOLUTION
a. Yes, both relationships are linear since the rate of change is constant in each case: $106 per credit hour for in-state students and $369 per credit hour for out-of-state students.
Cost of tuition ($)
7000 6000 5000 4000
Out-of-state
3000 2000
In-state
1000 0
b. Let N 5 number of credit hours, Ci 5 cost for an in-state student, and Co 5 cost for an out-of-state student. In each case if the number of credit hours is zero sN 5 0d , then the cost would be zero sCi 5 0 5 Co d . Hence both lines would pass through the origin (0, 0), making the vertical intercept 0 for both equations. So the results would be of the form Ci 5 106N
0
and Co 5 369N
5 10 15 Number of credit hours
Figure 2.41 Tuition for in-
state and out-of-state students at a midwestern university.
which are graphed in Figure 2.41. A reasonable domain would be 0 # N # 15. c. In both cases the tuition is directly proportional to the number of credit hours. The graphs verify this since both are straight lines going through the origin.
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Algebra Aerobics 2.8a 1. Construct an equation and draw the graph of the line that passes through the origin and has the given slope. a. m 5 21 b. m 5 0.5 2. For each of Tables 2.18 and 2.19, determine whether x and y are directly proportional to each other. Represent each relationship with an equation. a. x b. y x y 22 21 0 1 2
6 3 0 23 26
Table 2.18
0 1 2 3 4
5.
6.
5 8 11 14 17
7.
Table 2.19
3. In September 2006 the exchange rate was $1.00 U.S. to 0.79 euros, the common European currency. a. Find a linear function that converts U.S. dollars to euros. b. Find a linear function that converts U.S. dollars to euros with a service fee of $2.50. c. Which function represents a directly proportional relationship and why? 4. Suppose you go on a road trip, driving at a constant speed of 60 miles per hour. Create an equation relating
8.
distance d in miles and time traveled t in hours. Does it represent direct proportionality? What happens to d if the value for t doubles? If t triples? The total cost C for football tickets is directly proportional to the number of tickets purchased, N. If two tickets cost $50, construct the formula relating C and N. What would the total cost of 10 tickets be? Write a formula to describe each situation. a. y is directly proportional to x, and y is 4 when x is 12. b. d is directly proportional to t, and d is 300 when t is 50. Write a formula to describe the following: a. The diameter, d, of a circle is directly proportional to the circumference, C. b. The amount of income tax paid, T, is directly proportional to income, I. c. The tip amount t, is directly proportional to the cost of the meal, c. Assume that a is directly proportional to b. When a 5 10, b 5 15. a. Find a if b is 6. b. Find b if a is 4.
Horizontal and Vertical Lines The slope, m, of any horizontal line is 0. So the general form for the equation of a horizontal line is y 5 b 1 0x y5b
or just
For example, Table 2.20 and Figure 2.42 show points that satisfy the equation of the horizontal line y 5 1. If we calculate the slope between any two points in the table—for example, s22, 1d and (2, 1)—we get slope 5
0 121 5 50 22 2 2 24 y 4
x
y
24 22 0 2 4
1 1 1 1 1
Table 2.20
b
y=1 x
–4
0
Figure 2.42 Graph of the horizontal line y 5 1.
4
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For a vertical line the slope, m, is undefined, so we can’t use the standard y 5 b 1 mx format. The graph of a vertical line (as in Figure 2.43) fails the vertical line test, so y is not a function of x. However, every point on a vertical line does have the same horizontal coordinate, which equals the coordinate of the horizontal intercept. Therefore, the general equation for a vertical line is of the form x 5 c where c is a constant (the horizontal intercept) For example, Table 2.21 and Figure 2.43 show points that satisfy the equation of the vertical line x 5 1. 4
y
x=1
x
y
1 1 1 1 1
24 22 0 2 4
x
c
–4
4
–4
Table 2.21
Figure 2.43 Graph of the vertical
line x 5 1.
Note that if we tried to calculate the slope between two points, say s1, 24d and (1, 2), on the vertical line x 5 1 we would get slope 5
24 2 2 26 5 121 0
which is undefined.
The general equation of a horizontal line is y5b where b is a constant (the vertical intercept) and the slope is 0. The general equation of a vertical line is x5c where c is a constant (the horizontal intercept) and the slope is undefined.
EXAMPLE
4
Find the equation for each line in Figure 2.44.
y
y
5
–5
y
5
5
x
–5
–5 (a)
Figure 2.44 Four linear graphs.
5
–5 (b)
y
5
x
–5
5
5
–5 (c)
x
–5
5
–5 (d)
x
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SOLUTION
a. b. c. d.
y 5 23, a horizontal line x 5 3, a vertical line y 5 3x, a direct proportion, slope 5 3 y 5 22x 1 3, a line with slope 22 and y-intercept 3
Parallel and Perpendicular Lines Parallel lines have the same slope. So if the two equations y 5 b1 1 m1x and y 5 b2 1 m2x describe two parallel lines, then m1 5 m2. For example, the two lines y 5 2.0 2 0.5x and y 5 21.0 2 0.5x each have a slope of 20.5 and thus are parallel (see Figure 2.45).
?
SOMETHING TO THINK ABOUT 4
Describe the equation for any line perpendicular to the horizontal line y 5 b.
y
y
10
y = 3 – 2x y = 2.0 – 0.5x
x –4
4
x –10
10
y = –2 + (1/2)x y = –1.0 – 0.5x –4
Figure 2.45 Two parallel lines have the same slope.
–10
Figure 2.46 Two perpendicular
lines have slopes that are negative reciprocals.
Two lines are perpendicular if their slopes are negative reciprocals. If y 5 b1 1 m1x and y 5 b2 1 m2x describe two perpendicular lines, then m1 5 21/m2. For example, in Figure 2.46 the two lines y 5 3 2 2x and y 5 22 1 12 x have slopes of 22 and 12 , 1 respectively. Since 22 is the negative reciprocal of 12 (i.e., 2 1 5 21 4 12 5 s2 d 21 ? 21 5 22), the two lines are perpendicular. Why does this relationship hold for perpendicular lines? Consider a line whose slope is given by v/h. Now imagine rotating the line 90 degrees clockwise to y generate a second line perpendicular to the first 4 (Figure 2.47). What would the slope of this new Original line line be? The positive vertical change, v, becomes a 3 v positive horizontal change. The positive h horizontal change, h, becomes a negative vertical 2 –h change. The slope of the original line is v/h, and v the slope of the line rotated 90 degrees clockwise Rotated 90° is 2h/v. Note that 2h/v 5 21/sv/hd , which is 1 the original slope inverted and multiplied by 21. In general, the slope of a perpendicular line is 0 x 0 1 2 3 4 the negative reciprocal of the slope of the original line. If the slope of a line is m1, then the slope, m2, Figure 2.47 Perpendicular lines m 2 5 21/m 1. of a line perpendicular to it is 21/m1.
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This is true for any pair of perpendicular lines for which slopes exist. It does not work for horizontal and vertical lines since vertical lines have undefined slopes. Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals of each other.
EXAMPLE
5
SOLUTION
Determine from neither. a. y 5 2 1 7x b. y 5 6 2 x c. y 5 5 1 3x d. y 5 3x 1 13
the equations which pairs of lines are parallel, perpendicular, or and y 5 7x 1 3 and y 5 6 1 x and y 5 5 2 3x and 3y 1 x 5 2
a. The two lines are parallel since they share the same slope, 7. b. The two lines are perpendicular since the negative reciprocal of 21 (the slope of first line) equals 2s1/s21dd 5 2s21d 5 1, the slope of the second line. c. The lines are neither parallel nor perpendicular. d. The lines are perpendicular. The slope of the first line is 3. If we solve the second equation for y, we get 3y 1 x 5 2 3y 5 2 2 x y 5 2/3 2 (1/3)x So the slope of the second line is 2s1/3d , the negative reciprocal of 3.
Algebra Aerobics 2.8b 1. In each case, find an equation for the horizontal line that passes through the given point. a. s3, 25d b. s5, 23d c. s23, 5d 2. In each case, find an equation for the vertical line that passes through the given point. a. s3, 25d b. s5, 23d c. s23, 5d 3. Construct the equation of the line that passes through the points. a. s0, 27d , s3, 27d , and s350, 27d b. s24.3, 0d , s24.3, 8d and s24.3, 21000d 4. Find the equation of the line that is parallel to y 5 4 2 x and that passes through the origin. 5. Find the equation of the line that is parallel to W 5 360C 1 2500 and passes through the point where C 5 4 and W 5 1000. 6. Find the slope of a line perpendicular to each of the following. a. y 5 4 2 3x c. y 5 3.1x 2 5.8 b. y 5 x d. y 5 235 x 1 1
7. a. Find an equation for the line that is perpendicular to y 5 2x 2 4 and passes through s3, 25d . b. Find the equations of two other lines that are perpendicular to y 5 2x 2 4 but do not pass through the point s3, 25d . c. How do the three lines from parts (a) and (b) that are perpendicular to y 5 2x 2 4 relate to each other? d. Check your answers by graphing the equations. 8. Find the slope of the line Ax 1 By 5 C assuming that y is a function of x. (Hint: Solve the equation for y.) 9. Use the result of the previous exercise to determine the slope of each line described by the following linear equations (again assuming y is a function of x). a. 2x 1 3y 5 5 d. x 5 25 b. 3x 2 4y 5 12 e. x 2 3y 5 5 c. 2x 2 y 5 4 f. y 5 4 10. Solve the equation 2x 1 3y 5 5 for y, identify the slope, then find an equation for the line that is parallel to the line 2x 1 3y 5 5 and passes through the point (0, 4).
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11. Solve the equation 3x 1 4y 5 27 for y, identify the slope, then find an equation for the line that is perpendicular to the line of 3x 1 4y 5 27 and passes through (0, 3). 12. Solve the equation 4x 2 y 5 6 for y, identify the slope, then find an equation for the line that is perpendicular to the line 4x 2 y 5 6 and passes through s2, 23d . 13. Determine whether each equation could represent a vertical line, a horizontal line, or neither. a. x 1 1.5 5 0 b. 2x 1 3y 5 0 c. y 2 5 5 0
14. Write an equation for the line perpendicular to 2x 1 3y 5 6: a. That has vertical intercept of 5 b. That passes through the point (26, 1) 15. Write an equation for the line parallel to 2x 2 y 5 7: a. That has a vertical intercept of 9 b. That passes through (4, 3)
Piecewise Linear Functions Some functions are not linear throughout but are made up of linear segments. They are called piecewise linear functions. For example, we could define a function f(x) where: for x # 1 for x . 1
or, more compactly,
f(x) 5
2 1 x for x # 1 3 for x . 1
3
f(x) 5 2 1 x f(x) 5 3
The graph of f(x) in Figure 2.48 clearly shows the two distinct linear segments. y 4
x –5
5
–4
Figure 2.48 Graph of a
piecewise linear function. EXAMPLE
6
Consider the amount of gas in your car during a road trip. You start out with 20 gallons and drive for 3 hours, leaving you with 14 gallons in the tank. You stop for lunch for an hour and then drive for 4 more hours, leaving you with 6 gallons. a. Construct a piecewise linear function for the amount of gas in the tank as a function of time in hours. b. Graph the results.
SOLUTION
a. Let t 5 time (in hours). For 0 # t , 3, the average rate of change in gasoline over time is (14 –20) gallons/(3 hr) 5 (26 gallons)/(3 hr) 5 22 gallons/hr; that is, you are consuming 2 gallons per hour. The initial amount of gas is 20 gallons, so G(t), the amount of gas in the tank at time t, is given by G(t) 5 20 2 2t for 0 # t , 3 While you are at lunch for an hour, you consume no gasoline, so the amount of gas stays constant at 14 gallons. So G(t) 5 14
for 3 # t , 4
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At the end of lunch, t 5 4 and G(t) 5 14. You continue to drive for 4 more hours, ending up with 6 gallons. You are still consuming 2 gallons per hour since (6–14) gallons/4 hr 5 22 gallons/hr. So your equation will be of the form G(t) 5 b 2 2t. Substituting in t 5 4 and G(t) 5 14, we have 14 5 b 2 2 ? 4 ➯ b 5 22. So G(t) 5 22 2 2t for 4 # t , 8 Writing G(t) more compactly, we have 20 2 2tif 0 # t , 3 if 3 # t , 4 Gstd 5 • 14 22 2 2tif 4 # t , 8 b. The graph of G(t) is shown in Figure 2.49.
Gasoline (in gallons)
25 20 15 10 5 0
t 0
2
4 Time (in hours)
6
8
Figure 2.49 Number of gallons left in the
car’s tank. The absolute value function The absolute value of x, written as uxu, strips x of its sign. That means we consider only the magnitude of x, so uxu is never negative.1 • If x is positive (or 0), then uxu 5 x. • If x is negative, then uxu 5 2x. For example, u25u 5 2 (25) 5 5. We can construct the absolute value function using piecewise notation.
3
The absolute value function x for x $ 0 If f(x) 5 ux u, then f(x) 5 2x for x , 0
EXAMPLE
7
If f(x) 5 uxu, then: a. What is f(6)? f(0)? f(26)? b. Graph the function f(x) for x between 26 and 6. c. What is the slope of the line segment when x . 0? When x , 0?
1 Graphing calculators and spreadsheet programs usually have an absolute value function. It is often named abs where abs sxd 5 u x u . So abss23d 5 u 3 u 5 3.
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SOLUTION
a. f(6) 5 6; f(0) 5 0; f(26) 5 u26u 5 6 b. See Figure 2.50. 6
3
–6
–3
0
x 6
3
Figure 2.50 Graph of the absolute value function f(x) 5 uxu.
c. When x . 0, the slope is 1; when x , 0, the slope is 21. Absolute value inequalities (in one variable) are frequently used in describing an allowable range above or below a certain amount. EXAMPLE
8
Range in values of poll results Poll figures are often given with a margin of error. For example, in January 2007 a CNN poll said that 63% of Americans felt that the economy was in good condition, with a sample error of 63 points. Construct an absolute value inequality that describes the range of percentages P that are possible within this poll. Restate this condition without using absolute values, and display it on a number line.
SOLUTION
uP 2 63u # 3; that is, the poll takers are confident that the difference between the estimated percentage, 63%, and the actual percentage, P, is less than or equal to 3%. Equivalently we could write 60 # P # 66; that is, the actual percentage P is somewhere between 60% and 66% (see Figure 2.51). –3% 60%
+3% 63%
66%
Figure 2.51 Range of error around 63%
is 63 percentage points. Absolute value functions are useful in describing distances between objects. EXAMPLE
9
Distance between a cell phone and cell tower You are a passenger in a car, talking on a cell phone. The car is traveling at 60 mph along a straight highway and the nearest cell phone tower is 6 miles away. a. How long will it take you to reach the cell phone tower (assuming it is right by the road)? b. Construct a linear function D(t) that describes your distance to the cell tower (in miles $0) or from the cell tower (in miles , 0), where t is the number of hours traveled. c. Graph your function using a reasonable domain for t. d. What does uD(t)u represent? Graph uD(t)u on a separate grid and compare it with the graph of D(t).
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SOLUTION
117
a. Traveling at 60 miles per hour is equivalent traveling at 1 mile per minute. So traveling 6 miles from the start will take you 6 minutes or 0.1 hr to reach the cell tower. b. At the starting time t 5 0 hours, the distance to the nearest cell phone tower is 6 miles, so D(0) 5 6 miles. Thus, the vertical intercept is at (0, 6). After t 5 0.1 hr you are at the cell tower, so the distance between you and the cell tower is 0. Thus, D(0.1) 5 0 miles and hence the horizontal intercept is (0.1, 0). The slope of the line is (0 – 6)/(0.1 – 0) 5 260. So the distance function is D(t) 5 6 2 60t, where t is in hours and D(t) is in miles (to or from the cell tower). c. A reasonable domain might be 0 # t # 0.2 hours. See Figure 2.52.
Distance (miles)
Distance to, and Then from Cell Tower 8 6 4 2 0 –2 –4 –6 –8
Passing cell tower
t 0.2
0.1
Time (hours)
Figure 2.52 Graph of D(t) 5 6 2 60t.
d. uD(t)u describes the absolute value of the distance between you and the cell tower, indicating that the direction of travel no longer matters. Whether you are driving toward or away from the tower, the absolute value of the distance is always positive (or 0). For example, uD(0.2)u 5 u6 2 60 ? 0.2u 5 u6 2 12u 5 u26u 5 6 miles, which means that after 0.2 hours (or 12 minutes) you are 6 miles away from the tower. See Figure 2.53.
Distance (miles)
Distance Between Car and Cell Tower 8 6 4 2 0 –2 –4 –6
Passing cell tower
t 0.2
0.1
Time (hours)
Figure 2.53 Graph of uD(t)u 5 u5 2 25tu.
Step functions Some piecewise linear functions are called step functions because their graphs look like the steps of a staircase. Each “step” is part of a horizontal line. EXAMPLE
10
A step function: Minimum wages The federal government establishes a national minimum wage per hour. Table 2.22 shows the value of the minimum wage over the years 1990 to 2007. Federal Minimum Wage for 1990–2007 Year New Minimum Wage Set
Minimum Wage (per hour)
1990 1991 1996 1997 2007
$3.80 $4.25 $4.75 $5.15 $5.85
Table 2.22
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a. Construct a step function M(x), where M(x) is the minimum wage at year x. What is the domain? b. What is M(1992)? M(2006)? c. Graph the step function. d. Why do you think there was a lot of discussion in 2006 about raising the minimum wage? (Note: Individual states can set a higher minimum for their workers.) SOLUTION
3.80for 1990 # x , 1991 4.25for 1991 # x , 1996 a. M(x) 5 e 4.75for 1996 # x , 1997 5.15for 1997 # x , 2007 5.85for 2007 # x , 2008 The domain is 1990 # x , 2008. b. M(1992) 5 $4.25/hr; M(2006) 5 $5.15/hr c. See Figure 2.54. Federal Minimum Wage 7
Minimum ($/hr)
6 5 4 3 2 1 0 1990
1995
2000 Year
2005
2010
Figure 2.54 Step function for federal minimum wage between 1990
and 2007. d. By the end of 2006 the federal minimum wage had stayed the same ($5.15/hr) for 9 years. Inflation always erodes the purchasing power of the dollar, so $5.15 in 2006 bought a lot less than $5.15 in 1997. So many felt an increase in the minimum wage was way overdue.
Algebra Aerobics 2.8c 1. Construct the graphs of the following piecewise linear functions. (Be sure to indicate whether each endpoint is included on or excluded from the graph.) 1for 0 , x # 1 a. f sxd 5 • 0for 1 , x # 2 21for 2 , x # 3 x 1 3for 24 # x , 0 b. g sxd 5 e 2 2 xfor 0 # x # 4 2. Construct piecewise linear functions Q(t) and C(r) to describe the graphs in Figures 2.55 and 2.56.
3
t 5
–5
–7
Figure 2.55 Graph of Q(t).
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6 4 2
r –4
–2
2
4
–2
–4
Figure 2.56 Graph of C(r).
119
from home to a coffee shop, stopped for a cup, and then walked home at a faster pace. b. For the scenario in part (a), sketch a piecewise linear graph that shows the distance between you and the coffee shop over time. 8. The federal funds rate is the short-term interest rate charged by the Federal Reserve for overnight loans to other federal banks. This is one of the major tools the Federal Reserve Board uses to stimulate the economy (with a rate decrease) or control inflation (with a rate increase). Table 2.23 shows the week of each rate change during 2006. Federal Funds Rates During 2006
3. Evaluate the following: a. u22u c. u3 25u b. u6u
e. 2u3u ? u25u
d. u3u 2 u5u
4. Given the function g(x) 5 ux 2 3u: a. What is g(23)? g(0)? g(3)? g(6)? b. Sketch the graph of g(x) 5 ux 2 3u for 26 # x # 6. c. Compare the graph of g(x) 5 ux 2 3u with the graph of f(x) 5 uxu. d. Write g(x) using piecewise linear notation. 5. Rewrite the following expressions without using an absolute value sign, and then describe in words the result. a. ut 25u # 2 b. uQ 2 75u , 6 6. The optimal water temperature for trout is 558F, but they can survive water temperatures that are 208 above or below that. Write an absolute value inequality that describes the temperature values, T, that lie within the trout survival temperature range. Then write an equivalent expression without the absolute value sign. 7. a. Sketch a piecewise linear graph of the total distance you would travel if you walked at a constant speed
Week in 2006 When Rate Was Changed
Federal Funds Rate (%)
Week 1 Week 5 Week 13 Week 19 Week 26
4.25 4.50 4.75 5.00 5.25
Table 2.23
a. What was the federal funds rate in week 4? In week 52? What was the longest period in 2006 over which the federal funds rate remained the same? b. What appears to be the typical percentage increase used by the Board? Do the increases occur at regular intervals? c. During 2006 was the Federal Reserve Board more concerned about stimulating economic growth or curbing inflation? d. Construct a piecewise linear function to describe the federal funds rate during 2006. e. Sketch a graph of your function.
Exercises for Section 2.8 1. Using the general formula y 5 mx that describes direct proportionality, find the value of m if: a. y is directly proportional to x and y 5 2 when x 5 10. b. y is directly proportional to x and y 5 0.1 when x 5 0.2. c. y is directly proportional to x and y 5 1 when x 5 14. 2. For each part, construct an equation and then use it to solve the problem. a. Pressure P is directly proportional to temperature T, and P is 20 lb per square inch when T is 60 degrees Kelvin. What is the pressure when the temperature is 80 degrees Kelvin? b. Earnings E are directly proportional to the time T worked, and E is $46 when T is 2 hours. How long has a person worked if she earned $471.50?
c. The number of centimeters of water depth W produced by melting snow is directly proportional to the number of centimeters of snow depth S. If W is 15.9 cm when S is 150 cm, then how many centimeters of water depth are produced by a 100-cm depth of melting snow? 3. In the accompanying table y is directly proportional to x. Number of CDs purchased (x) Cost of CDs (y)
3 42.69
4 56.92
5
a. Find the formula relating y and x, then determine the missing value in the table. b. Interpret the coefficient of x in this situation.
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11. The accompanying figure shows the quantity of books (in millions) shipped by publishers in the United States between 2001 and 2005. Construct the equation of a horizontal line that would be a reasonable model for these data. Number of Books Shipped by Publishers 5000
5. For each of the following linear functions, determine the independent and dependent variables and then construct an equation for each function. a. Sales tax is 6.5% of the purchase price. b. The height of a tree is directly proportional to the amount of sunlight it receives. c. The average salary for full-time employees of American domestic industries has been growing at an annual rate of $1300/year since 1985, when the average salary was $25,000.
Number of books (in millions)
4. The electrical resistance R (in ohms) of a wire is directly proportional to its length l (in feet). a. If 250 feet of wire has a resistance of 1.2 ohms, find the resistance for 150 ft of wire. b. Interpret the coefficient of l in this context.
6. On the scale of a map 1 inch represents a distance of 35 miles. a. What is the distance between two places that are 4.5 inches apart on the map? b. Construct an equation that converts inches on the map to miles in the real world.
Source: U.S. Bureau of the Census. Statistical Abstract of the United States, 2006.
7. Find a function that represents the relationship between distance, d, and time, t, of a moving object using the data in the accompanying table. Is d directly proportional to t? Which is a more likely choice for the object, a person jogging or a moving car? t (hours)
d (miles)
0 1 2 3 4
0 5 10 15 20
x
w
y
z
0 1 2 3 4
1 2 5 10 17
0.0 2.5 5.0 7.5 10.0
0 213 223 21 243
9. Find the slope of the line through the pair of points, then determine the equation. a. (2, 3) and (5, 3) c. (23, 8) and (23, 4) b. (24, 27) and (12, 27) d. (2, 23) and (2, 21) 10. Describe the graphs of the following equations. a. y 5 22 c. x 5 25 e. y 5 324 d. y 5 4x
f. y 5 23
3000 2000 1000 0 2001
2002
2003
2004
2005
Year
12. An employee for an aeronautical corporation had a starting salary of $25,000/year. After working there for 10 years and not receiving any raises, he decides to seek employment elsewhere. Graph the employee’s salary as a function of time for the time he was employed with this corporation. What is the domain? What is the range? 13. For each of the given points write equations for three lines that all pass through the point such that one of the three lines is horizontal, one is vertical, and one has slope 2. a. (1, 24) b. (2, 0) 14. Consider the function ƒ(x) 5 4. a. What is ƒ(0)? ƒ(30)? ƒ(212.6)? b. Describe the graph of this function. c. Describe the slope of this function’s graph.
8. Determine which (if any) of the following variables (w, y, or z) is directly proportional to x:
b. x 5 22
4000
15. A football player who weighs 175 pounds is instructed at the end of spring training that he has to put on 30 pounds before reporting for fall training. a. If fall training begins 3 months later, at what (monthly) rate must he gain weight? b. Suppose that he eats a lot and takes several nutritional supplements to gain weight, but due to his metabolism he still weighs 175 pounds throughout the summer and at the beginning of fall training. Sketch a graph of his weight versus time for those 3 months. 16. a. Write an equation for the line parallel to y 5 2 1 4x that passes through the point (3, 7). b. Find an equation for the line perpendicular to y 5 2 1 4x that passes through the point (3, 7). 17. a. Write an equation for the line parallel to y 5 4 2 x that passes through the point (3, 7). b. Find an equation for the line perpendicular to y 5 4 2 x that passes through the point (3, 7). 18. Construct the equation of a line that goes through the origin and is parallel to the graph of given equation. a. y 5 6 b. x 5 23 c. y 5 2x 1 3
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19. Construct the equation of a line that goes through the origin and is perpendicular to the given equation. a. y 5 6 b. x 5 23 c. y 5 2x 1 3 20. Which lines are parallel to each other? Which lines are perpendicular to each other? a. y 5 13 x 1 2 c. y 5 22x 1 10 e. 2y 1 4x 5 212 b. y 5 3x 2 4 d. y 5 23x 2 2 f. y 2 3x 5 7 21. Because different scales may be used on the horizontal and vertical axes, it is often difficult to tell if two lines are perpendicular to each other. In parts (a) and (b), determine the equations of each pair of lines and show whether or not the paired lines are perpendicular to each other. a. b. 24
y
12
–5
5
x
y
–5
5
–16
x
–8
24. Find the equation of the line in the form y 5 mx 1 b for each of the following sets of conditions. Show your work. a. Slope is $1400/year and line passes through the point (10 yr, $12,000). b. Line is parallel to 2y 2 7x 5 y 1 4 and passes through the point (21, 2). c. Equation is 1.48x 2 2.00y 1 4.36 5 0. d. Line is horizontal and passes through (1.0, 7.2). e. Line is vertical and passes through (275, 1029). f. Line is perpendicular to y 5 22x 1 7 and passes through (5, 2). 25. In the equation Ax 1 By 5 C: a. Solve for y so as to rewrite the equation in the form y 5 mx 1 b. b. Identify the slope. c. What is the slope of any line parallel to Ax 1 By 5 C? d. What is the slope of any line perpendicular to Ax 1 By 5 C? 26. Use the results of Exercise 25, parts (c) and (d), to find the slope of any line that is parallel and then one that is perpendicular to the given lines. a. 5x 1 8y 5 37 b. 7x 1 16y 5 214 c. 30x 1 47y 5 0 27. Construct the graphs of the following piecewise linear functions. Be sure to indicate whether an endpoint is included in or excluded from the graph. 2for 23 , x # 0 a. f sxd 5 e 1for0 , x # 3 x 1 3for 24 , x # 0 b. g sxd 5 e 2 2 xfor0 , x # 4
22. In each part construct the equations of two lines that: a. Are parallel to each other b. Intersect at the same point on the y-axis c. Both go through the origin d. Are perpendicular to each other 23. For each of the accompanying graphs you don’t need to do any calculations or determine the actual equations. Rather, using just the graphs, determine if the slopes for each pair of lines are the same. Are the slopes both positive or both negative, or is one negative and one positive? Do the lines have the same y-intercept?
28. Construct piecewise linear functions for the following graphs. a. Graph of f(x) y 6 5 4 3
4
y
y
2
4
1 0
x
–4
4
x 0
1
2
3
4
5
6
x
–4
4
b. Graph of g(x) y
–4
5
–4 Graph C
Graph A 4
y
y 4
x x
x
–4
4
–5
5
4
–4 Graph B
–4 Graph D
121
–5
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29. Given the following graph of g(x): a. Construct a piecewise linear description of g(x). b. Construct another description using absolute values. c. Describe the relationship between this graph and the graph of f (x) 5 uxu.
d. What is the distance between them one hour before they meet? An hour after they meet? Interpret both values in context. (Hint: If they are traveling toward each other, the distance between them is considered positive. Once they have met and are traveling away from each other, the distance between them is considered negative.) e. Now construct an absolute value function that describes the (positive) distance between A and B at any point, and graph your result for 0 # t # 8 hours.
9
6
3
x –3
–6
0
3
6
30. a. Normal human body temperature is often cited as 98.68F. However, any temperature that is within 18F more or less than that is still considered normal. Construct an absolute value inequality that describes normal body temperatures T that lie within that range. Then rewrite the expression without the absolute value sign. b. The speed limit is set at 65 mph on a highway, but police do not normally ticket you if you go less than 5 miles above or below that limit. Construct an absolute value inequality that describes the speeds S at which you can safely travel without getting a ticket. Rewrite the expression without using the absolute value sign. 31. Assume two individuals, A and B, are traveling by car and initially are 400 miles apart. They travel toward each other, pass and then continue on. a. If A is traveling at 60 miles per hour, and B is traveling at 40 mph, write two functions, dA(t) and dB(t), that describe the distance (in miles) that A and B each has traveled over time t (in hours). b. Now construct a function for the distance DAB (t) between A and B at time t (in hours). Graph the function for 0 # t # 8 hours. Starting point for A
Starting point for B
400 miles dA
DAB
dB
c. At what time will A and B cross paths? At that point, how many miles has each traveled?
32. The greatest integer function y 5 [x] is defined as the greatest integer # x (i.e., it rounds x down to the nearest integer below it). a. What is [2] ? [2.5] ? [2.9999999] ? b. Sketch a graph of the greatest integer function for 0 # x , 5. Be sure to indicate whether each endpoint is included or excluded. [Note: A bank employee embezzled hundreds of thousands of dollars by inserting software to round down transactions (such as generating interest on an account) to the nearest cent, and siphoning the round-off differences into his account. He was eventually caught.] 33. The following table shows U.S. first-class stamp prices (per ounce) over time. Year
Price for First-Class Stamp
2001 2002 2006
34 cents 37 cents 39 cents
a. Construct a step function describing stamp prices for 2001–2006. b. Graph the function. Be sure to specify whether each of the endpoints is included or excluded. c. In 2007 the price of a first-class stamp was raised to 41 cents. How would the function domain and the graph change? 34. Sketch a graph for each of the following situations. a. The amount in your savings account over a month, where you direct-deposit your paycheck each week and make one withdrawal during the month. b. The amount of money in an ATM machine over one day, where the ATM is stocked with dollars at the beginning of the day, and then ATM withdrawals of various sizes are made.
2.9 Constructing Linear Models of Data According to Edward Tufte in Data Analysis of Politics and Policy, “Fitting lines to relationships is the major tool of data analysis.” Of course, when we work with actual data searching for an underlying linear relationship, the data points will rarely fall exactly in a straight line. However, we can model the trends in the data with a linear equation. Linear relationships are of particular importance not because most relationships are linear, but because straight lines are easily drawn and analyzed. A human can fit a
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123
straight line by eye to a scatter plot almost as well as a computer. This paramount convenience of linear equations as well as their relative ease of manipulation and interpretation means that lines are often used as first approximations to patterns in data.
Fitting a Line to Data: The Kalama Study
KALAMA
Children’s heights were measured monthly over several years as a part of a study of nutrition in developing countries. Table 2.24 and Figure 2.57 show data collected on the mean heights of 161 children in Kalama, Egypt. Mean Heights of Kalama Children 84.0
Age (months)
Height (cm)
18 19 20 21 22 23 24 25 26 27 28 29
76.1 77.0 78.1 78.2 78.8 79.7 79.9 81.1 81.2 81.8 82.8 83.5
83.0 82.0 Height (cm)
DATA
81.0 80.0 79.0 78.0 77.0 76.0 18
Table 2.24 Source: D. S. Moore and G. P. McCabe, Introduction to the Practice of Statistics. Copyright © 1989 by W. H. Freeman and Company. Used with permission.
20
22 24 26 Age (months)
28
30
Figure 2.57 Mean heights of children in Kalama, Egypt.
Sketching a line through the data Although the data points do not lie exactly on a straight line, the overall pattern seems clearly linear. Rather than generating a line through two of the data points, try eyeballing a line that approximates all the points. A ruler or a piece of black thread laid down through the dots will give you a pretty accurate fit. In the Extended Exploration on education and earnings following this chapter, we will use technology to find a “best-fit” line. Figure 2.58 on the next page shows a line sketched that approximates the data points. This line does not necessarily pass through any of the original points. Finding the slope Estimating the coordinates of two points on the line, say (20, 77.5) and (26, 81.5), we can calculate the slope, m, or rate of change, as s81.5 2 77.5d cm s26 2 20d months 4.0 cm 5 6 months < 0.67 cm/month
m5
So our model predicts that for each additional month an “average” Kalama child will grow about 0.67 centimeter.
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84.0 83.0 82.0 (26, 81.5) Height (cm)
124
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81.0 80.0 79.0 78.0 (20, 77.5) 77.0 76.0 18
20
22 24 26 Age (months)
28
30
Figure 2.58 Estimated coordinates of two points on the line.
Constructing the equation Since the slope of our linear model is 0.67 cm/month, then our equation is of the form H 5 b 1 0.67A
(1)
where A 5 age in months and H 5 mean height in centimeters. How can we find b, the vertical intercept? We have to resist the temptation to estimate b directly from the graph. As is frequently the case in social science graphs, both the horizontal and the vertical axes are cropped. Because the horizontal axis is cropped, we can’t read the vertical intercept off the graph. We’ll have to calculate it. Since the line passes through (20, 77.5) we can substitute (20, 77.5) in Equation (1) simplify solve for b
77.5 5 b 1 (0.67)(20) 77.5 5 b 1 13.4 b 5 64.1
Having found b, we complete the linear model: H 5 64.1 1 0.67A where A 5 age in months and H 5 height in centimeters. It offers a compact summary of the data. What is the domain of this model? In other words, for what inputs does our model apply? The data were collected on children age 18 to 29 months. We don’t know its predictive value outside these ages, so the domain consists of all values of A for which 18 # A # 29 The vertical intercept may not be in the domain Although the H-intercept is necessary to write the equation for the line, it lies outside of the domain. Compare Figure 2.58 with Figure 2.59. They show graphs of the same equation, H 5 64.1 1 0.67A. In Figure 2.58 both axes are cropped, while Figure 2.59 includes the origin (0, 0). In Figure 2.59 the vertical intercept is visible, and the shaded area between the dotted lines indicates the region that applies to our model. So a word of warning when reading graphs: Always look carefully to see if the axes have been cropped.
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125
100.0
Height (cm)
64.1, the H–intercept
50.0
0.00
0
10
20 30 Age (months)
40
Figure 2.59 Graph of H 5 64.1 1 0.67A
that includes the origin (0, 0). Shaded area shows the region that models the Kalama data.
Reinitializing the Independent Variable When we model real data, it often makes sense to reinitialize the independent variable in order to have a reasonable vertical intercept. This is especially true for time series, as shown in the following example, where the independent variable is the year. EXAMPLE
1
SOLUTION
Time series How can we find an equation that models the trend in smoking in the United States? The American Lung Association website provided the data reproduced in Table 2.25 and graphed in Figure 2.60. Although in some states smoking has increased, the overall trend is a steady decline in the percentage of adult smokers in the United States between 1965 and 2005.
45.0 40.0 35.0
1965 1974 1979 1983 1985 1990 1995 2000 2005
41.9 37.0 33.3 31.9 29.9 25.3 24.6 23.1 20.9
Table 2.25
30.0 Percentage
Year
Percentage of Adults Who Smoke
25.0 20.0 15.0 10.0 5.0 1965
1975
1985 Year
1995
2005
Figure 2.60 Percentage of adults who smoke.
The relationship appears reasonably linear. So the equation of a best-fit line could provide a fairly accurate description of the data. Since the horizontal axis is cropped, starting at the year 1965, the real vertical intercept would occur 1965 units to the left,
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at 0 A.D.! If you drew a big enough graph, you’d find that the vertical intercept would occur at approximately (0, 1220). This nonsensical extension of the model outside its known values would say that in 0 A.D., 1220% of the adult population smoked. A better strategy would be to define the independent variable as the number of years since 1965. Table 2.26 shows the reinitialized values for the independent variable, and Figure 2.61 gives a sketched-in best-fit line. P 45.0
No. of % of Years Adults Year Since 1965 Who Smoke 0 9 14 18 20 25 30 35 40
Table 2.26
41.9 37.0 33.3 31.9 29.9 25.3 24.6 23.1 20.9
35.0 Percentage
1965 1974 1979 1983 1985 1990 1995 2000 2005
(0, 42)
40.0
30.0
(25, 27)
25.0 20.0 15.0 10.0 5.0 0
0
10
20 Years since 1965
30
N 40
Figure 2.61 Percentage of adult smokers since 1965 with estimated
best-fit line.
We can estimate the coordinates of two points, (0, 42) and (25, 27), on our best-fit line. Using them, we have vertical intercept 5 42
and slope 5
42 2 27 15 5 5 20.6 0 2 25 225
If we let N 5 the number of years since 1965 and P 5 percentage of adult smokers, then the equation for our best-fit line is P 5 42 2 0.6N where the domain is 0 # N # 35 (see Figure 2.61). This model says that starting in 1965, when about 42% of U.S. adults smoked, the percentage of the adult smokers has declined on average by 0.6 percentage points a year for 35 years. What this model doesn’t tell us is that (according to the U.S. Bureau of the Census) the total number of smokers during this time has remained fairly constant, at 50 million.
Interpolation and Extrapolation: Making Predictions We can use this linear model on smokers to make predictions. We can interpolate or estimate new values between known ones. For example, in our smoking example we have no data for the year 1970. Using our equation we can estimate that in 1970 (when N 5 5), P 5 42 2 (0.6 ? 5) 5 39% of adults smoked. Like any other point on the best-fit line, this prediction is only an estimate and may, of course, be different from the actual percentage of smokers (see Figure 2.62). We can also use our model to extrapolate or to predict beyond known values. For example, our model predicts that in 2010 (when N 5 45), P 5 42 2 (0.6 ? 45) 5 15% of adults will smoke. Extrapolation much beyond known values is risky. For 2035 (where N 5 70) our model predicts that 0% will smoke, which seems unlikely. After 2035 our model would give the impossible answer that a negative percentage of adults will smoke.
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50.0
127
P Interpolated point at (5, 39)
40.0
Percentage
30.0 20.0 Extrapolated point at (45, 15) 10.0 0
20
40
60
80
N
–10.0 –20.0
Years since 1965
Figure 2.62 Interpolation and extrapolation of percentage
of smokers.
Algebra Aerobics 2.9 1. Figure 2.63 shows the total number of U.S. college graduates (age 25 or older) between 1960 and 2005. Number of College Graduates 30
2. Figure 2.64 shows the percentage of adults who (according to the U.S. Census Bureau) had access to the Internet either at home or at work between 1997 and 2003. Percent of U.S. Adults with Internet Access 100%
20
Percent with access
Number (millions)
25
15 10 5 0 1960
1970
1980
1990
2000
2010
Year
Figure 2.63 Total number of U.S. college graduates
80% 60% 40% 20% 0% 1996
1998
2000 Year
2002
2004
Figure 2.64 U.S. adults with Internet access.
over time. Source: U.S. National Center for Education Statistics, Digest of Education Statistics, annual.
a. Estimate from the scatter plot the number of college graduates in 1960 and in 2005. b. Since the data look fairly linear, sketch a line that would model the growth in U.S. college graduates. c. Estimate the coordinates of two points on your line and use them to calculate the slope. d. If x 5 number of years since 1960 and y 5 total number (in millions) of U.S. college graduates, what would the coordinates of your two points in part (c) be in terms of x and y? e. Construct a linear equation using the x and y values defined in part (d). f. What does your model tell you about the number of college graduates in the United States?
a. Since the data appear roughly linear, sketch a bestfit line. (This line need not pass through any of the three data points.) b. Reinitialize the years so that 1996 becomes year 0. Then identify the coordinates of any two points that lie on the line that you drew. Use these coordinates to find the slope of the line. What does this tell you about the percentage of adults with Internet access? c. Give the approximate vertical intercept of the line that you drew (using the reinitialized value for the year). d. Write an equation for your line. e. Use your equation to estimate the number of adults with Internet access in 1998 and in 2002. f. What would you expect to happen to the percentage after 2003? Do you think your linear model will be a good predictor after 2003?
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Exercises for Section 2.9 “Extended Exploration: Looking for Links between Education and Earnings,” which follows this chapter, has many additional exercises that involve finding best-fit lines using technology. Graphing program recommended (or optional) for Exercises 3, 4, 13, and 15. 1. Match each equation with the appropriate table. a. y 5 3x 1 2 b. y 5 12 x 1 2 c. y 5 1.5x 1 2 A. B. C. x y x y x y 0 2 4 6 8
2 3 4 5 6
0 2 4 6 8
2 5 8 11 14
0 2 4 6 8
10
15
x
0 2
4
6
8
y
15
6
0 2
4
–5
8
–20
–5
–10
–30
–10
–15
Graph A
y
x
0
x
Graph B
10 20 30 40
Graph C
3. (Graphing program recommended.) Identify which of the following data tables represent exact and which approximate linear relationships. For the one(s) that are exactly linear, construct the corresponding equation(s). For the one(s) that are approximately linear, generate the equation of a best-fit line; that is, plot the points, draw in a line approximating the data, pick two points on the line (not necessarily from your data) to generate the slope, and then construct the equation. a. x y
22 26.5
21 25.0
0 23.5
1 22.0
2 20.5
1998 172
2000 174
Source: Centers for Disease Control and Prevention, National Center for Health-Related Statistics, 2005.
b. The amount of tax owed on a purchase price. $2.00 $0.12
Kilograms Pounds
5
5
–10
1996 152
$5.00 $0.30
$10.00 $0.60
$12.00 $0.72
c. The number of pounds in a given number of kilograms.
10
10
Year 1994 Cocaine-related 143 emergency room episodes (in thousands)
Price Tax
2 8 14 20 26
2. Match each of the equations with the appropriate graph. a. y 5 10 2 2x b. y 5 10 2 5x c. y 5 10 2 0.5x y
a. The number of cocaine-related emergency room episodes.
3 1.0
1 2.2
5 11
10 22
20 44
Use the linear equations found in parts (a), (b), and (c) to approximate the values for: d. The number of cocaine-related emergency room episodes predicted for 2004. How does your prediction compare with the actual value in 2004 of 131 thousand cases? e. The amount of tax owed on $7.79 and $25.75 purchase prices. f. The number of pounds in 15 kg and 150 kg. 5. Determine which data represent exactly linear and which approximately linear relationships. For the approximately linear data, sketch a line that looks like a best fit to the data. In each case generate the equation of a line that you think would best model the data. a. The number of solar energy units consumed (in quadrillions of British thermal units, called Btus) Year Solar consumption (in quadrillions of Btus)
1998 0.07
1999 0.07
2000 0.07
2001 0.07
Source: U.S. Bureau of the Census, Statistical Abstract.
b. Gross farm output (in billions of dollars)
b. 21 8.5
0 6.5
1 3.0
2 1.2
3 21.5
4 22
c. N P
0 35
15 80
23 104
45 170
56 203
79 272
Farm output (billions of dollars)
250
t Q
200 150 100 50 0 1994
1996
2000
1998
2002
2004
Year
4. (Graphing program recommended.) Plot the data in each of the following data tables. Determine which data are exactly linear and which are approximately linear. For those that are approximately linear, sketch a line that looks like a best fit to the data. In each case generate the equation of a line that you think would best model the data.
Source: U.S. Census Bureau, Statistical Abstract.
6. In 1995 the United States consumed 464 million gallons of wine and in 2004, about 667 million gallons. a. Assuming the growth was linear, create a function that could model the trend in wine consumption.
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b. Estimate the amount of wine consumed in 2005. The actual amount was about 703 million gallons worth of wine. How accurate was your approximation? (Source: U.S. Department of Agriculture, Economic Research Service, 2007.)
7. The percentage of medical degrees awarded to women in the United States between 1970 and 2002 is shown in the accompanying graph. Percentage of M.D. Degrees Awarded to Women
30 20 10
1980
1985 1990 Year
1995
2000
2005
Infant Mortality versus Literacy Rate
Source: U.S. National Center for Education Statistics, “Digest of Education Statistics,” annual; Statistical Abstract of the United States.
The data show that since 1970 the percentage of female doctors has been rising. a. Sketch a line that best represents the data points. Use your line to estimate the rate of change of the percentage of M.D. degrees awarded to women. b. If you extrapolate your line, estimate when 100% of doctors’ degrees will be awarded to women. c. It seems extremely unlikely that 100% of medical degrees will ever be granted to women. Comment on what is likely to happen to the rate of growth of women’s degrees in medicine; sketch a likely graph for the continuation of the data into this century.
Infant mortality (deaths per 1000 live births)
Percentage
40
1975
a. Sketch a line through the graph of the data that best represents female infant mortality rates. Does the line seem to be a reasonable model for the data? What is the approximate slope of the line through these points? Show your work. b. Sketch a similar line through the male mortality rates. Is it a reasonable approximation? Estimate its slope. Show your work. c. List at least two important conclusions from the graph. 9. The accompanying scatter plot shows the relationship between literacy rate (the percentage of D A T A the population who can read and write) and infant mortality rate (infant deaths per 1000 live births) for 91 countries. The raw data are contained in the Excel or graph link file NATIONS and are described at the end of the Excel file. (You might wish to identify the outlier, the country with about a 20% literacy rate and a low infant mortality rate of about 40 per 1000 live births.) Construct a linear model. Show all your work and clearly identify the variables and units. Interpret your results.
50
0 1970
129
160 140 120 100 80 60 40 20 0 0
20
40 60 Literacy rate (%)
80
100
10. The accompanying graph shows data for the men’s Olympic 16-pound shot put. Men's Olympic 16-lb Shot Put Winning Results 80 70 Feet thrown
60
8. The accompanying graph shows the mortality rates (in deaths per 1000) for male and female infants in the United States from 1980 to 2000.
50 40 30 20
U.S. Infant Mortality Rates
10
Number of deaths (per thousand)
16
0 1920
Male Female
14
10 8 6 4 2 1985
1990 Year
1995
1960 Year
1980
2000 2010
Source: 2006 World Almanac and Book of Facts. Note: There were no Olympics in 1940 and 1944 due to World War II.
12
0 1980
1940
2000
Source: Centers for Disease Control and Prevention, www.cdc.gov.
a. The shot put results are roughly linear between 1920 and 1972. Sketch a best-fit line for those years. Estimate the coordinates of two points on the line to calculate the slope. Interpret the slope in this context. b. What is happening to the winning shot put results after 1972? Estimate the slope of the best-fit line for the years after 1972. c. Letting x 5 years since 1920, construct a piecewise linear function S (x) that describes the winning Olympic shot put results (in feet thrown) between 1920 and 2004.
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11. The accompanying chart shows the percentage of U.S. households that have at least one cell phone.
c. If a first-time mother is beyond the age you specified in part (b), is she more or less likely to develop breast cancer than a childless woman? d. Sketch a line that looks like a best fit to the data, estimate the coordinates of two points on the line, and use them to calculate the slope. e. Interpret the slope in this context. f. Construct a linear model for these data, identifying your independent and dependent variables.
Percent of All U.S. Households with Cell Phones Percent of households
80 70 60 50 40 30 20 10 0 1990
2000
1995
2005
Year
Source: 2006 World Almanac and Book of Facts Forrester Reports.
a. A linear model seems reasonable between 1990 and 2003. Sketch a best-fit line for those dates, then pick two points on your line and calculate the slope. Interpret the slope in this context. b. What appears seems to be happening after 2003? What might be the explanation? c. Letting x 5 years since 1990, construct a piecewise linear function P(x) that would describe the percentage of households with cell phones between 1990 and 2005.
13. (Graphing program recommended.) The given data show that health care is becoming more expensive and is taking a bigger share of the U.S. gross domestic product (GDP). The GDP is the market value of all goods and services that have been bought for final use. Year U.S. health care costs as a percentage of GDP Cost per person, $
7.1
141
341
8.9
12.2
13.3
15.3
1052 2691 4675 5671
a. Graph health care costs as a percentage of GDP versus year, with time on the horizontal axis. Measure time in years since 1960. Draw a straight line by eye that appears to be the closest fit to the data. Figure out the slope of your line and create a function H(t) for health care’s percentage of the GDP as a function of t, years since 1960. b. What does your formula predict for health care as a percentage of GDP for the year 2010? c. Why do you think the health care cost per person has gone up so much more dramatically than the health care percentage of the GDP?
1.6
1.4
1.2
1.0
14. a. From the accompanying chart showing sport utility vehicle (SUV) sales, estimate what the rate of increase in sales has been from 1990 to 2004. (Hint: Convert the chart into an equivalent scatter plot.)
0.8
0.6
Sport Utility Vehicle Sales in the United States, 1990–2004
0.2
15
20 25 30 35 Age at which woman has first child
40
4,000,000 3,000,000 2,000,000 1,000,000 2004
2003
2002
2001
2000
1999
1998
1997
1996
1995
1994
1991
1993
0
1992
a. If a woman has her first child at age 18, approximately what is her risk of developing cancer relative to a woman who has never borne a child? b. At roughly what age are the chances the same that a woman will develop breast cancer whether or not she has a child?
5,000,000
1990
Source: J. Cairns, Cancer: Science and Society (San Francisco: W. H. Freeman, 1978), p. 49.
Number of sales
0.4
0
5.1
Source: U.S. Health Care Financing Administration and Centers for Medicare and Medicaid Services.
12. The accompanying graph shows the relationship between the age of a woman when she has her first child and her lifetime risk of getting breast cancer relative to a childless woman.
The life-long risk of a mother developing breast cancer relative to the risk for childless women
1960 1970 1980 1990 2000 2003
Source: Ward’s Communications.
b. Estimate a linear formula to represent sales in years since 1990. c. Do you think the popularity of SUVs will continue to grow at the same rate? Why or why not?
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16. A veterinarian’s office displayed the following table comparing dog age (in dog years) to human age (in human years). The chart shows that the relationship is fairly linear. Comparative Ages of Dogs and Humans 120 Human years
15. (Graphing program optional.) The Gas Guzzler Tax is imposed on manufacturers on the sale of new-model cars (not minivans, sport utility vehicles, or pickup trucks) whose fuel economy fails to meet certain statutory regulations, to discourage the production of fuel-inefficient vehicles. The tax is collected by the IRS and paid by the manufacturer. The table shows the amount of tax that the manufacturer must pay for a vehicle’s miles per gallon fuel efficiency. Gas Guzzler Tax MPG
Tax per Car
12.5 13.5 14.5 15.5 16.5 17.5 18.5 19.5 20.5 21.5 22.5
$6400 $5400 $4500 $3700 $3700 $2600 $2100 $1700 $1300 $1000 $0
131
100 80 60 40 20 0
0
10 Dog years
20
a. Draw a line that looks like a best fit to the data. b. Estimate the coordinates and label two points on the line. Use them to find the slope. Interpret the slope in this context. c. Using H for human age and D for dog age, identify which you are using as the independent and which you are using as the dependent variable. d. Generate the equation of your line. e. Use the linear model to determine the “human age” of a dog that is 17 dog years old. f. Middle age in humans is 45–59 years. Use your model equation to find the corresponding middle age in dog years. g. What is the domain for your model?
Source: http://www.epa.gov/otaq/ cert/factshts/fefact 0.1.pdf.
a. Plot the data, verify that they are roughly linear, and add a line of best fit. b. Choose two points on the line, find the slope, and then form a linear equation with x as the fuel efficiency in mpg and y as the tax in dollars. c. What is the rate of change of the amount of tax imposed on fuel-inefficient vehicles? Interpret the units.
CHAPTER SUMMARY The Average Rate of Change The average rate of change of y with respect to x is change in y change in x units of y units of x For example, the units might be dollars/year (read as “dollars per year”) or pounds/person (read as “pounds per person”). The average rate of change between two points is the slope of the straight line connecting the points. Given two points sx1, y1 d and sx2, y2 d , The units of the average rate of change 5
average rate of change 5 5
change in y change in x y y 2 y1 5 2 5 slope x x2 2 x1
If the slope, or average rate of change, of y with respect to x is positive, then the graph of the relationship rises when read from left to right. This means that as x increases in value, y increases in value.
If the slope is negative, the graph falls when read from left to right. As x increases, y decreases. If the slope is zero, the graph is flat. As x increases, there is no change in y.
Linear Functions A linear function has a constant average rate of change. It can be described by an equation of the form Output 5 initial value 1 5 rate of change ? 3 input 3 5 x y m b or y 5 b 1 mx where b is the vertical intercept and m is the slope, or rate of change of y with respect to x. The Graph of a Linear Function The graph of the linear function y 5 b 1 mx is a straight line. The y-intercept, b, tells us where the line crosses the y-axis.
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The slope, m, tells us how fast the line is climbing or falling. The larger the magnitude (or absolute value) of m, the steeper the graph. If the slope, m, is positive, then the line climbs from left to right. If m is negative, the line falls from left to right.
Special Cases of Linear Functions Direct proportionality: y is directly proportional to (or varies directly with) x if y 5 mx where the constant m 2 0 This equation represents a linear function in which the y-intercept is 0, so the graph passes through (0, 0), the origin. Horizontal line: A line of the form y 5 b, with slope 0. Vertical line: A line of the form x 5 c, with slope undefined.
Parallel lines: Two lines that have the same slope. Perpendicular lines: Two lines whose slopes are negative reciprocals. Piecewise linear function: A function constructed from different linear segments. Some examples are the absolute value function, f (x) 5 uxu, and step functions, whose linear segments are all horizontal.
Fitting Lines to Data We can visually position a line to fit data whose graph exhibits a linear pattern. The equation of a best-fit line offers an approximate but compact description of the data. Lines are of special importance in describing patterns in data because they are easily drawn and manipulated and give a quick first approximation of trends.
CHECK YOUR UNDERSTANDING I. Is each of the statements in Problems 1–30 true or false? Give an explanation for your answer. In Problems 1–4 assume that y is a function of x. 1. The graph of y 2 5x 5 5 is decreasing. 2. The graph of 2x 1 3y 5 212 has a negative slope and negative vertical intercept. 3. The graph of x 2 3y 1 9 5 0 is steeper than the graph of 3x 2 y 1 9 5 0. 4. The accompanying figure is the graph of 5x 2 3y 5 22. 5
y
8. The average rate of change of a variable M between the years 1990 and 2000 is the slope of the line joining two points of the form sM1, 1990d and sM2, 2000d . 9. To calculate the average rate of change of a variable over an interval, you must have two distinct data points. 10. A set of data points of the form (x, y) that do not fall on a straight line will generate varying average rates of change depending on the choice of endpoints. 11. If the average rate of change of women’s salaries from 2003 to 2007 is $1000/year, then women’s salaries increased by exactly $1000 between 2003 and 2007.
x –5
7. The average rate of change between two points st1, Q1 d and st2, Q2 d is the same as the slope of the line joining these two points.
5
12. If the average rate of change is positive, the acceleration (or rate of change of the average rate of change) may be positive, negative, or zero. 13. If the average rate of change is constant, then the acceleration (or rate of change of the average rate of change) is zero.
–5
5. Health care costs between the years 1990 and 2007 would most likely show a positive average rate of change. 6. If we choose any two points on the graph in the accompanying figure, the average rate of change between them would be positive.
14. The average rate of change between sW1, D1 d and can be written as either sW2, D2 d sW1 2 W2 d/ sD1 2 D2 d or sW2 2 W1 d/sD2 2 D1 d . 15. If we choose any two distinct points on the line in the accompanying figure, the average rate of change between them would be the same negative number.
y
D
3
16
x –1
5
t –4 –3
8 –4
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16. The average rate of change between st1, Q1 d and st2, Q2 d can be written as either sQ1 2 Q2 d/st1 2 t2 d or sQ2 2 Q1 d/st2 2 t1 d . 17. On a linear graph, it does not matter which two distinct points on the line you use to calculate the slope.
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30. All linear functions can be written in the form y 5 b 1 mx. II. In Problems 31–40, give an example of a function or functions with the specified properties. Express your answer using equations, and specify the independent and dependent variables.
18. If the distance a sprinter runs (measured in meters) is a function of the time (measured in minutes), then the units of the average rate of change are minutes per meter.
31. Linear and decreasing with positive vertical intercept
19. Every linear function crosses the horizontal axis exactly one time.
33. Linear and with positive horizontal intercept and negative vertical intercept
20. If a linear function in the form y 5 mx 1 b has slope m, then increasing the x value by one unit changes the y value by m units.
34. Linear and does not pass through the first quadrant (where x . 0 and y . 0) 35. Linear with average rate of change of 37 minutes/lap
21. If the units of the dependent variable y are pounds and the units of the independent variable x are square feet, then the units of the slope are pounds per square foot.
36. Linear describing the value of a stock that is currently at $19.25 per share and is increasing exactly $0.25 per quarter
22. If the average rate of change between any two data points is increasing as you move from left to right, then the function describing the data is linear and is increasing.
37. Two linear functions that are parallel, such that moving one of the functions horizontally to the right two units gives the graph of the other
23. The function in the accompanying figure has a slope that is increasing as you move from left to right. M 20
39. Two linear functions where the slope of one is m and the slope of the second is 21/m, where m is a negative number
III. Is each of the statements in Problems 41–53 true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample.
D 20
41. If the average rate of change between any two points of a data set is constant, then the data are linear.
–20
24. The slope of the function f(x) is of greater magnitude than the slope of the function g(x) in the accompanying figures. y
10
42. If the slope of a linear function is negative, then the average rate of change decreases. 43. For any two distinct points, there is a linear function whose graph passes through them. 44. To write the equation of a specific linear function, one needs to know only the slope.
y
45. Function A in the accompanying figure is increasing at a faster rate than function B.
f(x)
x –10
38. Four distinct linear functions all passing through the point (0, 4)
40. Five data points for which the average rates of change between consecutive points are positive and are increasing at a decreasing rate
–20
2
32. Linear and horizontal with vertical intercept 0
10
x –2
g(x)
2
Q A
–2
–10
B
25. You can calculate the slope of a line that goes through any two points on the plane. 26. If f(x) 5 y is decreasing throughout, then the y values decrease as the x values decrease. 27. Having a slope of zero is the same as having an undefined slope. 28. Vertical lines are linear functions. 29. Two nonvertical lines that are perpendicular must have slopes of opposite sign.
t
46. The graph of a linear function is a straight line. 47. Vertical lines are not linear functions because they cannot be written in the form y 5 b 1 mx, as they have undefined slope.
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48. There exist linear functions that slant upward moving from left to right but have negative slope. 49. A constant average rate of change means that the slope of the graph of a function is zero. 50. All linear functions in x and y describe a relationship where y is directly proportional to x. 51. The function h(t) 5 ut 2 2 u is always positive. 52. The function h(t) 5 ut 2 2 u can be written as a piecewise linear function. 53. The graph to the right shows a step function.
7 6 5 4 3 2 1 0 0
1
2
3
4
CHAPTER 2 REVIEW: PUTTING IT ALL TOGETHER 1. According to the National Association of Realtors, the median price for a single-family home rose from $147,300 in 2000 to $217,300 in 2006. Describe the change in the following ways. a. The absolute change in dollars b. The percent increase c. The annual average rate of change of the price with respect to year (Note: Be sure to include units in all your answers.)
3. Given the following graph of the function h(t), identify any interval(s) over which: Graph of h(t) 6
D
C
2
B A
E F
O
–2
–4
2
4
6
8
t 10
G –2 –4
a. The function is positive; is negative; is zero b. The slope is positive; is negative; is zero
y
y
16 12 8 4 0
2. a. According to Apple Computer, sales of its iPod (the world’s best-selling digital audio player) soared from 304,000 in the third fiscal quarter of 2003 to 8,111,000 in the third fiscal quarter of 2006. What was the average rate of change in iPods sold per quarter? What was the average rate of change per month? b. iPod sales reached an all-time high of 14,043,000 in Apple’s first 2006 fiscal quarter (which included December 2005). What was the average rate of change in the number of iPods sold between the first and third fiscal quarters of 2006? What might be the reason for the decline?
4
4. Which line has the steeper slope? Explain why.
x 0
1
2
3
4
32 28 24 20 16 12 8 4 0
x 0
1
2
3
4
Graph B
Graph A
5. You are traveling abroad and realize that American, British, and French clothing sizes are different. The accompanying table shows the correspondence among female clothing sizes in these different countries. Women’s Clothing Sizes U.S.
Britain
France
4 6 8 10 12 14
10 12 14 16 18 20
38 40 42 44 46 48
a. Are the British and French sizes both linear functions of the U.S. sizes? Why or why not? b. Write a sentence describing the relationship between British sizes and U.S. sizes. c. Construct an equation that describes the relationship between French and U.S. sizes. 6. A bathtub that initially holds 50 gallons of water starts draining at 10 gallons per minute. a. Construct a function W(t) for the volume of water (in gallons) in the tub after t minutes.
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b. Graph the function and label the axes. Evaluate W(0) and W(5) and describe what they represent in this context. c. How would the original function and its graph change if the initial volume were 60 gallons? Call the new function U(t). d. How would the original function and graph change if the drain rate were 12 gallons per minute? Call the new function V(t). e. Add the graphs of U(t) and V(t) to the original graph of W(t). 7. Cell phones have produced a seismic cultural shift. No other recent invention has incited so much praise—and criticism. In 2000 there were 109.4 million cell phone subscriptions in the United States; since then subscriptions steadily increased to reach 207.9 million in 2005. (Note: Some people had more than one subscription.) In 2005 about 66% of the U.S. population had a cell phone. a. How many Americans were without a cell phone in 2005 (when the U.S. population was 296.4 million)? b. What was the average rate of change in millions of cell phone subscriptions per year between 2000 and 2005? c. Construct a linear function C(t) for cell phone subscriptions (in millions) for t 5 years from 2000. d. If U.S. cell phone subscriptions continue to increase at the same rate, how many will there be in 2010? Does your result sound plausible? (See Section 2.9, Exercise 11.)
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iii. 3x 2 y 5 0 iv. 8x 2 5 5 2y Assuming x is graphed on the horizontal axis, which equation(s) have graphs that: a. Are decreasing? b. Have a positive vertical intercept? c. Pass through the origin? d. Is (are) the steepest? 10. Match one or more of the following graphs with the following descriptions. Be sure to note the scale on each axis. a. Represents a constant rate of change of 3 b. Has a vertical intercept of 10 c. Is parallel to the line y 5 5 2 2x d. Has a steeper slope than the line C y
B
30
x
20
–20
8. The following graph shows average salaries for major league baseball players from 1990 to 2006.
A
–30
D
C
Average Salaries of Major League Baseball Players
Average salary (millions)
3.5
11. Gasoline prices spiked during the summer of 2006. The following table gives the weekly national average prices for regular gasoline as reported by the Department of Energy for three different weeks.
3.0 2.5 2.0
Regular Gas Prices in 2006
1.5 1.0 0.5 0.0 1990
1992 1994
1996 1998 Year
2000
2002
Week 31 (in August)
Week 43 (in October)
$2.22/gal
$3.04/gal
$2.22/gal
2004 2006
a. On the graph, sketch a straight line that is an approximate mathematical model for the data. b. Specify the coordinates of two points on your line and describe what they represent in terms of your model. c. Calculate the slope and then interpret it in terms of the salary in millions of dollars and the time interval. d. Let S(x) be a linear approximation of the salaries where x 5 years since 1990. What is the slope of the graph of S(x)? The vertical intercept? What is the equation for S(x)? 9. Consider the following linear equations. i. 3y 1 2x 2 15 5 0 ii. y 1 4x 5 1
Week 1 (in January)
a. Calculate the average rates of change for gasoline prices (in $/gal/week) between weeks 1 and 31; between weeks 31 and 43; between weeks 1 and 43. b. Write a 60-second summary about the price of gas during 2006. (Note: Unfortunately in the summer of 2007, gas prices once again went over $3.00 per gallon.) 12. You plan a trip to Toronto and find that on July 5, 2006, the exchange rate for Canadian money is CA$1 5 US$0.899. Although this exchange rate favors the U.S. visitor to Canada, you are not delighted to discover that the Canadians have a 14% sales tax. a. If a taxable item costs CA$50, how much do you actually have to pay in Canadian dollars?
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b. What is this worth in U.S. dollars? c. Construct a function that describes the total cost Cus (in U.S.) of purchasing a taxable item with a price of Pca (in Canadian dollars). d. Is the cost in U.S. dollars, Cus, directly proportional to the Canadian price, Pca? 13. Create a linear function for each condition listed below, using (at most) the numbers 22 and 3 and the variables Q and t. Assume t is the independent variable. The function’s graph is: a. Increasing b. Decreasing c. Horizontal d. Steeper than that of Q 5 5 2 t e. Parallel to Q 5 3t 2 4 f. Perpendicular to Q 5 (1/2)t 1 6 14. Given the following graph:
A
10
y
B
x 10
–5
–5
16. Generate the equation for a line under each of the following conditions. a. The line goes through the points (21, 24) and (4, 6). b. The line is parallel to the line in part (a) and goes through the point (0, 5). c. The line is perpendicular to the line in part (a) and goes through the origin. 17. In 2005 a record 51.5% of paper consumed in the United States (51.3 million tons) was recycled. The American Forest and Paper Association states that its goal is to have 55% recovery by 2012. a. Plot two data points that represent the percentage of paper recycled at t number of years since 2005. (You may want to crop the vertical axis to start at 50%.) Connect the two points with a line and calculate its slope. b. Assuming linear growth, the line represents a model, R(t), for the percentage of paper recycled at t years since 2005. Find the equation for R(t). c. Find R(0), R(5), and R(20). What does R(20) represent? 18. Sketch a graph that could represent each of the following situations. Be sure to label your axes. a. The volume of water in a kettle being filled at a constant rate. b. The height of an airplane flying at 30,000 feet for 3 hours. c. The unlimited demand for a certain product, such that if the company produces 10,000 of them, they can sell them for virtually any price. (Use the convention of economists, placing price on the vertical axis and quantity on the horizontal.) d. The price of subway tokens that periodically increase over time. e. The distance from your destination over time when you travel three subway stops to get to your destination. 19. Generate an equation for each line A, B, C, D, and E shown in the accompanying graph.
a. Find the equations of line A and line B. b. Show that line A is (or is not) perpendicular to line B. E
15. According to the Environmental Protection Agency’s “National Coastal Conditions Report II,” more than half of the U.S. population live in the narrow coastal fringes. Increasing population in these areas contributes to degradation (by runoff, sewage spills, construction, and overfishing) of the same resources that make the coasts desirable. In 2003 about 153 million people lived in coastal counties. This coastal population is currently increasing by an average of 3600 people per day. a. What is the average rate of change in millions of people per year? b. Let x 5 years since 2003 and construct a linear function P(x) for the coast population (in millions). c. Graph the function P(x) over a reasonable domain. d. What does your model give for the projected population in 2008?
B
10
y
D
C
A
x 10
–10
–10
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20. U.S. life expectancy has been steadily increasing (see DATA Exercise 15 in Section 2.1 and data file for LIFEXEC). The following graph shows the roughly linear trends for both males and females over more than three decades. Trends in U.S. Life Expectancy by Sex, 1970–2005
c. Why might a telephone company be interested in a ceiling function? 23. The table below gives historical data on voter turnout as a percentage of voting-age population (18 years and older) in U.S. presidential elections since 1960. Year
Voting-Age Population
Turnout
%Turnout of Voting-Age Pop.
2004 2000 1996 1992 1988 1984 1980 1976 1972 1968 1964 1960
221,256,931 205,815,000 196,511,000 189,529,000 182,778,000 174,466,000 164,597,000 152,309,190 140,776,000 120,328,186 114,090,000 109,159,000
122,294,978 105,586,274 96,456,345 104,405,155 91,594,693 92,652,680 86,515,210 81,555,789 77,718,554 73,211,875 70,644,592 68,838,204
55.3% 51.3% 49.1% 55.1% 50.1% 53.1% 52.6% 53.5% 55.2% 60.8% 61.9% 63.1%
75 70 65
Male Female
60 1970
1980
1990
2000
2010
Source: U.S. Bureau of the Census.
a. According to these data, what is the life expectancy of a female born in 1970? Of a male born in 2005? b. Sketch a straight line to create a linear approximation of the data for your gender. Males and females will have different lines. c. Identify two points on your line and calculate the slope. What does your slope mean in this context? The male slope is greater than the female slope. What does that imply? d. Generate a linear function for your line. (Hint: Let the independent variable be the number of years since 1970.) e. What would your model predict for your sex’s life expectancy in 2010? How close does your model come to Census Bureau current predictions of life expectancies in 2010 of 75.6 years for males and 81.4 for females? 21. For the function g(x) 5 uxu 1 2: a. Generate a table of values for g(x) using integer values of x between 23 and 3. b. Use the table to sketch a graph of g(x). c. How does this graph compare with the graph of f(x) 5 uxu? 22. In Exercise 32 in Section 2.8 we met the greatest integer function f(x) 5 [x], where [x] is the greatest integer #x; (i.e., you round down to the nearest integer). So [21.5] 5 22 and [3.99] 5 3. This function is sometimes written using the notation J x K and called the floor function. There is a similar ceiling function, g(x) 5 <x =, where <x = is the smallest integer $x; (i.e., here you round up to the nearest integer). So < 21.5= 5 21 and < 3.000001= 5 4. a. Complete the following table. x
22
21.5
21
20.5
0
0.5
1
1.5
2
Floor f(x) 5 J x K Ceiling g(x) 5 <x=
Using selected data from the table, graphs, and dramatically persuasive language, provide convincing arguments in a 60-second summary that during the years 1960 to 2004: i. Voter turnout has plummeted. ii. Voter turnout has soared. 24. a. Construct the equation of a line y1 whose graph is steeper than that of 8x 1 2y 5 6 and does not pass through quadrant I (where both x . 0 and y . 0). Plot the lines for both equations on the same graph. b. Construct another equation of a line y2 that is perpendicular to the original equation, 8x 1 2y 6, and does go through quadrant I. Add the plot of this line to the graph in part (a). 25. Sweden has kept meticulous records of its population for many years. The following graph shows the mortality rate (as a percentage) for female and male children in Sweden over a 250-year period. 250 Years of Mortality in Sweden for Children Between 0 and 5 Years 50 Female Male Percentage of children dying
Life expectancy (in years)
85 80
137
40
30
20
10
0
b. Plot the ceiling function and the floor function on two separate graphs, being sure to specify whether the endpoints of each line segment are included or excluded.
1750
1800
1850
1900 Year
Source: Statistics Sweden.
1950
2000
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a. Describe the overall trend, and the similarities and differences between the female and male graphs. b. Assuming the graphs are roughly linear between 1800 and 1950, draw a line approximating the female childhood mortality rate during this time. What is the slope of the line and what does it tell you in this context? c. Construct a linear function to model the deaths over this period. d. What is happening in the years after 1950?
that might have caused carbon dioxide discharges to increase at a lower rate? c. After 1943 the average rate of change steadily increases. What does that tell you about the increase of carbon dioxide? 27. Examine the following graph of a function y = f(x). y 4
Note: The large spike between 1750 and 1800 represents the more than 300,000 Swedish children (out of a total population of about 2 million) who died from smallpox, the most feared disease of that century.
E
3
26. According to the U.S. Environmental Protection Agency, carbon dioxide makes up 84.6% of the total U.S. greenhouse gas emissions. Carbon dioxide, or CO2, arises from the combustion of coal, oil, and gas. Greenhouse gases trap heat on our planet, causing it to become warmer. Effects of this phenomenon are already being seen in increased melting of glaciers and permafrost. If this continues unchecked, water levels will rise, causing flooding in coastal communities, where the majority of the U.S. population lives. (See Exercise 15.)
2
B
1
1
2.50
Point A B C D E
1.50 1.00
2
3
4
x
a. Using the points A, B, C, D, and E labeled on the graph, fill in the rest of the accompanying table.
Average Rate of Change of Carbon Dioxide per Year (over prior time interval)
2.00
D
C 0
The following graph shows the average rate of change in the concentration of atmospheric CO2 over time.
x-Coordinate of Point 0 1
Average Rate of y-Coordinate Change Between Two of Point Adjacent Points 4 1
n.a. 23
0.50
2025
2000
1975
1950
1925
1900
1875
1850
1825
1800
0.00
1775
Average change in CO2 concentration per year (in parts per million per year)
A
Year
a. Is the average rate of change always positive? If so, does that imply that the amount of atmospheric carbon dioxide is always increasing? Explain your answer. b. What does it mean when the average rate of change was decreasing (though not negative) between about 1915 and 1943? What was happening in the world during this period
b. Over what x interval is the function increasing? What is happening to the average rate of change between the points in that interval? c. Over what x interval is the function decreasing? What is happening to the absolute value of the average rate of change of the points in that interval? What does this mean? (Recall that the absolute value of a slope gives the steepness of the line.) d. What is the concavity of the entire function? How do the changes in the steepness of the curve help confirm the concavity of the function graph?
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E X P L O R AT I O N 2 . 1 Having It Your Way
Objective • construct arguments supporting opposing points of view from the same data Material/Equipment • excerpts from the Student Statistical Portrait of the University of Massachusetts, Boston (in the Appendix) or from the equivalent for the student body at your institution • computer with spreadsheet program and printer or graphing calculator with projection system (all optional) • graph paper and/or overhead transparencies (and overhead projector) Procedure Working in Small Groups Examine the data and graphs from the Student Statistical Portrait of the University of Massachusetts, Boston, or from your own institution. Explore how you would use the data to construct arguments that support at least two different points of view. Decide on the arguments you are going to make and divide up tasks among your team members. Rules of the Game • Your arguments needn’t be lengthy, but you need to use graphs and numbers to support your position. You may use only legitimate numbers, but you are free to pick and choose those that support your case. If you construct your own graphs, you may, of course, use whatever scaling you wish on the axes. • For any data that represent a time series, as part of your argument pick two appropriate endpoints and calculate the associated average rate of change. • Use “loaded” vocabulary (e.g., “surged ahead,” “declined drastically”). This is your chance to be outrageously biased, write absurdly flamboyant prose, and commit egregious sins of omission. • Decide as a group how to present your results to the class. Some students enjoy realistic “role playing” in their presentations and have added creative touches such as mock protesters complete with picket signs. Suggested Topics Your instructor might ask your group to construct one or both sides of the arguments on one topic. If you’re using data from your own institution, answer the questions provided by your instructor. Using the Student Statistical Portrait from the University of Massachusetts, Boston (data located in the Appendix) 1. Use the table “Undergraduate Admissions Summary” to construct a persuasive case for each situation. a. You are the Provost, the chief academic officer of the university, arguing in front of the Board of Trustees that the university is becoming more appealing to students. b. You are a student activist arguing that the university is becoming less appealing to students.
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2. Use the table “Trends in New Student Race/Ethnicity in the College of Liberal Arts” to make a convincing case for each of the following. a. You are the Affirmative Action Officer arguing that her office has done a terrific job. b. You are a reporter for the student newspaper criticizing the university for its neglect of minority students. 3. Use the table “SAT Scores of New Freshmen by College/Program” to “prove” each of the contradictory viewpoints. a. You are the Dean of the College of Liberal Arts arguing that you have brighter freshmen than those in the College of Management. b. You are the Dean of the College of Management arguing that your freshmen are superior. Exploration-Linked Homework With your partner or group prepare a short class presentation of your arguments, using, if possible, overhead transparencies or a projection panel. Then write individual 60-second summaries to hand in.
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EXPLORATION 2.2A Looking at Lines with the Course Software Objective • find patterns in the graphs of linear equations of the form y 5 mx 1 b Equipment • computer with course software “L1: m & b Sliders” in Linear Functions Procedure In each part try working first in pairs, comparing your observations and taking notes. Your instructor may then wish to bring the whole class back together to discuss everyone’s results. Part I: Exploring the Effect of m and b on the Graph of y 5 mx 1 b Open the program Linear Functions and click on the button “L1: m & b Sliders.” 1. What is the effect of m on the graph of the equation? Fix a value for b. Construct four graphs with the same value for b but with different values for m. Continue to vary m, jotting down your observations about the effect on the line when m is positive, negative, or equal to zero. Do you think your conclusions work for values of m that are not on the slider? Choose a new value for b and repeat your experiment. Are your observations still valid? Compare your observations with those of your partner. 2. What is the effect of b on the graph of the equation? Fix a value for m. Construct four graphs with the same value for m but with different values for b. What is the effect on the graph of changing b? Record your observations. Would your conclusions still hold for values of b that are not on the slider? Choose a new value for m and repeat your experiment. Are your observations still valid? Compare your observations with those of your partner. 3. Write a 60-second summary on the effect of m and b on the graph of y 5 mx 1 b. Part II: Constructing Lines under Certain Constraints 1. Construct the following sets of lines still using “L1: m & b Sliders.” Be sure to write down the equations for the lines you construct. What generalizations can you make about the lines in each case? Are the slopes of the lines related in some way? Are the vertical intercepts of the lines related? Construct any line. Then construct another line that has a steeper slope, and then construct one that has a shallower slope. Construct three parallel lines. Construct three lines with the same y-intercept, the point where the line crosses the y-axis. Construct a pair of lines that are horizontal. Construct a pair of lines that go through the origin. Construct a pair of lines that are perpendicular to each other. 2. Write a 60-second summary of what you have learned about the equations of lines.
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EXPLORATION 2.2B Looking at Lines with a Graphing Calculator
Objective • find patterns in the graphs of linear equations of the form y 5 mx 1 b Material/Equipment • graphing calculator (instructions for the TI-83 and TI-84 families of calculators are available in the Graphing Calculator Manual) Procedure Getting Started Set your calculator to the integer window setting. For the TI-83 or TI-84 do the following:
WINDOW FORMAT Xmin=47 Xmax=47 Xscl=10 Ymin=31 Ymax=31 Yscl=10 Xres=1 1. Press ZOOM and select [6:ZStandard]. 2. Press ZOOM and select [8:ZInteger], ENTER. 3. Press WINDOW to see whether the settings are the same as the duplicated screen image. Working in Pairs In each part try working first in pairs, comparing your observations and taking notes. Your instructor may then wish to bring the whole class back together to discuss everyone’s results. Part I: Exploring the Effect of m and b on the Graph of y 5 mx 1 b 1. What is the effect of m on the graph of the equation y 5 mx? a. Case 1: m . 0 Enter the following functions into your calculator and then plot the graphs. To get started, try m 5 1, 2, 5. Try a few other values of m where m . 0. Y1 5 x Y2 5 2x Y3 5 5x Y4 5 c Y5 5 Y6 5 Compare your observations with those of your partner. In your notebook describe the effect of multiplying x by a positive value for m in the equation y 5 mx.
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Exporation 2.2B
143
b. Case 2: m , 0 Begin by comparing the graphs of the lines when m 5 1 and m 5 21. Then experiment with other negative values for m and compare the graphs of the equations. Y1 5 x Y2 5 2x Y3 5 c Y4 5 Y5 5 Y6 5 Alter your description in part 1(a) to describe the effect of multiplying x by any real number m for y 5 mx (remember to also explore what happens when m 5 0). 2. What is the effect of b on the graph of an equation y 5 mx 1 b? a. Enter the following into your calculator and then plot the graphs. To get started, try m 5 1 and b 5 0, 20, 220. Try other values for b as well. Y1 5 x Y2 5 x 1 20 Y3 5 x 2 20 Y4 5 c Y5 5 Y6 5 b. Discuss with your partner the effect of adding any number b to x for y 5 x 1 b. (Hint: Use “trace” to find where the graph crosses the y-axis.) Record your comments in your notebook. c. Choose another value for m and repeat the exercise. Are your observations still valid? 3. Write a 60-second summary on the effect of m and b on the graph of y 5 mx 1 b. Part II: Constructing Lines under Certain Constraints 1. Construct the following sets of lines using your graphing calculator. Be sure to write down the equations for the lines you construct. What generalizations can you make about the lines in each case? Are the slopes of the lines related in any way? Are the vertical intercepts related? Which graph is the steepest? Construct three parallel lines. Construct three lines with the same y-intercept. Construct a pair of lines that are horizontal. Construct a pair of lines that go through the origin. Construct a pair of lines that are perpendicular to each other. 2. Write a 60-second summary of what you have learned about the equations of lines.
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LOOKING FOR LINKS BETWEEN EDUCATION AND EARNINGS OVERVIEW Is learning the key to earning? Does going to school pay off? In this extended exploration, you use a large data set from the U.S. Bureau of the Census to examine ways in which education and earnings may be related. Technology is used to fit lines to data, and you learn how to interpret the resulting linear models, called regression lines. You can explore further by finding evidence to support or disprove conjectures, examining questions raised by the analysis, and posing your own questions. In this exploration, you will • analyze U.S. Census data • use regression lines to summarize data • make conjectures about the relationship between education and earnings in the United States • find evidence to support or refute your conjectures • examine the distinction between correlation and causation
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Using U.S. Census Data
DATA
Does more education mean higher earnings? The answer may seem obvious. We may reasonably expect that having more education gives access to higher-paying jobs. Is this indeed the case? We explore how a social scientist might start to answer these questions using a random sample from U.S. Census data. Our data set, called FAM1000, provides information on 1000 individuals and their families. The Bureau of the Census, as mandated by the Constitution, conducts a nationwide census every 10 years. To collect more up-to-date information, the Census Bureau also conducts a monthly survey of American households for the Bureau of Labor Statistics called the Current Population Survey, or CPS. The CPS is the largest survey taken between census years. The CPS is based on data collected each month from approximately 50,000 households. Questions are asked about race, education, housing, number of people in the household, earnings, and employment status.1 The March surveys are the most extensive. Our sample of census data, FAM1000, was extracted from the March 2006 Current Population Survey. It contains information about 1000 individuals randomly chosen from those 16 and older who worked at least 1 week in 2005. You can use the FAM1000 data and the related software, called FAM1000 Census Graphs, to follow the discussion in the text and/or conduct your own case study. The full FAM1000 data set is in the Excel file FAM1000, and condensed versions are in the graph link files FAM1000 A to H. These data files and related software can be downloaded from the web at www.wiley.com/college/kimeclark or on your class Wiley Plus site. The software provides easy-to-use interactive tools for analyzing the FAM1000 data. Table 1 on page 147 is a data dictionary with short definitions for each data category. Think of the data as a large array of rows and columns of facts. Each row represents all the information obtained from one particular respondent about his or her family. Each column contains the coded answers of all the respondents to one particular question. Try deciphering the information contained in the first row of the 1000 rows in the FAM1000 data set in Table 2.
age sex region cencity marstat famsize edu occup hrswork wkswork yrft pearnings
42
2
4
1
0
3
12
4
40
52
1
$25,000
ptotinc faminc race hispanic $25,559 $46,814 5 0
Table 2
Referring to the data dictionary, we learn that the respondent is a 42-year-old female who lives in a city somewhere in the West. She is married with two other people in her family and she has a high school degree. In 2005 she worked in sales or a related occupation for 40 hours a week, 52 weeks of the year. Her personal earnings from work were $25,000, but her personal total income, was $25,559. That means she had an additional source of unearned income, such as dividends from stocks or bonds, or interest on a savings account. Her family income totaled $46,814. She did not identify with any of the racial or Hispanic categories listed on the census form. 1
The results of the survey are used to estimate numerous economic and demographic variables, such as the size of the labor force, the employment rate, earnings, and education levels. The results are widely quoted in the popular press and are published monthly in The Monthly Labor Review and Employment and Earnings, irregularly in Current Population Reports and Special Labor Force Reports, and yearly in Statistical Abstract of the United States and The Economic Report of the President.
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147
Data Dictionary for March 2006 Current Population Survey Variable
Definition
age sex
Age Sex
region
Census region
cencity
Residence location
Unit of Measurement (code and allowable range) Range is 16 to 85 1 5 Male 2 5 Female 1 5 Northeast 2 5 Midwest 3 5 South 4 5 West 1 5 Metropolitan 2 5 Nonmetropolitan 3 5 Not Identified 0 5 Presently married 1 5 Presently not married
marstat
Marital status
famsize
Family size
Range is 1 to 39
educ
Years of education
8 5 8 or fewer years of education 10 5 No high school degree, 9–12 years of education 12 5 High school degree 13 5 Some college 14 5 Associate’s degree 16 5 Bachelor’s degree 18 5 Master’s degree 20 5 Doctorate 22 5 Professional degree (e.g., MD)
occup
Occupation group of respondent
0 5 Not in universe, or children 1 5 Management, business, and financial occupations 2 5 Professional and related occupations 3 5 Service occupations 4 5 Sales and related occupations 5 5 Office and administrative support occupations 6 5 Farming, fishing, and forestry occupations
Variable
Definition
occup
Occupation group of respondent
hrswork
Usual hours worked per week Weeks worked in 2005 Employed full-time year-round
wkswork yrft
pearnings ptotinc
Total personal earnings from work Personal total income
faminc
Family income
race
Race of respondent
hispanic
Hispanic heritage
Unit of Measurement (code and allowable range) 7 5 Construction and extraction occupations 8 5 Installation, maintenance, and repair occupations 9 5 Production occupations 10 5 Transportation and material moving occupations 11 5 Armed Forces Range is 1 to 99 Range is 1 to 52 1 5 full-time year-round 2 5 part-time year-round 3 5 full-time part of the year 4 5 part-time part of the year Range is $0 to $650,000 Range is negative $1,000,000 to $10,000,000 Range is negative $400,000 to $24,000,000 1 5 White 2 5 Black 3 5 American Indian, Alaskan Native Only 4 5 Asian or Pacific Islander 5 5 Other 0 5 Not in universe 1 5 Mexican 2 5 Puerto Rican 3 5 Cuban 4 5 Central/South American 5 5 Other Spanish
Table 1 Source: Adapted from the U.S. Bureau of the Census, Current Population Survey, March 2006, by Jie Chen, Computing Services, University of Massachusetts, Boston.
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Summarizing the Data: Regression Lines Is There a Relationship between Education and Earnings? In the physical sciences, the relationship among variables is often quite direct; if you hang a weight on a spring, it is clear, even if the exact relationship is not known, that the amount the spring stretches is definitely dependent on the heaviness of the weight. Further, it is reasonably clear that the weight is the only important variable; the temperature and the phase of the moon, for example, can safely be neglected. In the social and life sciences it is usually difficult to tell whether one variable truly depends on another. For example, it is certainly plausible that a person’s earnings depend in part on how much formal education he or she has had, since we may suspect that having more education gives access to higher-paying jobs, but many other factors also play a role. Some of these factors, such as the person’s age or type of work, are measured in the FAM1000 data set; others, such as family background or good luck, may not have been measured or even be measurable. Despite this complexity, we attempt to determine as much as we can by first looking at the relationship between earnings and education alone. We start with a scatter plot of education and personal earnings from the FAM1000 data set. If we hypothesize that earnings depend on education, then the convention is to graph education on the horizontal axis. Each ordered pair of data values gives a point with coordinates of the form (education, personal earnings) Take a moment to examine the scatter plot in Figure 1. Each point refers to respondents with a particular level of education and income and has two coordinates. The first coordinate gives the years of education past grade 8 (So zero represents an 200,000
175,000
Personal earnings (dollars)
150,000
125,000
100,000
75,000
50,000
25,000
0
2
4
6
8
10
12
14
16
Education (years past grade 8)
Figure 1 Personal earnings vs. education. In attempting to pick a reasonable scale to display personal earnings on this graph, the vertical axis was cropped at $200,000, which meant excluding from the display the points for three individuals who each earned more than $200,000 in wages.
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eighth-grade education or less) and the second gives the personal earnings. For example, the points at the very top of the graph represent individuals who make $200,000 in personal wages. One of these points represents those with 16 years (8 years past eighth grade) of education and the other, those with 22 years (14 years past eighth grade) of education. The coordinates for these points are (8, $200,000) and (14, $200,000). We can refer back to the original data set to find out more information about these points, as well as the outliers that are not shown on the graph. Note that some dots represent more than one individual (i.e., for a particular level of education there might be many people with the same income). How might we think of the relationship between these two variables? Clearly, personal earnings are not a function of education in the mathematical sense since people who have the same amount of education earn widely different amounts. The scatter plot obviously fails the vertical line test. But suppose that, to form a simple description of these data, we were to insist on finding a simple functional description. And suppose we insist that this simple relationship be a linear function. In Chapter 2, we informally fit linear functions to data. A formal mathematical procedure called regression analysis lets us determine what linear function is the “best” approximation to the data; the resulting “best-fit” line is called a regression line and is similar in spirit to reporting only the mean of a set of single-variable data, rather than the entire data set. It can be a useful and powerful method of summarizing a set of data. We can measure how well a line represents a data set by summing the vertical distances squared between the line and the data points. The regression line is the line that makes this sum as small as possible. The calculations necessary to compute this line are tedious, although not difficult, and are easily carried out by computer software and graphing calculators. We can use the FAM1000 Census Graphs or Excel program to find regression lines. The techniques for determining regression lines are beyond the scope of this course. In Figure 2, we show the FAM1000 data set along with a regression line determined from the data points. The equation of the line is personal earnings 5 4188 1 5693 ? yrs. of educ. past grade 8 200,000
175,000
150,000 Personal earnings (dollars)
If you are interested in a standard technique for generating regression lines, a summary of the method used in the course software is provided in the reading "Linear Regression Summary."
149
125,000
100,000
75,000
50,000
25,000
0
2
4
6
8
10
12
14
Education (years past grade 8)
Figure 2 Regression line for personal earnings
vs. education.
16
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Since personal earnings are in dollars, the units for the term 4188 must be dollars, and the units for 5693 must be dollars per year of education. This line is certainly more concise than the original set of data points. Looking at the graph, you may judge with your eyes to what extent the line is a good description of the original data set. From the equation, the vertical intercept is 4188. Thus, this model predicts that individual personal earnings for those with an eighth-grade education or less will be $4188. The number 5693 represents the slope of the regression line, or the average rate of change of personal earnings with respect to years of education. Thus, this model predicts that for each additional year of education, individual personal earnings increase by $5693. We emphasize that, although we can construct an approximate linear model for any data set, this does not mean that we really believe that the data are truly represented by a linear relationship. In the same way, we may report the median to summarize a set of data, without believing that the data values are at the median. In both cases, there are features of the original data set that may or may not be important and that we do not report. The data points are widely scattered about the line, for reasons that are clearly not captured by the linear model. We can eliminate the clutter by grouping together all people with the same years of education and plotting the median personal earnings of each group. The result is the graph in Figure 3, which includes a regression line for the new data. For each year of education, only a single median earnings point has been graphed. For instance, the point corresponding to 12 years of education past grade 8 has a vertical value of approximately $75,000; hence the median of the personal earnings of everyone in the FAM1000 data with 20 years of education is about $75,000. The pattern is now clearer: An upward trend to the right is more obvious in this graph. 125,000 Median personal earnings ($)
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100,000 75,000 50,000 25,000
0
2
4
6
8
10
12
14
16
Education (years past grade 8)
Figure 3 Median personal earnings vs. education.
Every time we construct a simplified representation of an original data set, we should ask ourselves what information has been suppressed. In Figure 3 we have suppressed the spread of data in the vertical direction. For example, there are only 36 people with an eighth-grade education or less but 175 with 16 years of education. Yet each of these sets is represented by a single point. We can fit a line to the graph in Figure 3 using the same method of linear regression. The equation of this straight line is median personal earnings 5 22237 1 6592 ? yrs. of educ. past grade 8 Here, 22237 is the vertical intercept and 6592 is the slope or rate of change of median personal earnings with respect to education. This model predicts that for each additional year of education, median personal earnings increase by $6592. Note that this linear model predicts median personal earnings for the group, not personal earnings for an individual. The vertical intercept of this line is negative even though all earnings in the original data set are positive. The linear model is clearly inaccurate for those with an eighth-grade education or less.
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151
Figure 4 shows two regression lines: One represents the fit to the medians and the other represents the fit to all of the data. Both of these straight lines are reasonable answers to the question “What straight line best describes the relationship between education and earnings?” and the difference between them indicates the uncertainty in answering such a question. We may argue that the benefit in earnings for each year of education is $5693, or $6592, or something between these values.
Personal earnings (dollars)
125,000 100,000
Median points cc = 0.94
75,000
All points cc = 0.40
50,000 25,000
0
2
4
6
8
10
12
14
16
Education (years past grade 8)
Figure 4 Regression lines for personal earnings vs.
education.
Regression Line: How Good a Fit?
The programs R1–R4 and R7 in Linear Regression can help you visualize the links among scatter plots, best-fit lines, and correlation coefficients.
The reading “The Correlation Coefficient” explains how to calculate and interpret the correlation coefficient.
Once we have determined a line that approximates our data, we must ask, “How good a fit is our regression line?” To help answer this, statisticians calculate a quantity called the correlation coefficient. This number can be computed by statistical software, and we have included it on our graphs and labeled it “cc.”2 The correlation coefficient is always between 21 (negative association with no scatter; the data points fit exactly on a line with a negative slope) and 1 (positive association with no scatter; the data points fit exactly on a line with a positive slope). The closer the absolute value of the correlation coefficient is to 1, the better the fit and the stronger the linear association between the variables. A small correlation coefficient (with absolute value close to zero) indicates that the variables do not depend linearly on each other. This may be because there is no relationship between them, or because there is a relationship that is something more complicated than linear. In future chapters we discuss many possible nonlinear functional relationships. There is no definitive answer to the question of when a correlation coefficient is “good enough” to say that the linear regression line is a good fit to the data. A fit to the graph of medians (or means) generally gives a higher correlation coefficient than a fit to the original data set because the scatter has been smoothed out. (See the cc’s in Figure 4.) When in doubt, plot all the data along with the linear model and use your best judgment. The correlation coefficient is only a tool that may help you decide among different possible models or interpretations.
2 We use the label “cc” for the correlation coefficient in the text and software to minimize confusion. In a statistics course the correlation coefficient is usually referred to as Pearson’s r or just r.
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EXAMPLE
1
Interpreting the correlation coefficient Figure 5 contains the data and regression line for mean personal earnings vs. education. 125,000 Mean personal earnings ($)
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100,000 75,000 50,000 25,000
0
2
4
6
8
10
12
14
16
Education (years past grade 8)
Figure 5 Mean personal earnings vs. education.
The equation of the regression line is mean personal earnings 5 5562 1 5890 ? yrs. of educ. past grade 8 cc = 0.98. a. Interpret the slope of this new regression line. b. Compare this regression line with that for median personal earnings vs. education previously cited in the text: median personal earnings 5 22237 1 6592 ? yrs. of educ. past grade 8 cc 5 0.94 What do these equations predict for median and mean personal earnings for 12 years of education (high school)? For 18 years of education (master’s degree)? SOLUTION
a. The slope of 5890 suggests that each additional year of education corresponds to a $5890 yearly increase in mean personal earnings. b. Both equations have cc’s close to 1 (one is 0.98, the other 0.94), so both lines are good fits. Using the regression line for means, 12 years of education (or 4 years past grade 8), we get: mean personal earnings 5 $5562 1 $5890 (4) 5 $29,122 Rounding to nearest ten 5 $29,120 Using the regression line for means for 18 years of education (or 10 years past grade 8), we get: mean personal earnings 5 $5562 1 $5890 (10) 5 $64,462 Rounding to nearest ten 5 $64,460 Using the regression lines for medians and rounding to the nearest ten, 12 years of education corresponds to $24,130 and 18 years to $63,680. Comparing predictions from the mean and median regression lines, we see that the regression line for means predicts higher earnings than the median regression line. This is reasonable since in general, when using income measures, the mean will often exceed the median since medians are not susceptible to outliners.
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EXAMPLE
2
153
Examine the four scatter plots in Figure 6. y
y 10
10
5
5
0 5 Graph A
10
x
0 5 Graph C
y
10
x
y
10
10
5
5
0 5 Graph B
10
x
0 5 Graph D
10
x
Figure 6 Four scatter plots.
a. Which graph shows a positive linear correlation between x and y? A negative linear correlation? Zero correlation? b. Which graph shows the closest linear correlation between x and y? SOLUTION
a. Graph D shows a positive linear correlation between x and y (when one variable increases, the other increases). Graph B shows a negative linear correlation (when one variable increases, the other decreases). Both graphs A and C show zero correlation between x and y. Even though graph C shows a pattern in the relationship between x and y, the pattern is not linear. b. Graph B. The correlation coefficient of its regression line would be close to 21, almost a perfect (negative) correlation.
Interpreting Regression Lines: Correlation versus Causation One is tempted to conclude that increased education causes increased earnings. This may be true, but the model we have used does not offer conclusive proof. This model can show how strong or weak a relationship exists between variables but does not answer the question “Why are the variables related?” We need to be cautious in how we interpret our findings. Regression lines show correlation, not causation. We say that two events are correlated when there is a statistical link. If we find a regression line with a correlation coefficient that is close to 1 in absolute value, a strong relationship is suggested. In our previous example, education is positively correlated with personal earnings. If education increases, personal earnings increase. Yet this does not prove that education causes an increase in personal earnings. The reverse might be true; that is, an increase in personal earnings might cause an increase in education. The correlation may be due
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to other factors altogether. It might occur purely by chance or be jointly caused by yet another variable. Perhaps both educational opportunities and earning levels are strongly affected by parental education or a history of family wealth. Thus a third variable, such as parental socioeconomic status, may better account for both more education and higher earnings. We call such a variable that may be affecting the results a hidden variable. Figure 7 shows a clear correlation between the number of radios and the proportion of insane people in England between 1924 and 1937. (People were required to have a license to own a radio.) 25
Mental defectives per 1000
1937 1935 1934
20 1932
1936
1933
1931 15 1930 1928 10 1924
1929
1927 1926 1925
5 0
2000 4000 6000 8000 Licensed radio receivers (1000s)
10,000
Figure 7 A curious correlation? Source: E A. Tufte, Data Analysis for Politics and Policy, p. 90. Copyright © 1974 by Prentice Hall, Inc., Upper Saddle River, NJ. Reprinted with permission.
Are you convinced that radios cause insanity? Or are both variables just increasing with the years? We tend to accept as reasonable the argument that an increase in education causes an increase in personal earnings, because the results seem intuitively possible and they match our preconceptions. But we balk when asked to believe that an increase in radios causes an increase in insanity. Yet the arguments are based on the same sort of statistical reasoning. The flaw in the reasoning is that statistics can show only that events occur together or are correlated, but statistics can never prove that one event causes another. Any time you are tempted to jump to the conclusion that one event causes another because they are correlated, think about the radios in England!
Raising More Questions When a strong link is found between variables, often the next step is to raise questions whose answers may provide more insight into the nature of the relationship. How can the evidence be strengthened? What if we used the mean instead of the median, restricted ourselves to year-round full-time workers, or used other income measures, such as total personal income or total family income? Will the relationship between education and income still hold? Are there other variables that affect the relationship?
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Do Earnings Depend on Age? We started our exploration by looking at how earnings depend on education, because it seems natural that more education might lead to more earnings. But it is equally plausible that a person’s income might depend on his or her age. People may generally earn more as they advance through their working careers, but their earnings usually drop when they eventually retire. We can examine the FAM1000 data to look for evidence to support this hypothesis. It’s hard to see much when we plot all the data points. This time we use mean personal earnings and plot it versus age in Figure 8. The graph seems to suggest that up until about age 50, as age increases, mean personal earnings increase. After age 50, as people move into middle age and retirement, mean personal earnings tend to decrease. So age does seem to affect personal earnings, in a way that is roughly consistent with our intuition. But the relationship appears to be nonlinear, and so linear regression may not be a very effective tool to explore this dependence. 70,000 Mean personal earnings ($)
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60,000 50,000 40,000 30,000 20,000 10,000 0
10
20
30
40
50
60
70
80
90
Age (in years)
Figure 8 Mean personal wages vs. age.
The FAM1000 data set contains internal relationships that are not obvious on a first analysis. We might, for example, also investigate the relationship between education and age. Age may be acting as a hidden variable influencing the relationship between education and income. There are a few simple ways to attempt to minimize the effect of age. For example, we can restrict our analysis to individuals who are all roughly the same age. This sample still would include a very diverse collection of people. More sophisticated strategies involve statistical techniques such as multivariable analysis, a topic beyond the scope of this course.
Do Earnings Depend upon Gender? We can continue to look for relationships in the FAM1000 data set by using some of the other variables to sort the data in different ways. For example, we can look at whether the relationship between earnings and education is different for men and women. One way to do this is to compute the mean personal earnings for each year of education for men and women separately and restrict ourselves to those who work full-time year-round. If we put the data for men and women on the same graph (Figure 9), it is easier to make comparisons. We can see in Figure 9 that for those working full-time year-round, the mean personal earnings of men are consistently higher than the mean personal earnings of women. We can also examine the best-fit lines for mean personal earnings versus education for men and for women shown in Figure 10.
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Mean personal earnings ($)
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150,000 Men Women
125,000 100,000 75,000 50,000 25,000 0
2
4
6
8
10
12
14
16
150,000 125,000 Men cc = 0.95
100,000 75,000 50,000
Women cc = 0.67
25,000 0
4
2
6
8
10
12
14
16
Education (years past grade 8)
Education (years past grade 8)
Figure 9 Mean personal earnings vs. education for women and for men (working full-time yearround).
Figure 10 Regression lines for mean personal earnings vs. education for women and for men (working full-time year-round).
The linear model for mean personal earnings for men working full-time is given by Pmen 5 9965 1 7423E where E 5 years of education past grade 8 and Pmen 5 mean personal earnings for men. The correlation coefficient is 0.95. The rate of change of mean personal earnings with respect to education is approximately $7423/year. For males in this set, the mean personal earnings increase by roughly $7423 for each additional year of education. For women working full-time the comparable linear model is Pwomen 5 19,190 1 3000E where E 5 years of education past grade 8 and Pwomen 5 mean personal earnings for women. The correlation coefficient is 0.67. As you might predict from the relative status of men and women in the U.S. workforce, the rate of change for women is much lower. For women in this sample, the model predicts that the mean personal earnings increase by only $3,000 for each additional year of education. In Figure 10, we can see that the regression line for men is steeper than the one for women when plotted on the same grid. In Figure 9, the mean personal earnings for any particular number of years of education are consistently lower for women than for men. The disparity in mean personal earnings between men and women is most dramatic for those with 14 years of education beyond grade 8 or 22 years of education. Although the vertical intercept of the regression line for men’s personal earnings is below that for women, after 2 years of education beyond grade 8 the regression line for men lies above that for women. Going deeper What other variables have we ignored that we may want to consider in a more refined analysis of the impact of gender on personal earnings? We could, for example, consider type of job or amount of work. • Type of job. Do women and men make the same salaries when they hold the same types of jobs? We could compare only people within the same profession and ask whether the same level of education corresponds to the same level of personal earnings for women as for men. There are many more questions, such as: Are there more men than women in higher-paying professions? Do men and women have the same access to higher-paying jobs, given the same level of education? • Amount of work. In our analysis, we used women and men who were working fulltime to explore the impact of gender on earnings. Typically, part-time jobs pay less than full-time jobs, and more women hold part-time jobs than men. In addition, there are usually more women than men who are unemployed. What prediction would you make if all people in FAM1000 were used to study the impact of gender on earnings?
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How Good Are the Data? For this exploration earnings were defined and measured in a number of ways. What issues are raised by the way income in general is defined and measured? What groups of people may be undercounted in the U.S. Current Population Survey? What are some of the current controversies about how the U.S. Bureau of the Census collects the census data? You may want to search online and in the library for articles on the controversies surrounding the census. Who else collects data, and how can you determine if the data are reliable? As access to data is made easier and easier in our Internet society, the ability to assess the reliability and validity of the data becomes more and more important.
How Good Is the Analysis? What other factors might affect earnings that are not covered by our analysis? What are some limitations to using regression lines to summarize data? Are there hidden variables (such as a history of family wealth or parents’ socioeconomic status) that may affect both level of educational attainment and higher earnings? You may want to explore other methods of analysis that address some of these questions and could potentially reveal different patterns. The following readings at www.wiley.com/college/kimeclark are additional resources for examining the relationship between income and educational attainment in the United States. • The Bureau of Labor Statistics created a website where you can access all the data on earnings and education from the most recent Current Population Survey at http://data.bls.gov/PDQ/outside.jsp?survey=le. This site allows you to search according to categories, access historical data, and create graphs. • U.S. Census Bureau News, Census Bureau Data Underscore Value of College Degree. October 2006. • Bureau of Labor Statistics, Education Pays. U.S. Department of Labor. January 2007. • M. Maier, “Wealth, Income, and Poverty,” from The Data Game: Controversies in Social Science Statistics (New York: M. E. Sharpe, 1999). Reprinted with permission. • Income in the United States: 2002. Bureau of the Census, Current Population Reports, P60–221. September 2003. • The Big Payoff: Educational Attainment and Synthetic Estimates of Work-Life Earnings. Bureau of the Census, Current Population Reports, P23–210. July 2002.
Exploring on Your Own Your journey into exploratory data analysis is just beginning. You now have some tools to examine further the complex relationship between education and income. You may want to explore answers to the questions raised above or to your own questions and conjectures. For example, what other variables besides age do you think affect earnings? What other variables besides gender may affect the relationship between education and earnings? How would our analysis change if we used other income measures, such as personal total income or family income?
Working with Partners You may want to work with a partner so that you can discuss questions that may be worth pursuing and help each other interpret and analyze the findings. In addition, you can compare two regression lines more easily by using two computers or two graphing calculators.
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Generating Conjectures One way to start is to generate conjectures about the effects of other variables or different income measures on the relationship between education and earnings. (See the Data Dictionary in Table 1 for variables included in the FAM1000 Census data.) You can then generate and compare regression lines using the procedures described next.
Procedures for Finding Regression Lines 1. Finding regression lines: a. Using a computer: Open “F3: Regression with Multiple Subsets” in FAM1000 Census Graphs. This program allows you to find regression lines for education vs. earnings for different income variables and for different groups of people. Select (by clicking on the appropriate box) one of the four income variables: personal earnings, personal hourly wage, personal total income, or family income. Then select at least two regression lines that it would make sense to compare (e.g., men vs. women, white vs. non-white, two or more regions of the country). You should do some browsing through the various regression line options to pick those that are the most interesting. Print out your regression lines (on overhead transparencies if possible) so that you can present your findings to the class. DATA
b. Using graphing calculators and graph link files: FAM1000 graph link files A to H contain data for generating regression lines for several income variables as a function of years of education. The Graphing Calculator Manual contains descriptions and hardcopy of the files, as well as instructions for downloading and transferring them to TI-83 and TI-84 calculators. Decide on at least two regression lines that are interesting to compare. 2. For each of the regression lines you choose, work together with your partner to record the following information in your notebooks: The equation of the regression line What the variables represent A reasonable domain The subset of the data the line represents (men? non-whites?) The correlation coefficient Whether or not the line is a good fit and why The slope Interpretation of the slope (e.g., for each additional year of education, median personal earnings rise by such and such an amount)
Discussion/Analysis With your partner, explore ways of comparing the two regression lines. What do the correlation coefficients tell you about the strength of these relationships? How do the two slopes compare? Is one group better off? Is that group better off no matter how many years of education they have? What factors might be hidden or not taken into account? Were your original conjectures supported by your findings? What additional evidence could be used to support your analysis? Are your findings surprising in any way? If so, why? You may wish to continue researching questions raised by your analysis by returning to the original FAM1000 data or examining additional sources such as the related readings at www.wiley.com/college/kimeclark or the Current Population Survey website at www.census.gov/cps.
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159
Results Prepare a 60-second summary of your results. Discuss with your partner how to present your findings. What are the limitations of the data? What are the strengths and weaknesses of your analysis? What factors are hidden or not taken into account? What questions are raised?
EXERCISES Technology is required for generating scatter plots and regression lines in Exercises 11, 12, and 18. 1. a. Evaluate each of the following: k 0.65 k
k 20.68 k
k 20.07 k
k 0.70 k
b. List the absolute values in part (a) in ascending order from the smallest to the largest. 2. The accompanying figures show regression lines and corresponding correlation coefficients for four different scatter plots. Which of the lines describes the strongest linear relationship between the variables? Which of the lines describes the weakest linear relationship? y
y 12
12
10
10
8
8
6
6
4
4
2
2
a. Identify the slope of the regression line, the vertical intercept, and the correlation coefficient. b. What does the slope mean in this context? c. By what amount does this regression line predict that median personal earnings for those who live in the South change for 1 additional year of education? For 10 additional years of education? In Exercises 4 to 6, the data analyzed are from the FAM1000 data files and the equations can be generated using FAM1000 Census Graphs. Here, the income measure is personal total income, which D A T A includes personal earnings (from work) and other sources of unearned income, such as interest and dividends on investments. 4. The accompanying graph and regression line show median personal total income vs. years of education past grade 8.
x
x 0
2
4 6 8 cc = 0.57
10
0
12
2
4 6 8 cc = –0.48
10
12
Graph C
Graph A
y
y 12
12
10
10
8
8
6
6
4
4
2
4 6 8 cc = 0.92
10
12
cc = 0.91 100,000 75,000 50,000 25,000
2
4
6
8
10
12
14
16
Education (years past grade 8)
x
x 0
125,000
0
2
2
Median personal total income ($)
150,000
0
2
Graph B
4 6 8 cc = –0.97
10
Median personal total income 5 23687 1 7994 ? yrs. educ. past grade 8
12
Graph D
Four regression lines with their correlation coefficients. 3. The following equation represents the best-fit regression line for median personal earnings vs. years of education for the 298 people in the FAM1000 data set who live in the southern region of the United States. S 5 22105 1 6139E where S 5 median personal earnings in the South and E 5 years of education past grade 8, and cc 5 0.72.
a. What is the slope of the regression line? b. Interpret the slope in this context. c. By what amount does this regression line predict that median personal total income changes for 1 additional year of education? For 10 additional years of education? d. What features of the data are not well described by the regression line?
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5. The following equation represents a best-fit regression line for median personal total income of white males vs. years of education past grade eight:
Mean height of sons (inches)
74
median personal total incomewhite males 5 28850 1 10,773 ? yrs. educ. past grade 8 The correlation coefficient is 0.87 and the sample size is 452 white males. a. What is the rate of change of median personal total income with respect to years of education? b. Generate three points that lie on this regression line. Use two of these points to calculate the slope of the regression line. c. How does this slope relate to your answer to part (a)? d. Sketch the graph. 6. From the FAM1000 data, the best-fit regression line for median personal total income of white females vs. years of education past grade 8 is: median personal total incomewhite females 5 6760 1 4114 ? yrs. educ. past grade 8 The correlation coefficient is 0.93 and the sample size is 374 white females. a. Interpret the number 4114 in this equation. b. Generate a small table with three points that lie on this regression line. Use two of these points to calculate the slope of the regression line. c. How does this slope relate to your answer to part (a)? d. Sketch the graph. e. Describe some differences between median personal total income vs. education for white females and for white males (see Exercise 5). What are some of the limitations of the model in making this comparison?
S
72 70 68 66
64 0
60 62 64 66 68 70 72 74 Fathers' height (inches)
F
From Snedecor and Cochran, Statistical Methods, 8th ed. By permission of the Iowa State University Press. Copyright © 1967.
a. Interpret the number 0.516 in this context. b. Use the regression line to predict the mean height of sons whose fathers are 64 inches tall and of those whose fathers are 73 inches tall. c. Predict the height of a son who has the same height as his father. d. If there were over 1000 families, why are there only 17 data points on this graph? 8. The book Performing Arts—The Economic Dilemma studied the economics of concerts, operas, and ballets. It included the following scatter plot and corresponding regression line relating attendance per concert to the number of concerts given, for a major orchestra. What do you think were their conclusions? Relation Between Attendance per Concert and Number of Concerts, for a Major Orchestra 3,100 '39 3,000 '45 2,900 '46 2,800 Attendance per concert
7. The term “linear regression” was coined in 1903 by Karl Pearson as part of his efforts to understand the way physical characteristics are passed from generation to generation. He assembled and graphed measurements of the heights of fathers and their fully grown sons from more than a thousand families. The independent variable, F, was the height of the fathers. The dependent variable, S, was the mean height of the sons who all had fathers with the same height. The best-fit line for the data points had a slope of 0.516, which is much less than 1. If, on average, the sons grew to the same height as their fathers, the slope would equal 1. Tall fathers would have tall sons and short fathers would have equally short sons. Instead, the graph shows that whereas the sons of tall fathers are still tall, they are not (on average) as tall as their fathers. Similarly, the sons of short fathers are not as short as their fathers. Pearson termed this regression; the heights of sons regress back toward the height that is the mean for that population. The equation of this regression line is S 5 33.73 1 0.516F, where F 5 height of fathers in inches and S 5 mean height of sons in inches.
2,700
'37
'47
'40
'62 '64 '49
'52
2,600 '38
'55
'50
2,500 '56
2,400
'51
2,300 2,200 2,100
Equation of fitted line: (for 1945–1964) y = 3387–8.067x cc ≈ –0.566
'57
'63
'61
'58 0
50
60
70
80
90
100 110 120 130 140
Number of concerts
Source: William Baumol and William G. Bowen, Performing Arts— The Economic Dilemma. Reprinted with permission from Twentieth Century Fund.
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9. (Optional use of technology.) The accompanying DATA graph gives the mean annual cost for tuition and fees at public and private 4-year colleges in the EDUCOSTS United States since 1985. Mean Cost of 4-Year Colleges, Public and Private $30,000 Public Private
$25,000
Cost
$15,000 $10,000
Percentage of Adult Smokers (18 years and older)
$5,000 $0 1985
1990
1995
2000
2005
Year
Source: U.S. Bureau of the Census, Statistical Abstract of the United States: 2006.
It is clear that the cost of higher education is going up, but public education is still less expensive than private. The graph suggests that costs of both public and private education versus time can be roughly represented as straight lines. a. By hand, sketch two lines that best represent the data. Calculate the rate of change of education cost per year for public and for private education by estimating the coordinates of two points that lie on the line, and then estimating the slope. b. Construct an equation for each of your lines in part (a). (Set 1985 as year 0.) If you are using technology, generate two regression lines from the Excel or graph link data file EDUCOSTS and compare these equations with the ones you constructed. c. If the costs continue to rise at the same rates for both sorts of schools, what would be the respective costs for public and private education in the year 2010? Does this seem plausible to you? Why or why not? 10. Stroke is the third-leading cause of death in the United States, behind heart disease and cancer. The accompanying graph shows the average neuron loss for a typical ischemic stroke. Neuron Loss From a Typical Ischemic Stroke 2.5 Number of neurons in billions
a. Find the slope of the line by estimating the coordinates of two points on the line. Interpret the meaning of the slope in this context. b. Construct an equation for the line, where n 5 number of neurons in billions and d 5 duration of stroke in hours. c. The average human forebrain has about 22 billion neurons and the average stroke lasts about 10 hours. Find the percentage of neurons that are lost from a 10-hour stroke. 11. (Requires technology.) The accompanying table shows (for the years 1965 to 2005 and for people 18 D A T A and over) the total percentage of cigarette smokers, the percentage of males who are smokers, and the SMOKERS percentage of females who are smokers.
$20,000
2.0
1.5
1.0
0.5
0
161
4
8
12
Stroke duration in hours
Source: American Heart Association.
16
Year
Total Population
Males
Females
1965 1974 1979 1983 1985 1987 1988 1990 1991 1992 1993 1994 1995 2000 2003 2005
42.4 37.1 33.5 32.1 30.1 28.8 28.1 25.5 25.6 26.5 25.0 25.5 24.7 23.3 21.6 20.9
51.9 43.1 37.5 35.1 32.6 31.2 30.8 28.4 28.1 28.6 27.7 28.2 27.0 25.7 24.1 23.9
33.9 32.1 29.9 29.5 27.9 26.5 25.7 22.8 23.5 24.6 22.5 23.1 22.6 21.0 19.2 18.1
Source: U.S. Bureau of the Census, Statistical Abstract of the United States: 2006.
a. Construct a scatter plot of the percentage of all smokers 18 and older vs. time. i. Calculate the average rate of change from 1965 to 2005. ii. Calculate the average rate of change from 1990 to 2005. Be sure to specify the units in each case. b. On your graph, sketch an approximate regression line. By estimating coordinates of points on your regression line, calculate the average rate of change of the percentage of total smokers with respect to time. c. Using technology, generate a regression line for the percentage of all smokers 18 and older as a function of time. (Set 1965 as year 0.) Record the equation and the correlation coefficient. How good a fit is this regression line to the data? Compare the rate of change for your hand-generated regression line to the rate of change for the technology-generated regression line. d. Using technology, generate and record regression lines (and their associated correlation coefficients) for the percentages of both males and females who are smokers vs. time. e. Write a summary paragraph using the results from your graphs and calculations to describe the trends in smoking from 1965 to 2005.
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a. Sketch a line that best approximates the data by hand. Set 1972 as year 0 and compute an equation for this line. Interpret the slope of your line in this context. If using technology, generate a regression line from the data in the Excel or graph link file MARATHON and compare its equation with the equation you computed by hand. b. If the marathon times continue to change at the rate given in your linear model, predict the winning running time for the women’s marathon in 2010. Does that seem reasonable? If not, why not? c. The graph seems to flatten out after about 1986. Based on this trend, what would you predict for the winning running time for the women’s marathon in 2010? Does this prediction seem more realistic than your previous prediction? d. Write a short paragraph summarizing the trends in the Boston Marathon winning times for women.
12. (Requires technology.) The accompanying table shows DATA the calories per minute burned by a 154-pound person moving at speeds from 2.5 to 12 miles/hour (mph). CALORIES (Note: A fast walk is about 5 mph; faster than that is considered jogging or slow running.) Marathons, about 26 miles long, are now run in slightly over 2 hours, so that top distance runners are approaching a speed of 13 mph. Speed (mph)
Calories per Minute
2.5 3.0 3.5 4.0 4.5 5.0 5.5
3.0 3.7 4.2 5.5 7.0 8.3 10.1
Speed (mph)
Calories per Minute
6.0 7.0 8.0 9.0 10.0 11.0 12.0
12.0 14.0 15.6 17.5 19.6 21.7 24.5
a. Plot the data. b. Does it look as if the relationship between speed and calories per minute is linear? If so, generate a linear model. Identify the variables and a reasonable domain for the model, and interpret the slope and vertical intercept. How well does your line fit the data? c. Describe in your own words what the model tells you about the relationship between speed and calories per minute.
4.40 DATA
4.30
MARATHON
Boston Marathon Winning Times for Women 200 180 160 Winning time (min)
World Records for Men's Mile 4.50
4.20 Time (minutes)
13. (Optional use of technology.) The accompanying graph shows the winning running times in minutes for women in the Boston Marathon.
14. (Optional use of technology.) The accompanying graph shows the world record times for the men’s D A T A mile. As of January 2006, the 1999 record still stands. Note that several times the standing world MENSMILE record was broken more than once during a year.
4.10 4.00 3.90 3.80
140 120
3.70
100
3.60
80
3.50 1900
60
1920
1940
1960
1980
2000
Year 40
Source: www.runnersworld.com.
20 0 1970
1975
1980 1985 1990 Year
1995
2000
Source: www.bostonmarathon.org/BostonMarathon/ PastChampions.asp.
2005
2010
a. Generate a line that approximates the data (by hand or, if using technology, with the data in the Excel or graph link file MENSMILE. (Set 1910 or 1913 as year 0). If you are using technology, specify the correlation coefficient. Interpret the slope of your line in this context.
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b. If the world record times continue to change at the rate specified in your linear model, predict the record time for the men’s mile in 2010. Does your prediction seem reasonable? If not, why not? c. In what year would your linear model predict the world record to be 0 minutes? Since this is impossible, what do you think is a reasonable domain for your model? Describe what you think would happen in the years after those included in your domain. d. Write a short paragraph summarizing the trends in the world record times for the men’s mile. 15. The temperature at which water boils is affected by the difference in atmospheric pressure at different altitudes above sea level. The classic cookbook The Joy of Cooking by Irma S. Rombauer and Marion Rombauer Becker gives the data in the accompanying table (rounded to the nearest degree) on the boiling temperature of water at different altitudes above sea level.
163
c. Using your formula, find an altitude at which water can be made to boil at 328F, the freezing point of water at sea level. At what altitude would your formula predict this would happen? (Note that airplane cabins are pressurized to near sea-level atmospheric pressure conditions in order to avoid unhealthy conditions resulting from high altitude.) 16. (Optional use of technology.) The accompanying DATA graph shows the world distance records for the women’s long jump. Several times a new longLONGJUMP jump record was set more than once during a given year.
Women's Long Jump World Records 7.75 7.55 7.35
Boiling Temperature of Water Altitude (ft above sea level)
Temperature Boiling 8F
0 2,000 5,000 7,500 10,000 15,000 30,000
212 208 203 198 194 185 158
a. Use the accompanying table for the following: i. Plot boiling temperature in degrees Fahrenheit, 8F, vs. the altitude. Find a formula to describe the boiling temperature of water, in 8F, as a function of altitude. ii. According to your formula, what is the temperature at which water will boil where you live? Can you verify this? What other factors could influence the temperature at which water will boil? b. The highest point in the United States is Mount McKinley in Alaska, at 20,320 feet above sea level; the lowest point is Death Valley in California, at 285 feet below sea level. You can think of distances below sea level as negative altitudes from sea level. At what temperature in degrees Fahrenheit will water boil in each of these locations according to your formula?
Distance (m)
7.15 6.95 6.75 6.55 6.35 6.15 5.95 5.75 1950
1960
1970
1980
1990
Year
Note: As of January 2006, the 1988 long jump record still stands. Source: Data extracted from the website at http://www.uta.fi/~csmipe/sports/eng/mwr.html.
a. Generate a line that approximates the data (by hand or, if using technology, use the data from the Excel or graph link file LONGJUMP. (Set 1950 or 1954 as year 0). Interpret the slope of your line in this context. If you are using technology, specify the correlation coefficient. b. If the world record distances continue to change at the rate described in your linear model, predict the world record distance for the women’s long jump in the year 2010. c. What would your model predict for the record in 1943? How does this compare with the actual 1943 record of 6.25 meters? What do you think would be a reasonable domain for your model? What do you think the data would look like for years outside your specified domain?
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17. (Optional use of technology.) The accompanying graph shows the increasing number of motor vehicle registrations (cars and trucks) in the United States.
DATA
MOTOR
Year 1930 1940 1950 1960 1970 1980 1990 2000 2003 2004
Number of Motor Vehicle Registrations
Registrations (millions)
250
200
150
100
50 0 1940
1960
1980
2000
Year
Source: U.S. Federal Highway Administration, Highway Statistics, annual.
a. Using 1945 as the base year, find a linear equation that would be a reasonable model for the data. If using technology, use the data in the Excel or graph link file MOTOR. b. Interpret the slope of your line in this context. c. What would your model predict for the number of motor vehicle registrations in 2004? How does this compare with the actual data value of 228.3 million? d. Using your model, how many motor vehicles will be registered in the United States in 2010? Do you think this is a reasonable prediction? Why or why not? 18. (Requires technology.) Examine the following data on U.S. union membership from 1930 to 2004.
U.S. Union Membership, 1930–2004
DATA
UNION
Labor Force* (thousands) 29,424 32,376 45,222 54,234 70,920 90,564 103,905 120,786 122,481 123,564
Union Members† (thousands) 3,401 8,717 14,267 17,049 19,381 19,843 16,740 16,258 15,800 15,472
Percentage of Labor Force 11.6 26.9 31.5 31.4 27.3 21.9 16.1 13.5 12.9 12.5
*Does not include agricultural employment; from 1985, does not include self-employed or unemployed persons. †From 1930 to 1980, includes dues-paying members of traditional trade unions, regardless of employment status; from 1985, includes members of employee associations that engage in collective bargaining with employers. Source: Bureau of Labor Statistics, U.S. Dept. of Labor.
a. Graph the percentage of labor force in unions vs. time from 1950 to 2004. Measuring time in years since 1950, find a linear regression formula for these data using technology. b. When does your formula predict that only 10% of the labor force will be unionized? c. What data would you want to examine to understand why union membership is declining? 19. If a study shows that smoking and lung cancer have a high positive correlation, does this mean that smoking causes lung cancer? Explain your answer. 20. Parental income has been found to have a high positive correlation with their children’s academic success. What are two different ways you could interpret this finding?
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WHEN LINES MEET: LINEAR SYSTEMS OVERVIEW When is solar heating cheaper than conventional heating? Will you pay more tax under a flat tax or a graduated tax plan? We can answer such questions using a system of linear equations or inequalities. After reading this chapter, you should be able to • construct, graph, and interpret: systems of linear equations systems of linear inequalities piecewise linear functions • find a solution for: a system of two linear equations a system of two linear inequalities
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3.1 Systems of Linear Equations An Economic Comparison of Solar vs. Conventional Heating Systems On a planet with limited fuel resources, heating decisions involve both monetary and ecological considerations. Typical costs for three different kinds of heating systems for a small one-bedroom housing unit are given in Table 3.1. Typical Costs for Three Heating Systems Type of System
Installation Cost ($)
Operating Cost ($/yr)
Electric Gas Solar
5,000 12,000 30,000
1,100 700 150
Table 3.1
Solar heating is clearly the most costly to install and the least expensive to run. Electric heating, conversely, is the cheapest to install and the most expensive to run. Setting up a system By converting the information in Table 3.1 into equations, we can find out when the solar heating system begins to pay back its initially higher cost. If no allowance is made for inflation or changes in fuel price,1 the general equation for the total cost, C, is C 5 installation cost 1 (annual operating cost)(years of operation) If we let n equal the number of years of operation and use the data from Table 3.1, we can construct the following linear equations: Celectric 5 5000 1 1100n Cgas 5 12,000 1 700n Csolar 5 30,000 1 150n Together they form a system of linear equations. Table 3.2 gives the cost data at 5-year intervals, and Figure 3.1 shows the costs over a 40-year period for the three heating systems. Heating System Total Costs Year
Electric ($)
Gas ($)
Solar ($)
0 5 10 15 20 25 30 35 40
5,000 10,500 16,000 21,500 27,000 32,500 38,000 43,500 49,000
12,000 15,500 19,000 22,500 26,000 29,500 33,000 36,500 40,000
30,000 30,750 31,500 32,250 33,000 33,750 34,500 35,250 36,000
Table 3.2
1
A more sophisticated model might include many other factors, such as interest, repair costs, the cost of depleting fuel resources, risks of generating nuclear power, and what economists call opportunity costs.
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50
167
50 Electric
Electric
40
40 Solar
Cost ($1000s)
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30 Gas
20 10
30 20
P
Gas
10
0
5
10 15 20 25 30 35 40 Years
0
Figure 3.1 Comparison of home
5
10 15 20 25 30 35 40 Years
Figure 3.2 Gas versus electric.
heating costs.
Comparing costs Using Graphs Let’s compare the costs for gas and electric heat. Figure 3.2 shows the graphs of the equations for these two heating systems. The point of intersection, P, shows where the lines predict the same total cost for both gas and electricity, given a certain number of years of operation. From the graph, we can estimate the coordinates of the point P: P 5 (number of years of operation, cost) < s17, $24,000d The total cost of operation is about $24,000 for both gas and electric after about 17 years of operation. We can compare the relative costs of each system to the left and right of the point of intersection. Figure 3.2 shows that gas is less expensive than electricity to the right of the intersection point and more expensive than electricity to the left of the intersection point. Using Equations Where the gas and electric lines intersect, the coordinates satisfy both equations. At the point of intersection, the total cost of electric heat, Celectric, equals the total cost of gas heat, Cgas. Thus, the two expressions for the total cost can be set equal to each other to find the exact values for the coordinates of the intersection point: (1)
Cgas 5 12,000 1 700n
(2)
Celectric 5 Cgas
Set (1) equal to (2) substitute
Celectric 5 5000 1 1100n
5000 1 1100n 5 12,000 1 700n
subtract 5000 from each side subtract 700n from each side divide each side by 400
1100n 5 7000 1 700n 400n 5 7000 n 5 17.5 years
When n 5 17.5 years, the total cost for electric or gas heating is the same. The total cost can be found by substituting this value for n in Equation (1) or (2): Substitute 17.5 for n in Equation (1)
Celectric 5 5000 1 1100(17.5) 5 5000 1 19,250 Celectric 5 $24,250
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Since we claim that the pair of values (17.5, $24,250) satisfies both equations, we need to check, when n 5 17.5 years, that Cgas is also $24,250: Cgas 5 12,000 1 700(17.5)
Substitute 17.5 for n in Equation (2)
5 12,000 1 12,250 Cgas 5 $24,250 The coordinates (17.5, $24,250) satisfy both equations.
Solutions to a system When n 5 17.5 years, then Celectric 5 Cgas 5 $24,250. The point (17.5, $24,250) is called a solution to the system of these two equations. After 17.5 years, a total of $24,250 could have been spent on heat for either an electric or a gas heating system.
A Solution to a System of Equations A pair of real numbers is a solution to a system of equations in two variables if and only if the pair of numbers is a solution to each equation in the system.
1
Estimating solutions to systems from graphs Using the graphs in Figures 3.3 and 3.4, estimate when the cost will be the same for each of the following systems: a. Electric vs. solar heating b. Gas vs. solar heating You will be asked to find more accurate solutions in the exercises.
50
50 Electric
40
40 Solar 30 20
Cost ($1000s)
EXAMPLE
Cost ($1000s)
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Solar 30 20
Gas
10
10
0
5
10 15 20 25 30 35 40 Years
Figure 3.3 Electric versus solar
heating.
0
5
10 15 20 25 30 35 40 Years
Figure 3.4 Gas versus solar heating.
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a. The graphs of electric and solar costs intersect at approximately (26, $34,000). That means the costs of electric and solar heating are both approximately $34,000 at about 26 years of operation. b. Gas and solar heating costs are equivalent at approximately (33, $35,000), or in 33 years at a cost of $35,000.
Algebra Aerobics 3.1 1. Our model tells us that electric systems are cheaper than gas for the first 17.5 years of operation. Using Figure 3.1, estimate the interval over which gas is the cheapest of the three heating systems. When does solar heating become the cheapest system compared with the other two heating systems? 2. For the system of equations
b. How many solutions are there for the system of equations in part (a)? 4. Estimate the solution(s) for each of the systems of equations in Figure 3.5. 8
y
8
4x 1 3y 5 9 5x 1 2y 5 13 a. Determine whether (3, 21) is a solution. b. Show why (1, 4) is not a solution for this system. 3. a. Show that the following equations are equivalent: 4x 5 6 1 3y 12x 2 9y 5 18
–8
8
x
y
–8
–8
8
x
–8
Graph A
Graph B
Figure 3.5 Two graphs of systems of linear equations.
Exercises for Section 3.1 1. a. Determine whether (5, 210) is a solution for the following system of equations:
4. Estimate the solution to the system of linear equations graphed in the accompanying figure.
4x 2 3y 5 50 2x 1 2y 5 5
5
y
b. Explain why (210, 5) is not a solution for the system in part (a). 2. a. Determine whether (22, 3) is a solution for the following system of equations: 3x 1 y 5 23 x 1 2y 5 4
x –5
5
b. Explain why (3, 22) is not a solution for the system in part (a). 3. Explain what is meant by “a solution to a system of equations.”
–5
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5. a. Estimate the coordinates of the point of intersection on the accompanying graph. b. Describe what happens to the population of Pittsburgh in relation to the population of Las Vegas to the right and to the left of this intersection point.
800,000
Population
500,000 400,000 300,000
x2y55 2x 1 y 5 1
9. Create the system of equations that produced the accompanying graph. Estimate the solution for the system from the graph and then confirm using your equations.
200,000 100,000 0 1940 1950
b. x 1 y 5 9 2x 2 3y 5 22
a. Graph the system. Estimate the solution for the system and then find the exact solution. b. Check that your solution satisfies both of the original equations.
Pittsburgh Las Vegas
600,000
a. x 1 2y 5 1 x 1 4y 5 3
8. For the linear system e
Population Trends for Pittsburgh, PA and Las Vegas, NV
700,000
7. Construct a sketch of each system by hand and then estimate the solution(s) to the system (if any).
1960
1970
1980
1990
2000
2010
Year
10
y
6. a. Match each system of linear equations with the graph of the system.
10
y
–10
10
y
x 10
10
x –10
–10
10
Graph C
y
10
–10
10. The U.S. Bureau of Labor Statistics reports on the percent of all men in the civilian work force, and the corresponding percent of all women. Lines of best-fit on the accompanying graph are used to make predictions about the future of the labor force.
–10
Graph A 10
x –10
y
Projected Civilian Labor Participation by Sex
–10
x 10
x –10
90
10
80
Men
70
Graph B
–10 Graph D
Percent
60 –10
Women
50 40 30
i. y 5 2x 2 2 y 5 13 x 1 2 ii. y 2 2x 5 4 y 2 2x 5 24 iii. y 2 x 5 22 3y 2 x 5 6 iv. 3y 1 x 5 23 y 5 22x 1 4 b. Estimate the point of intersection if there is one. c. Verify that the point of intersection satisfies both equations.
20 10 0 1980
1990
2000
2010 Year
2020
2030
a. Estimate the point of intersection. b. Describe the meaning of the point of intersection.
2040
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3.2 Finding Solutions to Systems of Linear Equations In the heating example, we found a solution for a system of two equations by finding the point where the graphs of the two equations intersect. At the point of intersection, the equations have the same value for the independent variable (input) and the same value for the dependent variable (output).
Visualizing Solutions For a single linear equation, the graph of its solutions is one line, and every point on that line is a solution. So, one linear equation has an unlimited number of solutions. In a system of two linear equations, a solution must satisfy both equations. We can easily visualize what might happen. If we graph two different straight lines and the lines intersect (Figure 3.6), there is only one solution—at the intersection point. The coordinates of the point of intersection can be estimated by inspecting the graph. For example, in Figure 3.6 the two lines appear to cross near the point (21, 1.5). If the lines are parallel, they never intersect and there are no solutions (Figure 3.7). One solution
No solutions
y
y
y 8
8
8
8
–8
Infinitely many solutions
x
x –8
8
8
–8
–8
–8
x –8
Figure 3.6 Lines intersect at a
Figure 3.7 Parallel lines never
Figure 3.8 The two equations
single point.
intersect.
represent the same line. Every point on the line is a solution to both equations.
How else might the graphs of two lines be related? The two equations could represent the same line. Consider the following two equations: y 5 0.5x 1 2 3y 5 1.5x 1 6
(1) (2)
If we multiply each side of Equation (1) by 3, we obtain Equation (2). The two equations are equivalent, since any solution of one equation is also a solution of the other equation. The two equations represent the same line. There are an infinite number of points on that line, and they are all solutions to both equations (Figure 3.8).
The Number of Solutions for a Linear System On the graph of a system of two linear equations, a solution is a point where the two lines intersect. There can be One solution, if the lines intersect once No solution, if the lines are parallel and distinct Infinitely many solutions, if the two lines are the same
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Strategies for Finding Solutions A system of two linear equations can be solved in several ways, each of which will provide the same result. In each case a solution, if it exists, consists of either one or infinitely many pairs of numbers of the form (x, y). The form of the equations can help determine the most efficient strategy. Two of the most common methods are substitution and elimination. Substitution method When at least one of the equations is in (or can easily be converted to) function form, y 5 b 1 mx, we can use the substitution method. The point of intersection is a solution to both equations, so at that point the two equations have the same value for x and the same value for y. To find values for the intersection point, we can start by substituting the expression for y from one equation into the other equation. EXAMPLE
1
SOLUTION
When both equations are in function form A long-distance phone plan seen on TV costs $0.03 per minute plus a fixed charge of $0.39 per call. Your current service charges $0.05 per minute plus a fixed charge of $0.20 per call. During one call, after how many minutes would the cost be the same under the two plans? If we let n 5 the number of minutes in one call, then C1 5 0.39 1 0.03n C2 5 0.20 1 0.05n
models the cost of one call on the TV plan models the cost of one call on your current plan
This is the simplest case of substitution, where both equations are already in function form. To find out when the costs of the two plans are equal, that is, when C1 5 C2, we can substitute C2 for C1. C1 5 0.39 1 0.03n 0.20 1 0.05n 5 0.39 1 0.03n 0.05n 2 0.03n 5 0.39 2 0.20 0.02n 5 0.19 n 5 0.19/ 0.02 5 9.5 minutes
Given the equation for C1 substitute C2 for C1 solve for n
When n 5 9.5, the costs of the two plans are equal. To find the cost, we can substitute n 5 9.5 into the equation for C1 or C2. Using C1, we have C1 5 0.39 1 0.03(9.5) 5 0.39 1 0.285 5 $0.675 So for a call lasting 9.5 minutes, the cost on either plan is approximately 68 cents. Double-Checking the Solution. into C2.
We can check our answer by substituting n 5 9.5
C2 5 0.20 1 0.05(9.5) 5 0.20 1 0.475 5 $0.675 This confirms our original calculation. EXAMPLE
2
When only one equation is in function form a. Find the point (if any) where the graphs of the following two linear equations intersect: 6x 1 7y 5 25 y 5 15 1 2x b. Graph the two equations on the same grid, labeling their intersection point.
(1) (2)
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a. In Equation (1) substitute the expression for y from Equation (2): 6x 1 7(15 1 2x) 5 25 6x 1 105 1 14x 5 25 20x 5 280 x 5 24
Simplify
We can use one of the original equations to find the value for y when x 5 24. Using Equation (2), y 5 15 1 2x y 5 15 1 2(24) y 5 15 2 8 y57
Substitute 24 for x multiply
(2)
Try double-checking your answer in Equation (1). b. Figure 3.9 shows a graph of the two equations and the intersection point at (24, 7). y 20
y = 15 + 2x
(–4, 7)
6x + 7y = 25
x –8
8
–20
Figure 3.9 Graphs of 6x 1 7y 5 25 and y 5 15 1 2x.
Algebra Aerobics 3.2a 1. Solve for the indicated variable. a. 2x 1 y 5 7 for y b. 3x 1 5y 5 6 for y c. x 2 2y 5 21 for x 2. Determine the number of solutions without solving the system. Justify your answer. a. y 5 3x 2 5 b. y 5 2x 2 4 y 5 3x 1 8 y 5 3x 2 4 3. Solve the following systems of equations using the substitution method. a. y 5 x 1 4 b. y 5 21700 1 2100x y 5 22x 1 7 y 5 4700 1 1300x
c. F 5 C F 5 32 1 95C [Part (c) was a question on the TV program Who Wants to Be a Millionaire?] 4. Solve the following systems of equations using the substitution method. a. y 5 x 1 3 c. x 5 2y 2 5 5y 2 2x 5 21 4y 2 3x 5 9 b. z 5 3w 1 1 d. r 2 2s 5 5 9w 1 4z 5 11 3r 2 10s 5 13
Elimination method Another method, called elimination, can be useful when neither equation is in function form. The strategy is to modify the equations (through multiplication or rearrangement) so that adding (or subtracting) the modified equations eliminates one variable.
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EXAMPLE
3
SOLUTION
Assume you have $10,000 to invest in an “up market” when the economy is booming. You want to split your investment between conservative bonds and riskier stocks. The bonds will stay fixed in value but return a guaranteed 7% per year in dividends. The stocks pay no dividends, but your return is from the increase in stock value, predicted to be 14% per year. Overall you want a 12% or $1200 return on your $10,000 at the end of one year. a. How much should you invest in bonds and how much in stocks? b. What if the economy has a drastic downturn, as it did between 1999 and 2002? Assuming you split your investment as recommended in part (a), what will your return be after one year if your stock value decreased by 10%? a. If B 5 $ invested in bonds and S 5 $ invested in stocks, then B 1 S 5 $10,000
(1)
The expected return on your investments after one year is (7% of B) 1 (14% of S) 5 $1200 0.07B 1 0.14S 5 $1200
or
(2)
We can solve the system by eliminating one variable, in this case B, from both equations. Given the two equations B 1 S 5 $10,000 0.07B 1 0.14S 5 $1200
(1) (2)
If we multiply both sides of Equation (1) by 0.07 we get an equivalent Equation (1)*, which has the same coefficient for B as Equation (2). We do this so that we can subtract the equations and eliminate B. Given subtract Equation (2) to eliminate B
0.07B 1 0.07S 5 $700 2(0.07B 1 0.14S 5 $1200)
(1)* (2)
20.07S 5 2$500
Dividing both sides by 20.07, we have S < $7143. So to achieve your goal of a $1200 return on $10,000 you would need to invest $7143 in stocks and $10,000 2 $7143 5 $2857 in bonds. b. If the economy turns sour at the end of the year and the value of your stock drops 10%, then your return 5 (7% of $2857 from bonds) 2 (10% of $7143 from stocks) < $200 2 $714 5 2$514 So you would lose over $500 on your $10,000 investment that year. How can you tell if your system has no or infinitely many intersection points? As we remarked at the beginning of this section, two parallel lines will never intersect and duplicate lines will intersect everywhere, creating infinitely many intersection points. How can you tell whether or not your system of equations falls into either category?
EXAMPLE
4
A system with no solution: Parallel lines a. Solve the following system of two linear equations: y 5 20,000 1 1500x 2y 2 3000x 5 50,000 b. Graph your results.
(1) (2)
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a. If we assume the two lines intersect, then the y values for Equations (1) and (2) are the same at some point. We can try to find this value for y by substituting the expression for y from Equation (1) into Equation (2): Simplify false statement
2(20,000 1 1500x) 2 3000x 5 50,000 40,000 1 3000x 2 3000x 5 50,000 40,000 5 50,000 (???)
What could this possibly mean? Where did we go wrong? If we return to the original set of equations and solve Equation (2) for y in terms of x, we can see why there is no solution for this system of equations. Add 3000x to both sides divide by 2
2y 2 3000x 5 50,000 2y 5 50,000 1 3000x y 5 25,000 1 1500x
(2) (2)*
Equations (1) and (2)* (rewritten form of the original Equation (2)) have the same slope of 1500 but different y-intercepts. y 5 20,000 1 1500x y 5 25,000 1 1500x
(1) (2)*
Written in this form, we can see that we have two parallel lines, so the lines never intersect (see Figure 3.10). Our initial premise, that the two lines intersected and thus the two y-values were equal at some point, was incorrect. b. Figure 3.10 shows the graphs of the two lines. y 35,000
y = 25,000 + 1500x
y = 20,000 + 1500x
x –10
0
10
Figure 3.10 Graphs of
y 5 25,000 1 1500x and y 5 20,000 1 1500x. EXAMPLE
5
A system with infinitely many solutions: Equivalent equations Solve the following system: 45x 5 2y 1 33 2y 1 90x 5 66
SOLUTION
(1) (2)
As always, there are multiple ways of solving the system. One strategy is to put both equations in function form: Solve Equation (1) for y add y to both sides add 245x to both sides Solve Equation (2) for y add 290x to both sides divide by 2
45x 5 2y 1 33 y 1 45x 5 33 y 5 245x 1 33 2y 1 90x 5 66 2y 5 290x 1 66 y 5 245x 1 33
(1)
(2)
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The two original equations really represent the same line, y 5 245x 1 33, so any of the infinitely many points on the line is a solution to the system (Figure 3.11).
y 400
x –10
10
–200
Figure 3.11 Graph of
y 5 245x 1 33.
Linear Systems in Economics: Supply and Demand Economists study the relationship between the price p of an item and the quantity q, the number of items produced. Economists traditionally place quantity q on the horizontal axis and price p on the vertical axis.2 From the consumer’s point of view, an increase in price decreases the quantity demanded. So the consumer’s demand curve would slope downward (see Figure 3.12).
p Demand curve
Supply curve
Price
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Equilibrium point
Quantity
q
Figure 3.12 Supply and demand curves.
From the manufacturer’s (or supplier’s) point of view, an increase in price is linked with an increase in the quantity they are willing to supply. So the manufacturer’s supply curve slopes upward. The intersection point between the demand and supply curves is called the equilibrium point. At this point supply equals demand, so both suppliers and consumers are happy with the quantity produced and the price charged.
2
This can be confusing since we usually think of quantity as a function of price.
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EXAMPLE
6
SOLUTION
177
Milk supply and demand curves Loren Tauer3 studied the U.S. supply and demand curves for milk. If q 5 billions of pounds of milk and p 5 dollars per cwt (where 1 cwt 5 100 lb), he estimated that the demand function for milk is p 5 55.9867 2 0.2882q and the supply function is p 5 0.0865q. a. Find the equilibrium point. b. What will happen if the price of milk is higher than the equilibrium price? a. The equilibrium point occurs where supply 5 demand 0.0865q 5 55.9867 2 0.2882q 0.3747q 5 55.9867 q < 149.42 billions of pounds of milk
substituting for p solving for q we have
If we use the supply function to find p, we have the supply function when q 5 149.42
p 5 0.0865q p 5 0.0865 ? 149.42 < $12.92 per cwt
If we use the demand function to find p we would also get the demand function when q 5 149.42
p 5 55.9867 2 0.2882q p 5 55.9867 2 s0.2882 ? 149.42d < $12.92
So the equilibrium point is (149.42, $12.92); that is, when the price is $12.92 per cwt, manufacturers are willing to produce and consumers are willing to buy 149.42 billion pounds of milk. b. If the price of milk rises above $12.92 per cwt to, say, p1, then, as shown in Figure 3.13, consumers would buy less than 149.42 billion pounds (amount q1) while manufacturers would be willing to produce more than 149.42 billion pounds (amount q2). p Demand
Supply
p1
q1
q2
q
Figure 3.13 At price p1,
consumers would buy q1 billion pounds of milk, but producers would manufacture q2 billion pounds.
So there would be a surplus of q2 2 q1 billion pounds of milk, which would drive down the price of milk toward the equilibrium point. 3
Loren W. Tauer, “The value of segmenting the milk market into bST-produced and non-bST produced milk,” Agribusiness 10(1): 3–12 as quoted in Edmond C. Tomastik, Calculus: Applications and Technology, 3rd ed. (Belmont, CA: Thomson Brooks/Cole, 2004).
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Algebra Aerobics 3.2b 1. Solve each system of equations using the method you think is most efficient. a. 2y 2 5x 5 21 c. t 5 3r 2 4 3y 1 5x 5 11 4t 1 6 5 7r b. 3x 1 2y 5 16 d. z 5 2000 1 0.4(x 2 10,000) 2x 2 3y 5 211 z 5 800 1 0.2x 2. Solve each system of equations. If technology is available, check your answers by graphing each system. a. y 5 2x 1 4 c. y 5 1500 1 350x y 5 2x 1 4 2y 5 700x 1 3500 b. 5y 1 30x 5 20 y 5 26x 1 4 3. Construct a system of two linear equations in two unknowns that has no solution. 4. Determine the number of solutions without solving the system. Explain your reasoning. a. 2x 1 5y 5 7 c. 2x 1 3y 5 1 3x 2 8y 5 21 4x 1 6y 5 2 b. 3x 1 y 5 6 d. 3x 1 y 5 8 6x 1 2y 5 5 3x 1 2y 5 8 5. Solve each of the following systems of equations. a. y 5 x 1 4 b. 0.5x 1 0.7y 5 10 y x 30x 1 50y 5 1000 2 1 3 5 3
6. A small paint dealer has determined that the demand function for interior white paint is 4p 1 3q 5 240, where p 5 dollars/gallon of paint and q 5 number of gallons. a. Find the demand for white paint when the price is $39.00 per gallon. b. If consumer demand is for 20 gallons of paint, what price would these consumers be willing to pay? c. Sketch the demand function, placing q on the horizontal axis and p on the vertical axis. d. The supply function for interior white paint is p 5 0.85q. Sketch the supply curve on the same graph as the demand curve. e. Find the equilibrium point and interpret its meaning. f. At a price of $39.00 per gallon of paint, is there a surplus or shortage of supply? 7. Fill in the missing coefficient of x such that there will be an infinite number of solutions to the system of equations: y 5 2x 1 4 ?? x 5 22y 1 8
Exercises for Section 3.2 1. Two companies offer starting employees incentives to stay with the company after they are trained for their new jobs. Company A offers an initial hourly wage of $7.00, then increases the hourly wage by $0.15 per month. Company B offers an initial hourly wage of $7.45, then increases the hourly wage by $0.10 per month. Wage Comparison Company A and Company B $9.50
Hourly wage (dollars)
$9.00
b. Estimate that hourly wage. c. Form two linear functions for the hourly wages in dollars of WA smd for company A and WB smd for company B after m months of employment. d. Does your estimated solution from part (a) satisfy both equations? If not, find the exact solution. e. What is the exact hourly wage when the two companies offer the same wage? f. Describe the circumstances under which you would rather work for company A. For company B. 2. In the text the following cost equations were given for gas and solar heating:
$8.50
$8.00
Cgas 5 12,000 1 700n
B
$7.50
Csolar 5 30,000 1 150n
A
$7.00 $6.50 0
2
8 4 6 Months of employment
10
12
a. Examine the accompanying graph. After how many months does it appear that the hourly wage will be the same for both companies?
where n represents the number of years since installation and the cost represents the total accumulated costs up to and including year n. a. Sketch the graph of this system of equations. b. What do the coefficients 700 and 150 represent on the graph, and what do they represent in terms of heating costs? c. What do the constant terms 12,000 and 30,000 represent on the graph? What does the difference between 12,000 and 30,000 say about the costs of gas vs. solar heating?
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d. Label the point on the graph where the total accumulated gas and solar heating costs are equal. Make a visual estimate of the coordinates, and interpret what the coordinates mean in terms of heating costs. e. Use the equations to find a better estimate for the intersection point. To simplify the computations, you may want to round values to two decimal places. Show your work. f. When is the total cost of solar heating more expensive than gas? When is the total cost of gas heating more expensive than solar? 3. Answer the questions in Exercise 2 (with suitable changes in wording) for the following cost equations for electric and solar heating: Celectric 5 5000 1 1100n
8. For each graph, construct the equations for each of the two lines in the system, and then solve the system using your equations. y
5. a. Solve the following system algebraically: S 5 20,000 1 2500n S 5 25,000 1 2000n b. Graph the system in part (a) and use the graph to estimate the solution to the system. Check your estimate with your answer in part (a). 6. Predict the number of solutions to each of the following systems. Give reasons for your answer. You don’t need to find any actual solutions. a. y 5 20,000 1 700x y 5 15,000 1 800x b. y 5 20,000 1 700x y 5 15,000 1 700x c. y 5 20,000 1 700x y 5 20,000 1 800x 7. For each system: a. Indicate whether the substitution or elimination method might be easier for finding a solution to the system of equations. i. y 5 13 x 1 6 iv. y 5 2x 2 3 1 4y 2 8x 5 212 y 5 3x 2 4 ii. 2x 2 y 5 5 5x 1 2y 5 8 iii. 3x 1 2y 5 2 x 5 7y 2 30
v. 23x 1 y 5 4 23x 1 y 5 22 vi. 3y 5 9 x 1 2y 5 11
b. Using your chosen method, find the solution(s), if any, of each system.
y
8
8
x –8
8
x –8
8
–8
–8
Graph A
Graph B
9. a. Solve the following system algebraically:
Csolar 5 30,000 1 150n 4. Consider the following job offers. At Acme Corporation, you are offered a starting salary of $20,000 per year, with raises of $2500 annually. At Boca Corporation, you are offered $25,000 to start and raises of $2000 annually. a. Find an equation to represent your salary, SA snd , after n years of employment with Acme. b. Find an equation to represent your salary, SB snd , after n years of employment with Boca. c. Create a table of values showing your salary at each of these corporations for integer values of n up to 12 years. d. In what year of employment would the two corporations pay you the same salary?
179
x 1 3y 5 6 5x 1 3y 5 26 b. Graph the system of equations in part (a) and estimate the solution to the system. Check your estimate with your answers in part (a). 10. Calculate the solution(s), if any, to each of the following systems of equations. Use any method you like. a. y 5 21 2 2x c. y 5 2200x 2 700 y 5 13 2 2x y 5 1300x 1 4700 b. t 5 23 1 4w d. 3x 5 5y 212w 1 3t 1 9 5 0 4y 2 3x 5 23 In some of the following examples you may wish to round off your answers: e. y 5 2200x 2 1800 h. 2x 1 3y 5 13 y 5 1300x 2 4700 3x 1 5y 5 21 f. y 5 4.2 2 1.62x i. xy 5 1 x 2y 1 3x 5 2 1.48x 2 2y 1 4.36 5 0 g. 4r 1 5s 5 10 (A nonlinear system! 2r 2 4s 5 23 Hint: Solve xy 5 1 for y and use substitution.) 11. Assume you have $2000 to invest for 1 year. You can make a safe investment that yields 4% interest a year or a risky investment that yields 8% a year. If you want to combine safe and risky investments to make $100 a year, how much of the $2000 should you invest at the 4% interest? How much at the 8% interest? (Hint: Set up a system of two equations in two variables, where one equation represents the total amount of money you have to invest and the other equation represents the total amount of money you want to make on your investments.) 12. Two investments in high-technology companies total $1000. If one investment earns 10% annual interest and the other earns 20%, find the amount of each investment if the total interest earned is $140 for the year (clearly in dot com days). 13. Solve the following systems: y a. 3x 1 2 5 1 b. 4x 1 y 5 9 x 2 y 5 43
y 5 2x
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14. For each of the following systems of equations, describe the graph of the system and determine if there is no solution, an infinite number of solutions, or exactly one solution. a. 2x 1 5y 5 210 b. 3x 1 4y 5 5 c. 2x 2 y 5 5 y 5 20.4x 2 2 3x 2 2y 5 5 6x 2 3y 5 4 15. If y 5 b 1 mx, solve for values for m and b by constructing two linear equations in m and b for the given sets of ordered pairs. a. When x 5 2, y 5 22 and when x 5 23, y 5 13. b. When x 5 10, y 5 38 and when x 5 1.5, y 5 24.5. 16. The following are formulas predicting future raises for four different groups of union employees. N represents the number of years from the start date of all the contracts. Each equation represents the salary that will be earned after N years. Group A: Group B: Group C: Group D:
Salary 5 30,000 1 1500N Salary 5 30,000 1 1800N Salary 5 27,000 1 1500N Salary 5 21,000 1 2100N
a. Will group A ever earn more per year than group B? Explain. b. Will group C ever catch up to group A? Explain. c. Which group will be making the highest yearly salary in 5 years? How much will that salary be? d. Will group D ever catch up to group C? If so, after how many years and at what salary? e. How much total salary would an individual in each group have earned 3 years after the contract?
a. Sketch both equations on the same graph. On your graph identify the supply equation and the demand equation. b. Find the equilibrium point and interpret its meaning. 20. For a certain model of DVD-VCR combo player, the following supply and demand equations relate price per player, p (in dollars) and number of players, q (in thousands).
a. b. c. d.
and
Demand
21. Explain what is meant by “two equivalent equations.” Give an example of two equivalent equations. 22. Construct a problem not found in this text that involves supply and demand where the situation can be modeled with a system of linear equations. Solve your system and verify your results by graphing the system. 23. A restaurant is located on ground that slopes up 1 foot for every 20 horizontal feet. The restaurant is required to build a wheelchair ramp starting from an entry platform that is 3 feet above ground. Current regulations require a wheelchair ramp to rise up 1 foot for every 12 horizontal feet, (See accompanying figure where H 5 height in feet, d 5 distance from entry in feet, and the origin is where the H-axis meets the ground.) Where will the new ramp intersect the ground? H
New ramp 1/12 slope Ground slope 1/20 3' Distance from entry platform
R 5 15.5x
What does 12.5 represent? What does 15.5 represent? Find the equilibrium point. What is the cost of producing x units at the equilibrium point? The revenue at the equilibrium point?
18. A large wholesale nursery sells shrubs to retail stores. The cost C(x) and revenue R(x) equations (in dollars) for x shrubs are C(x) 5 15x 1 12,000
Supply
p 5 185 2 5q a. Find the point of equilibrium. b. Interpret this result.
17. A small T-shirt company created the following cost and revenue equations for a line of T-shirts, where cost C is in dollars for producing x units and revenue R is in dollars from selling x units: C 5 12.5x 1 360
p 5 80 1 2q
and
d
24. A house attic as shown has a roofline with a slope of 50 up for every 120 of horizontal run; this is a slope of 5/12. Since the roofline starts at 49 above the floor, it is not possible to stand in much of the attic space. The owner wants to add a dormer with a 2/12 slope, starting at a 79 height, to increase the usable space.
Roofline height (feet)
Proposed dormer
R(x) 5 18x
a. Find the equilibrium point. b. Explain the meaning of the coordinates for the equilibrium point. 19. The supply and demand equations for a particular bicycle model relate price per bicycle, p (in dollars) and q, the number of units (in thousands). The two equations are p 5 250 1 40q
Supply
p 5 510 2 25q
Demand
Existing roof
7'
4'
Attic interior
18'
Floor distance (feet)
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a. With height and floor distance coordinates as shown on the house sketch, find a formula for the original roofline, using R for roof height and F for floor distance from the wall. Also find a formula for the dormer roofline, using D for dormer roof height and F for floor distance from the wall. b. Make a graph showing the R and D roof height lines for floor distances F from 09 to 189. c. At what height and floor distance will the dormer roofline intersect the existing roofline? d. How far do you need to measure along the horizontal floor distance to give 6960 of head room in the original roofline? What percent of the horizontal floor distance of 18 ft allows less than 6960 of head room?
30. While totally solar energy–powered home energy systems are quite expensive to install, passive solar systems are much more economical. Many passive solar features can be incorporated at the time of construction with a small additional initial cost to a conventional system. These features enable energy costs to be one-half to one-third of the costs in conventional homes. Below is the cost analysis from the case study Esperanza del Sol.4 Cost of installation of a conventional system: Additional cost to install passive solar features: Annual energy costs for conventional system: Annual energy costs of hybrid system with additional passive solar features:
25. Solve the following system of three equations in three variables, using the steps outlined below: 2x 1 3y 2 z 5 11
(1)
5x 2 2y 1 3z 5 35
(2)
x 2 5y 1 4z 5 18
(3)
a. Use Equations (1) and (2) to eliminate one variable, creating a new Equation (4) in two variables. b. Use Equations (1) and (3) to eliminate the same variable as in part (a). You should end up with a new Equation (5) that has the same variables as Equation (4). c. Equations (4) and (5) represent a system of two equations in two variables. Solve the system. d. Find the corresponding value for the variable eliminated in part (a). e. Check your work by making sure your solution works in all three original equations. 26. Using the strategy described in Exercise 25, solve the following system: 2a 2 3b 1 c 5 4.5
(1)
a 2 2b 1 2c 5 0
(2)
3a 2 b 1 2c 5 0.5
(3)
27. a. Construct a system of linear equations in two variables that has no solution. b. Construct a system of linear equations in two variables that has exactly one solution. c. Solve the system of equations you constructed in part (b) by using two different algebraic strategies and by graphing the system of equations. Do your answers all agree? 28. Nenuphar wants to invest a total of $30,000 into two savings accounts, one paying 6% per year in interest and the other paying 9% per year in interest (a more risky investment). If after 1 year she wants the total interest from both accounts to be $2100, how much should she invest in each account?
181
$10,000 $150 $740 $540
a. Write the cost equation for the conventional system. b. Write the cost equation for the passive hybrid solar system. c. When would the total cost of the passive hybrid solar system be the same as the conventional system? d. After 5 years, what would be the total energy cost of the passive hybrid solar system? What would be the total cost of the conventional system? 31. a. Construct a system of linear equations where both of the following conditions are met: The coordinates of the point of intersection are (2, 5). One of the lines has a slope of 24 and the other line has a slope of 3.5. b. Graph the system of equations you found in part (a). Verify that the coordinates of the point of intersection are the same as the coordinates specified in part (a). 32. A husband drives a heavily loaded truck that can go only 55 mph on a 650-mile turnpike trip. His wife leaves on the same trip 2 hours later in the family car averaging 70 mph. Recall that distance traveled 5 speed ? time traveled. a. Derive an expression for the distance, Dh, the husband travels in t hours since he started. b. How many hours has the wife been traveling if the husband has traveled t hours st $ 2d ? c. Derive an expression for the distance, Dw, that the wife will have traveled while the husband has been traveling for t hours st $ 2d . d. Graph distance vs. time for husband and wife on the same axes. e. Calculate when and where the wife will overtake the husband. f. Suppose the husband and wife wanted to arrive at a restaurant at the same time, and the restaurant is 325 miles from home. How much later should she leave, assuming he still travels at 55 mph and she at 70 mph?
29. When will the following system of equations have no solution? Justify your answer. y 5 m 1x 1 b1 y 5 m 2x 1 b2
4
Adapted from Buildings for a Sustainable America: Case Studies, American Solar Energy Society, Boulder, CO.
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33. Life-and-death travel problems are dealt with by air traffic computers and controllers who are trying to prevent collisions of planes traveling at various speeds in threedimensional space. To get a taste of what is involved, consider this situation: Airplanes A and B are traveling at the same altitude on the paths shown on the position plot.
y miles (North)
35. a. Examine the accompanying figure. Does the new supply curve represent an increase or decrease in supply. Why? (Again, try picking an arbitrary price and see if at that price, the supplier would want to increase or decrease production.) b. Sketch in a possible demand curve, and label the old and new equilibrium points. What does the shift from the old to the new equilibrium point mean for both consumers and suppliers? Old supply curve
Airplane A
50
Airplane B
30
P
(West)
–30
80
New supply curve
Price
x miles (East)
(South)
Qnew
Qold
a. Construct two equations that describe the positions of airplanes A and B in x- and y-coordinates. Use yA and yB to denote the north/south coordinates of airplanes A and B, respectively. b. What are the coordinates of the intersection of the airplanes’ paths? c. Airplane A travels at 2 miles/minute and airplane B travels at 6 miles/minute. Clearly, their paths will intersect if they each continue on the same course, but will they arrive at the intersection point at the same time? How far does plane A have to travel to the intersection point? How many minutes will it take to get there? How far does plane B have to travel to the intersection point? How many minutes will it take to get there? (Hint: Recall the rule of Pythagoras for finding the hypotenuse of a right triangle: a2 1 b2 5 c2, where c is the hypotenuse and a and b are the other sides.) Is this a safe situation? 34. a. Examine the accompanying figure, where the demand curve has been moved to the right. Does the new demand curve represent an increase or decrease in demand. Why? (Hint: Pick an arbitrary price, and see if consumers would want to buy more or less at that price.)
New demand curve
Price
P Old demand curve
Qnew
Qold Quantity
b. Sketch in a possible supply curve identifying the old and new equilibrium points. What does the shift from the old equilibrium point to the new mean for both consumers and suppliers?
Quantity
36. In studying populations (human or otherwise), the two primary factors affecting population size are the birth rate and the death rate. There is abundant evidence that, other things being equal, as the population density increases, the birth rate tends to decrease and the death rate tends to increase.5 a. Generate a rough sketch showing birth rate as a function of population density. Note that the units for population density on the horizontal axis are the number of individuals for a given area. The units on the vertical axis represent a rate, such as the number of individuals per 1000 people. Now add to your graph a rough sketch of the relationship between death rate and population density. In both cases assume the relationship is linear. b. At the intersection point of the two lines the growth of the population is zero. Why? (Note: We are ignoring all other factors, such as immigration.) The intersection point is called the equilibrium point. At this point the population is said to have stabilized, and the size of the population that corresponds to this point is called the equilibrium number. c. What happens to the equilibrium point if the overall death rate decreases, that is, at each value for population density the death rate is lower? Sketch a graph showing the birth rate and both the original and the changed death rates. Label the graph carefully. Describe the shift in the equilibrium point. d. What happens to the equilibrium point if the overall death rate increases? Analyze as in part (c). 37. Use the information in Exercise 36 to answer the following questions: a. What if the overall birth rate increases (that is, if at each population density level the birth rate is higher)? Sketch a 5
See E. O. Wilson and W. H. Bossert, A Primer oƒ Population Biology. Sunderland, MA: Sinauer Associates, 1971, p. 104.
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it was worth $850,000 as of 2007. Alas, one of her colleagues has not been so fortunate. He bought a house in that same year for $160,000. Not long after his family moved in, rumors began to circulate that the housing complex had been built on the site of a former toxic dump. Although never substantiated, the rumors adversely affected the value of his home, which has steadily decreased in value over the years and in 2007 was worth a meager $40,000. In what year would the two homes have been assessed at the same value?
graph showing the death rate and both the original and the changed birth rates. Be sure to label the graph carefully. Describe the shift in the equilibrium point. b. What happens if the overall birth rate decreases? Analyze as in part (a). 38. In Section 2.7, Exercise 19, you read of a math professor who purchased her condominium in Cambridge, MA, for $70,000 in 1977. Its assessed value has climbed at a steady rate so that
3.3 Reading between the Lines: Linear Inequalities Above and Below the Line Sometimes we are concerned with values that lie above or below a line, or that lie between two lines. To describe these regions we need some mathematical conventions. Terminology for describing regions The two linear functions y1 5 1 1 3x and y2 5 5 2 x are graphed in Figure 3.14. How would you describe the various striped regions? y 20
y1
x y2 6
–2 –5
Figure 3.14 Regions bounded by the lines y1 5 1 1 3x and y2 5 5 2 x.
A solid line indicates that the points on the line are included in the area. A dotted line indicates that the points on the line are not included in the area. So the verticalstriped region below the solid line y1 can be described as all points (x, y) that satisfy the inequality y # y1 or equivalently
Condition (1)
y # 1 1 3x
The equation y1 5 1 1 3x is a boundary line that is included in the region. The horizontally striped region above the dotted line y2 can be described as all points (x, y) that satisfy the inequality y2 , y or
Condition (2)
52x,y
So y2 5 5 2 x is a boundary line that is not included in the region. In the cross-hatched region, the y values must satisfy both conditions (1) and (2); that is, we must have y2 , y # y1 or equivalently
5 2 x , y # 1 1 3x
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This is called a compound inequality. We could describe this inequality by saying that y is greater than 5 2 x and less than or equal to 1 1 3x. So the region can be described as all points (x, y) that satisfy the compound inequality 5 2 x , y # 1 1 3x
Manipulating Inequalities Recall that any term may be added to or subtracted from both sides of an inequality without changing the direction of the inequality. The same holds for multiplying or dividing by a positive number. However, multiplying or dividing by a negative number reverses the inequality. For example, Given the previous inequality if we wanted to solve for x we could subtract 5 from both sides
52x,y
then multiply both sides by 21
x . 2y 1 5
2x , y 2 5
Note that subtracting 5 (or equivalently adding 25) preserved the inequality, but multiplying by 21 reversed the inequality. So “,” in the first two inequalities became “.” in the last inequality.
EXAMPLE
1
The U.S. Army recommends that sleeping bags, which will be used in temperatures between 2408 and 1408 Fahrenheit, have a thickness of 2.5 inches minus 0.025 times the number of degrees Fahrenheit. a. Construct a linear equation that models the recommended sleeping bag thickness as a function of degrees Fahrenheit. Identify the variables and domain of your function. b. Graph the model (displaying it over its full domain) and shade in the area where the sleeping bag is not warm enough for the given temperature. c. What symbolic expressions would describe the shaded area? d. If a manufacturer submitted a sleeping bag to the Army with a thickness of 2.75 inches, would it be suitable for i. 2208 Fahrenheit? ii. 08 Fahrenheit? iii. 1208 Fahrenheit?
SOLUTION
a. If we let F 5 number of degrees Fahrenheit and T 5 thickness of the sleeping bag in inches, then T 5 2.5 2 0.025F
where the domain is 2408 # F # 408
b. See Figure 3.15.
4
T (inches)
2 Not warm enough
F –40⬚
–20⬚
0⬚
20⬚
40⬚
Figure 3.15 T, sleeping bag thickness as
a function of F, degrees Fahrenheit.
When the thickness of the sleeping bag, T, is less than the recommended thickness, then the bag will not be warm enough.
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c. The shaded region can be described as 0 , thickness of sleeping bag , recommended thickness 0,
, 2.5 2 0.025F
T
where 240º # F # 40º. We could rephrase this to say that the region is bounded by four lines: T 5 0, T 5 2.5 2 0.025F, F 5 2408, and F 5 408. Note that the dotted line indicates that the line itself is not included. d. If the sleeping bag thickness T is 2.75, we can find the corresponding recommended temperature by solving our equation for F. 2.75 5 2.5 2 0.025F 0.25 5 20.025F F 5 2108
Substitute 2.75 for T simplify to get or
So the sleeping bag would not be thick enough for 2208F, since the point s2208, 2.75sd lies in the shaded area below the Army’s recommended values. It would be more than thick enough for 08F or 208F since the points s08, 2.75sd and s208, 2.75sd both lie above the shaded area.
Reading between the Lines EXAMPLE
2
The U.S. Department of Agriculture recommends healthy weight zones for adults based on their height. For men between 60 and 84 inches tall, the recommended lowest weight, Wlo (in lb), is Wlo 5 105 1 4.0H and the recommended highest weight, Whi (in lb), is Whi 5 125.4 1 4.6H where H is the number of inches above 60 inches (5 feet). a. Graph and label the two boundary equations and indicate the underweight, healthy, and overweight zones. b. Give a mathematical description of the healthy weight zone for men. c. Two men each weigh 180 lb. One is 5r 10s tall and the other 6r 1s tall. Is either within the healthy weight zone? d. A 5r 11s man weighs 135 lb. If he gains 2 lb a week, how long will it take him to reach the healthy range? a. See Figure 3.16. 250
Whi Overweight
W, weight (lb)
SOLUTION
Healthy
200
Wlo
150 100 Underweight 50 0
0
6 12 18 H, height above 60" (inches)
24
Figure 3.16 Graph of men’s healthy weight
zone between recommended low sWlo d and high sWhi d weights.
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b. Assuming that men’s heights generally run between 5 feet (600) and 7 feet (840), then H (the height in inches above 60 inches) is bounded by 0s # H # 24s The recommendations say that a man’s weight W (in lb) should be more than or equal to sWlo d and should be less than or equal to Whi. So we have the computed inequality Wlo # W # Whi 105 1 4.0H # W # 125.4 1 4.6H
or
c. For a man who is 5r 10s (or 70s ) tall, H 5 10s and his maximum recommended weight, Whi, is 125.4 1 s4.6 ? 10d 5 171.4 lb. So if he weighs 180 lb, he is overweight. For a man who is 6r 1s (or 73s ) tall, H 5 13s . His minimum recommended weight, Wlo, is 105 1 s4.0 ? 13d 5 157 lb. His maximum recommended weight, Whi, is 125.4 1 s4.6 ? 13d 5 185.2 lb. So his weight of 180 lb would place him in the healthy zone. d. If a man is 5r 11s, his recommended minimum weight, Wlo is 105 1 s4.0 ? 11d 5 149 lb. If he currently weighs 135 lb, he would need to gain at least 149 2 135 5 14 lb. If he gained 2 lb a week, it would take him 7 weeks to reach the minimum recommended weight of 149 lb.
EXAMPLE
3
SOLUTION
Given the inequalities 2x 2 3y # 12 and x 1 2y , 4: a. Solve each for y. b. On the same graph, plot the boundary line for each inequality (indicating whether it is solid or dotted) and then shade the region described by each inequality. c. Write a compound inequality describing the overlapping region.
a. To solve for y in the first inequality
2x 2 3y # 12
add 22x to both sides
23y # 22x 1 12
divide both sides by 23, reversing the inequality symbol
23y 22x 12 $ 1 23 23 23
then simplify Solving for y in the second inequality
y$
2 x24 3
x 1 2y , 4
add 2x to both sides
2y , 2x 1 4
divide both sides by 2
2x 4 2y , 1 2 2 2
then simplify
1 y,2 x12 2
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b. See Figure 3.17. 10
y
5 2 3
y = x–4 –10
x
0
–5
10
5
1 2
y =– x+2 –5
–10
Figure 3.17 The darkest shaded region lies between
y , 2 12 x 1 2 and y $ 23 x 2 4.
c. The overlapping region consists of all ordered pairs (x, y) such that 2 1 3 x 2 4 # y , 22 x 1 2. We could describe the compound inequality by saying “y is greater than or equal to 23 x 2 4 and less than 212 x 1 2.” EXAMPLE
4
Describe the shaded region in Figure 3.18. 6
y
Line 3 4 Line 2 Line 1 2 Line 4 –4
–2
0
x 2
4
–2
Figure 3.18 A shaded region with four boundary
lines.
SOLUTION
We need to find the equation for each of the four boundary lines. The two vertical lines are the easiest: Line 1 is the line x 5 23 (a solid line, so included in the region); Line 2 is the line x 5 2 (dotted, so excluded from the region). Line 3 is the horizontal line y 5 5 (dotted, so it is excluded). Line 4 has a vertical intercept of 1. It passes through (0, 1) and (24, 0) so its slope is (0 2 1) / (24 2 0) 5 1/4 or 0.25. So the equation for line 4 is y 5 11 0.25x. The region can be described as all pairs (x, y) that satisfy both compound inequalities: 23 # x , 2
and
1 1 0.25x # y , 5
Breakeven Points: Regions of Profit or Loss A simple model for the total cost C to a company producing n units of a product is C 5 fixed costs 1 scost per unitd ? n
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A corresponding model for the total revenue R is R 5 sprice per unitd ? n The breakeven point occurs when C 5 R, or the total cost is equal to the total revenue. When R . C, or total revenues exceed total cost, the company makes a profit and when R , C, the company loses money. EXAMPLE
5
SOLUTION
Cost versus revenue In 1996 two professors from Purdue University reported on their study of fertilizer plants in Indiana.6 They estimated that for a large-sized fertilizer manufacturing plant the fixed costs were about $450,000 and the additional cost to produce each ton of fertilizer was about $210. The fertilizer was sold at $270 per ton. a. Construct and graph two equations, one representing the total cost C(n), and the other the revenue R(n), where n is the number of tons of fertilizer produced. b. What is the breakeven point, where costs equal revenue? c. Shade between the lines to the right of the breakeven point. What does the region represent? a. C(n) 5 450,000 1 210n and R(n)5 270n. See Figure 3.19. $4,000,000 Revenue $3,000,000
Cost
$2,000,000
Breakeven point
$1,000,000
n 0
2000
4000
6000 8000 Tons of fertilizer
10000
12000
14000
Figure 3.19 Graph of revenue vs. cost with breakeven point.
b. The breakeven point is where or substituting for C(n) and R(n) solving for n we get
cost 5 revenue C(n) 5 R(n) 450,000 1 210n 5 270n 450,000 5 60n n 5 7500 tons of fertilizer
Substituting 7500 tons into the revenue equation, we have Rs7500d 5 270 ? 7500 5 $2,025,000 We could have substituted 7500 tons into the cost equation to get the same value. Cs7500d 5 450,000 1 210 ? 7500 5 $2,025,000 6
Duane S. Rogers and Jay T. Aldridge, “Economic impact of storage and handling regulation on retail fertilizer and pesticide plants,” Agribusiness 12(4): 327–337, as quoted in Edmond C. Tomastik, Calculus: Applications and Technology, 3rd ed. (Belmont CA: Thomson Brooks/Cole, 2004).
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So the breakeven point is (7500, $2,025,000). At this point the cost of producing 7500 tons of fertilizer and the revenue from selling 7500 tons both equal $2,025,000. c. In the shaded area between R(n) and C(n) to the right of the breakeven point, R(n) . C(n), so revenue exceeds costs, and the manufacturers are making money. Economists call this the region of profit.
Algebra Aerobics 3.3
8
y
–8
8
8
x
–8
8
Graph A
Graph C
y
–8
8
8
–8 Graph B
x
y
–8
8
–8 Graph D
Figure 3.20 Graphs of four linear inequalities. 7
x
–8
–8
8
y
x
4. A small company sells dulcimer7 music books on the Internet. Examine the graph in Figure 3.21 of the cost (C) and revenue (R) equations for selling n books. 1400
Revenue, R
1200 Cost, C
1000 Dollars
1. Solve graphically each set of conditions. a. y $ 2x 2 1 d. y $ 0 y$42x y , 20.5x 1 2 b. y $ 2x 2 1 y , 0.5x 1 2 y#32x e. y # 0 c. y $ 400 1 10x x$0 y $ 200 1 20x y $ 2100 1 x x$0 f. 0 # x # 200 y$0 y . 2x 2 400 y , 100 2 1.5x 2. Determine which of the following points (if any) satisfy the system of inequalities y . 2x 2 3 and y # 3x 1 8 a. (2, 3) c. (0, 8) e. (20, 28) b. (24, 7) d. (24, 26) f. (1, 21) 3. Determine each inequality that has the given graph in Figure 3.20.
800 600 400 200
n 0
25
50
75 100 125 Number of books
150
175
200
Figure 3.21 Cost and revenue for dulcimer books.
a. Estimate the breakeven point and interpret its meaning. b. Shade in the region corresponding to losses for the company. c. What are the fixed costs for selling dulcimer music books? d. Another company buys dulcimer music books for $3.00 each and sells them for $7.00 each. The fixed cost for this company is $400. Form a system of inequalities that represents when this company would make a profit. Use C1 for cost and R1 for revenue for n books. 5. Suppose that the two professors from Example 5 in this section estimated 6 years later that for a largesized fertilizer manufacturing plant the fixed costs were about $50,000 and the additional cost to produce each ton of fertilizer was about $235. However, market conditions and competition caused the company to continue to sell the fertilizer at $270 per ton. a. Form the cost function C(n) and revenue function R(n) for n tons of fertilizer. b. Find the breakeven point and interpret its meaning. c. Graph the two functions and shade in the region corresponding to profits for the company.
A mountain dulcimer is an Appalachian string instrument, usually with four strings, commonly played on the lap by strumming or plucking.
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Exercises for Section 3.3 A graphing program is required for Exercises 19 and 20 and recommended for Exercises 15, 16 & 17. Access to the Internet is required in Exercise 14(c). 1. On different grids, graph and shade in the areas described by the following linear inequalities. a. y , 2x 1 2 c. y , 4 b. y $ 23x 2 3 d. y $ 23 x 2 4 2. On different grids, graph and shade in the areas described by the following linear inequalities. a. x 2 y , 0 c. 3x 1 2y . 6 b. x # 22 d. 5x 2 2y # 10 3. Use inequalities to describe each shaded region. 5
y
5
y
x –5
5
x –5
–5 Graph A
5
–5 Graph B
4. Use inequalities to describe each shaded region. 5
y
5
y
x –5
5
x –5
–5 Graph A
5
9. Match each description in parts (a) to (e) with the appropriate compound inequality in parts (f) to (j). a. y is greater than 4 and less than x 2 3. b. y is greater than or equal to x 2 3 and less than 26. c. y is less than 2x 1 5 and greater than 26. d. y is greater than or equal to 2x 1 5 and less than 26. e. y is less than or equal to x 2 3 and greater than 2x 1 5. f. g. h. i. j.
2x 1 5 # y , 26 4 , y , x 23 2x 1 5 , y # x 2 3 x 2 3 # y , 26 26 , y , 2x 1 5
10. For the inequalities y . 4x 2 3 and y # 23x 1 4: a. Graph the two boundary lines and indicate with different stripes the two regions that satisfy the individual inequalities. b. Write the compound inequality for y. Indicate the doublehatched region on the graph that satisfies both inequalities. c. What is the point of intersection for the boundary lines? d. If x 5 3, are there any corresponding y values in the region defined in part (b)? e. Is the point (1, 4) part of the double-hatched region? f. Is the point (21, 4) part of the double-hatched region? 11. Examine the shaded region in the graph. a. Create equations for the boundary lines l 1 and l 2 using y as a function of x. b. Determine the compound inequality that created the shaded region.
–5 Graph B
5. On different grids, graph each inequality (shading in the appropriate area) and then determine whether or not the origin, the point (0, 0), satisfies the inequality. a. 22x 1 6y , 4 c. y . 3x 27 b. x $ 3 d. y 2 3 . x 1 2 6. Determine whether or not the point (21, 3) satisfies the inequality. a. x 2 3y . 6 c. y # 3 b. x , 3 d. y # 212 x 1 3
5
y
l2 l1 x –5
5
–5
12. Examine the shaded region in the graph. Determine the compound inequality of y in terms of x that created the shaded region.
7. Explain how you can tell if the region described by the inequality 3x 2 5y , 15 is above or below the boundary line of 3x 2 5y 5 15.
y
5
8. Shade the region bounded by the inequalities x 1 3y # 15 2x 1 y # 15
x –5
5
x$0 y$0
–5
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13. The Food and Drug Administration labels suntan products with a sun protection factor (SPF) typically between 2 and 45. Multiplying the SPF by the number of unprotected minutes you can stay in the sun without burning, you are supposed to get the increased number of safe sun minutes. For example, if you can stay unprotected in the sun for 30 minutes without burning and you apply a product with a SPF of 10, then supposedly you can sun safely for 30 ? 10 5 300 minutes or 5 hours. Assume that you can stay unprotected in the sun for 20 minutes without burning. a. Write an equation that gives the maximum safe sun time T as a function of S, the sun protection factor (SPF). b. Graph your equation. What is the suggested domain for S? c. Write an inequality that suggests times that would be unsafe to stay out in the sun. d. Shade in and label regions on the graph that indicate safe and unsafe regions. (Use two different shadings and remember to include boundaries.) e. How would the graph change if you could stay unprotected in the sun for 40 minutes? Note that you should be cautious about spending too much time in the sun. Factors such as water, wind, and sun intensity can diminish the effect of SPF products. 14. (Access to Internet required for part (c).) Doctors measure two kinds of cholesterol in the body: low-density lipoproteins, LDL, called “bad cholesterol” and highdensity lipoproteins, HDL, called “good cholesterol” because it helps to remove the bad cholesterol from the body. Rather than just measuring the total cholesterol, TC, many doctors use the ratio of TC/HDL to help control heart disease. General guidelines have suggested that men should have TC/HDL of 4.5 or below, and women should have TC/HDL of 4 or below. On the following graphs place HDL on the horizontal and TC on the vertical axis. a. Construct an equation for men that describes TC as a function of HDL assuming that the ratio of the two numbers is at the recommended maximum for men. Graph the function, using HDL values up to 75. Label on the graph higher-risk and lower-risk areas for heart disease for men. b. Construct a similar equation for women that describes TC as a function of HDL assuming that the ratio of the two numbers is at the recommended maximum for women. Graph the function, using HDL values up to 75. Label on the graph higher-risk and lower-risk areas for heart disease for women. c. Use the Internet to find the most current cholesterol guidelines. 15. (Graphing program recommended.) The blood alcohol concentration (BAC) limits for drivers vary from state to state, but for drivers under the age of 21 it is commonly set at
191
0.02. This level (depending upon weight and medication levels) may be exceeded after drinking only one 12-oz can of beer. The formula N 5 6.4 1 0.0625(W 2 100) gives the number of ounces of beer, N, that will produce a BAC legal limit of 0.02 for an average person of weight W. The formula works best for drivers weighing between 100 and 200 lb. a. Write an inequality that describes the condition of too much blood alcohol for drivers under 21 to legally drive. b. Graph the corresponding equation and label the areas that represent legally safe to drive, and not legally safe to drive conditions. c. How many ounces of beer is it legally safe for a 100-lb person to consume? A 150-lb person? A 200-lb person? d. Simplify your formula in part (b) to the standard y 5 mx 1 b form. e. Would you say that “6 oz of beer 1 1 oz for every 20 lb over 100 lb” is a legally safe rule to follow? 16. (Graphing program recommended.) The Ontario Association of Sport and Exercise Sciences recommends the minimum and maximum pulse rates P during aerobic activities, based on age A. The maximum recommended rate, Pmax, is 0.87(220 2 A). The minimum recommended pulse rate, Pmin, is 0.72(220 2 A). a. Convert these formulas to the y 5 mx 1 b form. b. Graph the formulas for ages 20 to 80 years. Label the regions of the graph that represent too high a pulse rate, the recommended pulse rate, and too low a pulse rate. c. What is the maximum pulse rate recommended for a 20-year-old? The minimum for an 80-year-old? d. Construct an inequality that describes too low a pulse rate for effective aerobic activity. e. Construct an inequality that describes the recommended pulse range. 17. (Graphing program recommended.) We saw in this section the U.S. Department of Agriculture recommendations for healthy weight zones for males based on height. There are comparable recommendations for women between 59 (or 600) and 6930 (or 750) tall. For women the recommended lowest weight Wlo (in lb) is Wlo 5 100 1 3.5H and the recommended highest weight Whi (in lb) is Whi 5 118.2 1 4.2H where H is the number of inches above 600 (or 5 feet). a. Graph and label the two equations and indicate the underweight, healthy weight, and overweight zones. b. Give a mathematical description of the healthy weight zone for women.
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c. Two women have the same weight of 130 lb. One is 5920 and the other is 5950. Does either one lie within the healthy weight zone? Why? d. A 5940 woman weighs 165 lb. If her doctor puts her on a weight-loss diet of 1.5 lb per week, how many weeks would it take for her to reach the healthy range? 18. Cotton and wool fabrics, unless they have been preshrunk, will shrink when washed and dried at high temperatures. If washed in cold water and drip-dried they will shrink a lot less, and if dry-cleaned, they will not shrink at all. A particular cotton fabric has been found to shrink 8% with a hot wash/dry, and 3% with a cold wash/drip dry.
21. Describe the shaded region in each graph with the appropriate inequalities. y
y 5
5
1
x –5
x
–5
5
5
–5
–5 Graph A
Graph B
a. Find a formula to express the hot wash length, H, as a function of the original length, L. Then find a formula for the cold wash length, C, as a function of the original length, L. What formula expresses dry-clean length, D, as a function of the original length?
22. Describe the shaded region. (Hint: Create a vertical line through point A, dividing the shaded region into two smaller triangular regions.)
b. Make a graph with original length, L, on the horizontal axis, up to 60 inches. Plot three lines showing cold wash length, C, hot wash length, H, and dry-clean length, D. Label the lines.
10
y
19. (Graphing program required.) Two professors from Purdue University reported that for a typical small-sized fertilizer plant in Indiana the fixed costs were $235,487 and it cost $206.68 to produce each ton of fertilizer. a. If the company planned to sell the fertilizer at $266.67 per ton, find the cost, C, and revenue, R, equations for x tons of fertilizer. b. Graph the cost and revenue equations on the same graph and calculate and interpret the breakeven point. c. Indicate the region where the company would make a profit and create the inequality to describe the profit region. 20. (Graphing program required.) A company manufactures a particular model of DVD player that sells to retailers for $85. It costs $55 to manufacture each DVD player, and the fixed manufacturing costs are $326,000. a. Create the revenue function R(x) for selling x number of DVD players. b. Create the cost function C(x) for manufacturing x DVD players. c. Plot the cost and revenue functions on the same graph. Estimate and interpret the breakeven point. d. Shade in the region where the company would make a profit. e. Shade in the region where the company would experience a loss. f. What is the inequality that represents the profit region?
(6, 9)
(0, 6)
c. If you buy trousers with an original inseam length of 320, how long will the inseam be after a hot wash? A cold wash? d. If you need 3 yards of fabric (a yard is 39) to make a wellfitted garment, how much would you have to buy if you plan to hot-wash the garment?
y3
y1 y2 (3, 3) A
x –10
10
23. A financial advisor has up to $30,000 to invest, with the stipulation that at least $5000 is to be placed in Treasury bonds and at most $15,000 in corporate bonds. a. Construct a set of inequalities that describes the relationship between buying corporate vs. Treasury bonds where the total amount invested must be less than or equal to $30,000. (Let C be the amount of money invested in corporate bonds, and T the amount invested in Treasury bonds.). b. Construct a feasible region of investment; that is, shade in the area on a graph that satisfies the spending constraints on both corporate and Treasury bonds. Label the horizontal axis “Amount invested in Treasury bonds” and the vertical axis “Amount invested in corporate bonds.” c. Find all of the intersection points (corner points) of the bounded investment feasibility region and interpret their meanings. 24. A Texas oil supplier sends out at most 10,000 barrels of oil per week. Two distributors need oil. Southern Oil needs at least 2000 barrels of oil per week and Regional Oil needs at least 5000 barrels of oil per week. a. Let S be the number of barrels of oil sent to Southern Oil and let R be the number of barrels sent to Regional Oil per week. Create a system of inequalities that describes all of the conditions. b. Graph the feasible region of the system. c. Choose a point inside the region that would satisfy the conditions and describe its meaning.
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3.4 Systems with Piecewise Linear Functions: Tax Plans Graduated vs. Flat Income Tax Income taxes may be based on either a flat or a graduated tax rate. With a flat tax rate, no matter what the income level, everyone is taxed at the same percentage. Flat taxes are often said to be unfair to those with lower incomes, because paying a fixed percentage of income is considered more burdensome to someone with a low income than to someone with a high income. A graduated tax rate means that people with higher incomes are taxed at higher rates. Such a tax is called progressive and is generally less popular with those who have high incomes. Whenever the issue appears on the ballot, the pros and cons of the graduated vs. flat tax rate are hotly debated in the news media and paid political broadcasting. Of the forty-one states with a broad-based income tax, thirty-five had a graduated income tax in 2004.
The New York Times article “How a Flat Tax Would Work for You and for Them” discusses the trade-offs in using a flat tax.
The New York Times article “A Taxation Policy to Make John Stuart Mill Weep” discusses the growing trend by Congress of taxing different sources of income differently.
EXAMPLE
1
The taxpayer For the taxpayer there are two primary questions in comparing the effects of flat and graduated tax schemes. For what income level will the taxes be the same under both plans? And, given a certain income level, how will taxes differ under the two plans? Taxes are influenced by many factors, such as filing status, exemptions, source of income, and deductions. For our comparisons of flat and graduated income tax plans, we examine one filing status and assume that exemptions and deductions have already been subtracted from income. Match each graph in Figure 3.22 with the appropriate description.
0
Dollars
Tax (in dollars)
Tax (in dollars)
Tax (in dollars)
a. Income taxes are a flat rate of 5% of your income. b. Income taxes are graduated, with a rate of 5% for first $100,000 of income and a rate of 8% for any additional income . $100,000. c. Sales taxes are 5% for 0 # purchases # $100,000 and a flat fee of $5000 for purchases . $100,000.
0
Graph A
Dollars
0
Graph B
Dollars Graph C
Figure 3.22 Graphs of different tax plans. SOLUTION
Graph A matches description (b). Graph B matches description (a). Graph C matches description (c). A flat tax model Flat taxes are a fixed percentage of income. If the flat tax rate is 15% (or 0.15 in decimal form), then flat taxes can be represented as f(i) 5 0.15i where i 5 income. This flat tax plan is represented in Figure 3.23.
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8000 Tax (in dollars)
Flat tax
6000 4000 2000
i 0
Income (in dollars)
50,000
Figure 3.23 Flat tax at a rate
of 15%.
A graduated tax model A Piecewise Linear Function Under a graduated income tax, the tax rate changes for different income levels. We can use a piecewise linear function to model a graduated tax. Let’s consider a graduated tax where the first $10,000 of income is taxed at 10% and any income over $10,000 is taxed at 20%. For example, if your income after deductions is $30,000, then your taxes under this plan are graduated tax 5 (10% of $10,000) 1 (20% of income over $10,000) 5 0.10($10,000) 1 0.20($30,000 2 $10,000) 5 0.10($10,000) 1 0.20($20,000) 5 $1000 1 $4000 5 $5000 If an income, i, is over $10,000, then under this plan, graduated tax 5 0.10($10,000) 1 0.20(i 2 $10,000) 5 $1000 1 0.20(i 2 $10,000) The Graph This graduated tax plan is represented in Table 3.3 and Figure 3.24. The graph of the graduated tax is the result of piecing together two different line segments that represent the two different formulas used to find taxes. The short segment represents taxes for low incomes between $0 and $10,000, and the longer, steeper segment represents taxes for higher incomes that are greater than $10,000. 8000
Taxes under Graduated Tax Plan Income after Deductions ($) 0 5,000 10,000 20,000 30,000 40,000 50,000 Table 3.3
Taxes ($) 0 500 1,000 1,000 1 2,000 5 3,000 1,000 1 4,000 5 5,000 1,000 1 6,000 5 7,000 1,000 1 8,000 5 9,000
6000 Taxes (in dollars)
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10/11/07
High income
4000
Low income
2000
i 50,000
0 Income (in dollars)
Figure 3.24 Graduated tax.
The Equations To find an algebraic expression for the graduated tax, we need to use different formulas for different levels of income. Functions that use different formulas for different intervals of the domain are called piecewise defined. Since each income
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determines a unique tax, we can define the graduated tax as a piecewise function, g, of income i as gsid 5 e
0.10i 1000 1 0.20si 2 10,000d
for i # $10,000 for i . $10,000
The value of i (the input or independent variable) determines which formula to use to evaluate the function. This function is called a piecewise linear function, since each piece consists of a different linear formula. The formula for incomes equal to or below $10,000 is different from the formula for incomes above $10,000. Evaluating Piecewise Functions To find g($8000), the value of g(i) when i 5 $8000, use the first formula in the definition since income, i, in this case is less than $10,000: For i # $10,000 substituting $8000 for i
g(i) 5 0.10i g($8000) 5 0.10($8000) 5 $800
To find g($40,000), we use the second formula in the definition, since in this case income is greater than $10,000: For i . $10,000 substituting $40,000 for i
g(i) 5 $1000 1 0.20(i 2 $10,000) g($40,000)5 $1000 1 0.20($40,000 2 $10,000) 5 $1000 1 0.20($30,000) 5 $1000 1 $6000 5 $7000
Comparing the Two Tax Models Using graphs In Figure 3.25 we compare the different tax plans by plotting the flat and graduated tax equations on the same graph. y 8000 Graduated tax 6000 Taxes (in dollars)
Kime07_C03pg165-210.qxd
Flat tax
4000 ($20,000, $3000) 2000
0
x 50,000
0 Income (in dollars)
Figure 3.25 Flat tax versus graduated tax.
The intersection points indicate the incomes at which the amount of tax is the same under both plans. From the graph, we can estimate the coordinates of the two points as (0, 0) and ($20,000, $3000). That is, under both plans, with $0 income you pay $0 taxes, and with approximately $20,000 in income you would pay approximately $3000 in taxes. Individual voters want to know what impact these different plans will have on their taxes. From the graph in Figure 3.25, we can see that to the left of the intersection point
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at ($20,000, $3000), the flat tax is greater than the graduated tax for the same income. To the right of this intersection point, the flat tax is less than the graduated tax for the same income. So for incomes less than $20,000, taxes are greater under the flat tax plan, and for incomes greater than $20,000, taxes will be less under the flat tax plan. Using equations To verify the accuracy of our estimates for the coordinates of the point(s) of intersection, we can set f (i) 5 g(i). We know that f (i) 5 0.15i. Which of the two expressions do we use for g(i)? The answer depends upon what value of income, i, we consider. For i # $10,000, we have g(i) 5 0.10i. f(i) 5 g(i) 0.15i 5 0.10i i50
If and i # $10,000, then This can only happen when
If i 5 0, both f(i) and g(i) equal 0; therefore, one intersection point is indeed (0, 0). For i . $10,000, we use g(i) 5 $1000 1 0.20(i 2 $10,000) and again set f(i) 5 g(i).
?
If and i . $10,000, then apply the distributive property multiply combine terms add 20.20i to each side divide by 20.05
SOMETHING TO THINK ABOUT
We have assumed that deductions are the same under both plans. If we drop this assumption, how could we make a flat tax plan less burdensome for people with low incomes?
f(i) 5 g(i) 0.15i 5 $1000 1 0.20(i 2 $10,000) 0.15i 5 $1000 1 0.20i 2 (0.20)($10,000) 0.15i 5 $1000 1 0.20i 2 $2000 0.15i 5 2$1000 1 0.20i 20.05i 5 2$1000 i 5 2$1000/(20.05) i 5 $20,000
So each plan results in the same tax for an income of $20,000. How much tax is required? We can substitute $20,000 for i into either the flat tax formula or the graduated tax formula for incomes over $10,000 and solve for the tax. Given the flat tax function, if i 5 $20,000, then
f(i) 5 0.15i f(i) 5 (0.15)($20,000) 5 $3000
We can check to make sure that when i 5 $20,000, the graduated tax, g(i), will also be $3000: If i . $10,000, then so, if i 5 $20,000, then perform operations
g(i) 5 $1000 1 0.20(i 2 $10,000) g(i) 5 $1000 1 0.20($20,000 2 $10,000) 5 $3000
These calculations confirm that the other intersection point is, as we estimated, ($20,000, $3000).
The Case of Massachusetts In the state of Massachusetts there has been an ongoing debate about whether or not to change from a flat tax to a graduated income tax8. In 1994 Massachusetts voters considered a proposal called Proposition 7. Proposition 7 would have replaced the flat tax rate (at the time of the proposal, 5.95%) with graduated income tax rates (called marginal rates) as shown in Table 3.4.
8
For a state-by-state comparison of current income tax rates see www.taxadmin.org/fta/rate/ind_inc.html. For a comparison of the recent state tax proposals in Tennessee see www.yourtax.org.
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Massachusetts Graduated Income Tax Proposal Marginal Rate In Exploration 3.1 you can analyze the impact of Proposition 7 on people using different filing statuses.
Filing Status Married/joint Married/separate Single Head of household
5.5%
8.8%
,$81,000 ,$40,500 ,$50,200 ,$60,100
9.8%
$81,0002$150,000 $40,5002$ 75,000 $50,2002$ 90,000 $60,1002$120,000
$150,0001 $ 75,0001 $ 90,0001 $120,0001
Table 3.4 Source: Office of the Secretary of State, Michael J. Connolly, Boston, 1994.
Questions for the taxpayer For what income and filing status would the taxes be equal under both plans? Who will pay less tax and who will pay more under the graduated income tax plan? We analyze the tax for a single person and leave the analyses of the other filing categories for you to do. Finding out who pays what The proposed graduated income tax is designed to tax at higher rates that portion of the individual’s income that exceeds a certain threshold. For example, for those who file as single people, the graduated tax rate means that earned income under $50,200 would be taxed at a rate of 5.5%. Any income between $50,200 and $90,000 would be taxed at 8.8%, and any income over $90,000 would be taxed at 9.8%. Using Equations. The tax for a single person earning $100,000 would be the sum of three different dollar amounts. 5 (0.055)($50,200)
5.5% on the first $50,200
5 $2761 8.8% on the next $39,800 (the portion of income between $50,200 and $90,000) 5 (0.088)($39,800) < $3502 9.8% on the remaining $10,000
5 (0.098)($10,000) 5 $980
The total graduated tax would be $2761 1 $3502 1 $980 5 $7243. Under the flat tax rate the same individual would pay 5.95% of $100,000 5 0.0595($100,000), or $5950. Table 3.5 shows the differences between the flat tax and the proposed graduated tax plan for single people at several different income levels. We can represent flat taxes for single people as a linear function f of income i, where f(i) 5 0.0595i Massachusetts Taxes: Flat Rate vs. Graduated Rate for Single People Income after Exemptions and Deductions ($) 0 25,000 50,000 75,000 100,000 Table 3.5
Current Flat Tax at 5.95% ($) 0 1488 2975 4463 5950
Graduated Tax under Proposition 7 ($) 0 1375 2750 4943 7243
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We can write the graduated tax for single people as a piecewise linear function g of income i, where 0.055i g sid 5 c$2761 1 0.088si 2 $50,200d $6263 1 0.098si 2 $90,000d
for 0 # i , $50,200 for $50,200 # i # $90,000 for i . $90,000
Note that $2761 in the second formula of the definition is the tax on the first $50,200 of income (5.5% of $50,200), and $6263 in the third formula of the definition is the sum of the taxes on the first $50,200 and the next $39,800 of income (5.5% of $50,200 1 8.8% of $39,800 < $2761 1 $3502 5 $6263). Using Graphs. For what income would single people pay the same tax under both plans? The flat tax and the graduated income tax for single people are compared in Figure 3.26. An intersection point on the graph indicates where taxes are equal. One intersection point occurs at (0, 0). That makes sense, since under either plan if you have zero income, you pay zero taxes. The second intersection point is at approximately ($58,000, $3500). That means for an income of approximately $58,000, the taxes are the same and are approximately $3500. In the Algebra Aerobics you are asked to describe what happens to the right and to the left of the intersection point.
10,000
8000 Tax (in dollars)
Graduated tax 6000 Flat tax 4000
2000
0
0
20,000 40,000 60,000 80,000 100,000 120,000 Income (in dollars)
Figure 3.26 Massachusetts graduated tax vs. flat tax.
Algebra Aerobics 3.4 1. Graph the following piecewise functions. 2x 2 1 for x # 0 a. f (x) 5 e 1 for x . 0 2x 2 1 4 for 0 # x # 5 2x 2 6 for x . 5 0 for 0 # x # 10 c. k (x) 5 e 3x 2 30 for x . 10 2. Use equations to describe the piecewise linear function on each graph in Figure 3.27. b. g (x) 5 e
y
y
6
8
–4
x
x 4 –4 Graph A
8
–3
3 –3 Graph B
Figure 3.27 Two piecewise linear functions.
6
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3. Evaluate each of the following piecewise defined functions at x 5 25, 0, 2, and 10. 3 for x # 1 a. Psxd 5 e 1 2 2x for x . 1 x 2 4 for x , 2 x 1 4 for x $ 2 4. Construct a new graduated tax function, if: a. The tax is 5% on the first $50,000 of income and 8% on any income in excess of $50,000. b. The tax is 6% on the first $30,000 of income and 9% on any income in excess of $30,000. b. Wsxd 5 e
199
5. Use the graph in Figure 3.26 to predict which tax is larger for each of the following incomes and by approximately how much. Then use the equations defining f and g on page 197 and 198 to check your predictions. a. $30,000 b. $60,000 c. $120,000 6. Use equations to describe the following graph. 500
y
x 500
Exercises for Section 3.4 Graphing program recommended for Exercise 12.
5. Construct a piecewise linear function for each of the accompanying graphs.
1. Consider the following function:
y
x#0 x.0
Evaluate f(210), f(22), f(0), f(2), and f(4). 2. Consider the following function: g sxd 5 e
3 4 1 2x
2x b. f sxd 5 c x 2x 1 4
5
if x # 0 if 0 , x # 2 if x . 2
5
5
–5
Graph A
1
1
2
x
3
1
3
x
Graph B
7. Create a graph for each piecewise function: 20 2 2x if 0 # x , 5 if 5 # x # 10 a. h sxd 5 • 10 10 1 2sx 2 10d if x . 10 b. k(x) 5
y
x –5
2
6. Create a graph for each piecewise function: 2x 1 1 if x # 0 a. f sxd 5 e 1 if x . 0 2x if x , 0 b. g (x) 5 e x if x $ 0
x –5
2
Graph A
if x # 2 if x . 2
y
2
1
3. Match each function with its graph. x 2x 1 4
3
x#1 x.1
Evaluate g(25), g(22), g(0), g(1), g(1.1), g(2), and g(10).
a. f sxd 5 e
y
3
5
0.15x if 0 # x # 20,000 • 3000 1 0.18sx 2 20,000d if 20,000 , x # 50,000 8400 1 0.23sx 2 50,000d if x . 50,000 8. Each of the following graphs shows the activity of a hiker relative to her base campsite. Describe her actions first with words and then with a piecewise function.
–5 30
Graph B
4. Construct a small table of values and graph the following piecewise linear functions. In each case specify the domain. 5 for x , 10 a. f (x) 5 e 215 1 2x for x $ 10 12t for 210 # t # 1 b. g (t) 5 e t for 1 , t , 10
y
30
25
25
20
20
Miles
2x 1 1 3x
Miles
f sxd 5 e
15 10
y
15 10
5
5
x
x 5
10 15 Hours Graph A
20
5
10 15 Hours Graph B
20
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9. Find out what kind of income tax (if any) your state has. Is it flat or graduated? Construct and graph a function that describes your state’s income tax for one filing status. Compute the income tax for various income levels. 10. a. Construct a graduated tax function where the tax is 10% on the first $30,000 of income, then 20% on any income in excess of $30,000. b. Construct a flat tax function where the tax is 15% of income. c. Calculate the tax for both the flat tax function from part (b) and the graduated tax function from part (a) for each of the following incomes: $10,000, $20,000, $30,000, $40,000, and $50,000. d. Graph the graduated and flat tax functions on the same grid and estimate the coordinates of the points of intersection. Interpret the points of intersection. 11. Fines for a particular speeding ticket are defined by the following piecewise function, where s is the speed in mph and F(s) is the fine in dollars. 0 50 1 5ss 2 45d Fssd 5 d 100 1 10ss 2 55d 200 1 20ss 2 65d
if s # 45 if 45 , s # 55 if 55 , s # 65 if s . 65
a. What is the implied posted speed limit for this situation? b. Create a table of values for the fine, beginning at 40 mph and incrementing by 5-mph steps up to 80 mph, and then graph F(s). c. Describe in words how a speeding fine is calculated. d. Explain what 5, 10, and 20 in the formulas for the respective sections of the piecewise function represent. e. Find F(30), F(57), and F(67). f. Graph F(s). 12. (Graphing program recommended). In 2004, Missouri had a graduated tax plan but it was considering adopting a flatrate tax of 4% on income after deductions. The tax rate for single people under the graduated plan is shown in the accompanying table. For what income levels would a single person pay less tax under the flat tax plan than under the graduated tax plan? Missouri State Tax for a Single Person Income after Deductions ($)
Marginal Tax Rate (%)
#1000 1001–2000 2001–3000 3001–4000 4001–5000 5001–6000 6001–7000 7001–8000 8001–9000 .9000
1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00
13. You are thinking about replacing your long-distance telephone service. A cell phone company charges a monthly fee of $40 for the first 450 minutes and then charges $0.45 for every minute after 450. Every call is considered a longdistance call. Your local phone company charges you a fee of $60 per month and then $0.05 per minute for every longdistance call. a. Assume you will be making only long-distance calls. Create two functions, C(x) for the cell phone plan and L(x) for the local telephone plan, where x is the number of long-distance minutes. After how many minutes would the two plans cost the same amount? b. Describe when it is more advantageous to use your cell phone for long-distance calls. c. Describe when it is more advantageous to use your local phone company to make long-distance calls. 14. The accompanying table, taken from a pediatrics text, provides a set of formulas for the approximate “average” weight and height of normal infants and children. Age
Weight (lb)
At birth 3–12 months 1–5 years 6–12 years
7 (age in months) 1 11 5 ? (age in years) 1 17 7 ? (age in years) 1 5
Age
Height (in)
At birth 1 year 2–12 years
20 30 2.5 ? (age in years) 1 30
Source: R. E. Behrman and V. C. Vaughan (eds.), Nelson Textbook of Pediatrics, 12th ed. (Philadelphia: W. B. Saunders, 1983), p. 19.
For children from birth to 12 years of age, construct and graph a piecewise linear function for each of the following (assuming age is the independent variable): a. Weight in pounds (How does this model compare to the model for female infants in Chapter 2, Section 5?) b. Height in inches Note that there are certain gaps in the table that need to be resolved in order to construct piecewise linear functions. For example, you will need to decide which weight formula to use for a child who is 512 years old and which height formula to use for a child who is 112 years old. 15. Heart health is a prime concern, because heart disease is the leading cause of death in the United States. Aerobic activities such as walking, jogging, and running are recommended for cardiovascular fitness, because they increase the heart’s strength and stamina. a. A typical training recommendation for a beginner is to walk at a moderate pace of about 3.5 miles/hour (or approximately 0.0583 miles/minute) for 20 minutes. Construct a function that describes the distance traveled
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Dbeginner , in miles, as a function of time, T, in minutes, for someone maintaining this pace. Construct a small table of values and graph the function using a reasonable domain. b. A more advanced training routine is to walk at a pace of 3.75 miles/ hour (or 0.0625 miles/minute) for 10 minutes and then jog at 5.25 miles/ hour (or 0.0875 miles/minute) for 10 minutes. Construct a piecewise linear function that gives the total distance, Dadvanced, as a function of time. T, in minutes. Generate a small table of values and plot the graph of this function on your graph in part (a). c. Do these two graphs intersect? If so, what do the intersection point(s) represent? 16. A graduated income tax is proposed in Borduria to replace an existing flat rate of 8% on all income. The new proposal states that persons will pay no tax on their first $20,000 of income, 5% on income over $20,000 and less than or equal to $100,000, and 10% on their income over $100,000. (Note: Borduria is a fictional totalitarian state in the Balkans that figures in the adventures of Tintin.) a. Construct a table of values that shows how much tax persons will pay under both the existing 8% flat tax and
b. c. d. e.
f.
g.
h.
201
the proposed new tax for each of the following incomes: $0, $20,000, $50,000, $100,000, $150,000, $200,000. Construct a graph of tax dollars vs. income for the 8% flat tax. On the same graph plot tax dollars vs. income for the proposed new graduated tax. Construct a function that describes tax dollars under the existing 8% tax as a function of income. Construct a piecewise function that describes tax dollars under the proposed new graduated tax rates as a function of income. Use your graph to estimate the income level for which the taxes are the same under both plans. What plan would benefit people with incomes below your estimate? What plan would benefit people with incomes above your estimate? Use your equations to find the coordinates that represent the point at which the taxes are the same for both plans. Label this point on your graph. If the median income in the state is $27,000 and the mean income is $35,000, do you think the new graduated tax would be voted in by the people? Explain your answer.
C H A P T E R S U M M A RY Systems of Linear Equations
Infinitely many solutions
y
A pair of real numbers is a solution to a system of linear equations in two variables if and only if the pair of numbers is a solution of each equation. A system of two linear equations in two variables may have one solution, no solutions, or infinitely many solutions.
8
x –8
8
One solution
y 8 –8
Systems of Linear Inequalities 8
–8
x
The solutions to a system of linear inequalities in two variables are pairs of real numbers that satisfy all the inequalities. The solutions typically form a region of the plane and may be represented by a compound inequality such as y1 , y # y2 5 2 x , y # 1 1 3x
–8
No solutions
The shaded area in the figure represents the solution area for this system.
y 8
20
y y2
x –8
8
y1 x 6
–2 –8
–5
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Piecewise Linear Functions Taxes (in dollars)
Functions that use different formulas for different intervals of the domain are called piecewise functions. Functions constructed out of pieces of several different linear functions are called piecewise linear functions. The accompanying figure shows the graph of a linear piecewise function g of income i, where g sid 5 e
y
8000
6000 High income 4000 Low income 2000
for i # $10,000 for i . $10,000
0.10i 1000 1 0.20si 2 10,000d
x
0
50,000
0 Income (in dollars)
C H E C K Y O U R U N D E R S TA N D I N G I. Are the statements in Problems 1–29 true or false? Give an explanation for your answer.
Questions 7 and 8 refer to the accompanying graphs of systems of two equations in two variables.
1. Assuming y is a function of x, the number pair (25, 3) is a solution to the following system of equations:
7. System A has a unique solution at approximately (50, 25).
e
2x 2 3y 5 21 . x 5 2y 1 3
8. System B appears to have no solution.
2 2. Assuming y is a function of x, the number pair Q16 5 , 5 R is a solution to the following system of equations:
Questions 9 and 10 refer to the supply and demand curves in the accompanying graph. Price 125
y 5 10 2 3x e 2x 2 y 5 6 x2 3. The system d
5 52 y
B
100
A
75
x 1 2y 5 1 3
50
is not a linear system of
(3, 50)
25 Quantity
equations.
0
x 5 21 3 4. The system of linear equations d has no 6y x 23 2 5 5 5 5 solution(s). 2y 2
5. A system of linear equations in two variables either has a pair of numbers that is a unique solution or has no solution.
1
3
4
5
6
9. Graph B is the demand curve because as price increases, the quantity demanded decreases. 10. If three items are produced and sold at a price of $50, the quantity supplied will be equal to the quantity demanded. 11. The shaded region in the accompanying graph appears to satisfy the linear inequality 12 2 x , 3y.
6. A pair of numbers can be a solution to a system of linear equations in two variables if the pair is a solution to at least one of the equations. 100
2
y 5
y
y
10 15
x
–5 –2
x
50
–10
x 0
25 System A
10
–10
50 System B
12. The darkest shaded region in the graph on the next page appears to satisfy the system of linear inequalities c
y#42 y#2
x 3
.
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Check Your Understanding
y
Questions 21–23 refer to the accompanying graph of linear approximations for the body mass index (BMI) of girls age 7–15. BU gives the minimal recommended BMI and BN the maximum recommended BMI. B0 is the dividing line between being somewhat overweight and being obese.
5
15
x
–5
The National Center For Health Statistics calculates the BMI as
–2
BMI 5 c
13. The linear inequalities y # 5x 2 3 and y , 5x 2 3 have exactly the same solutions.
S
40
(1000, 34) 30
D
20
B0
Obese
BN Body mass index (BMI)
p
weight slbd 703 d ? c d height sind height sind
30
Questions 14 and 15 refer to the accompanying graph of supply and demand, where q 5 quantity and p 5 price: 50
203
Over 20 Normal
10
BU
Under
10
q
0
0
500 1000 1500 2000 2500
7
14. The equilibrium point (1000, 34) means that at a price of $1000, suppliers will supply 34 items and consumers will demand 34 items. 15. If the price is less than the equilibrium price, the quantity supplied is less than the quantity demanded and therefore competition will cause the price to increase toward the equilibrium price. Questions 16–20 refer to the accompanying graph, where R represents total revenue and C represents total costs. 20,000
9
11 12 Age (years)
10
13
14
15
Sources: National Center for Health Statistics, National Center for Chronic Disease Prevention and Health Promotion (2000), www.cdc.gov/growthcharts.
21. A 15-year-old girl who weighs 105 pounds and is 55 inches tall has a BMI that is within the normal zone. 22. A 10-year-old girl with BMI 5 12 is underweight for her height and age. 23. A mathematical description of the normal BMI zone for girls age 7–15 is BU # BMI # BN .
$
R
Questions 24 and 25 refer to the following system of linear 2x 1 3 # y inequalities: e y,51x
C
24. The boundary line y 5 5 1 x is not included in the solution region.
17,500 15,000 Dollars
8
12,500 10,000 7500
25. The pair of numbers (0, 0) is a solution to the system.
5000 2500
n 0
200
400
600
Questions 26 and 27 refer to the accompanying graph of the piecewise function y 5 f(x).
800 1000
y
Number of items
5
16. The estimated breakeven point is approximately (500, $10,000). 17. The profit is zero at the breakeven point.
–5
x
18. If 200 units are produced and sold, the total revenue is larger than the total cost.
5
19. If the revenue per unit increases, and hence, R, the revenue line, becomes steeper, the breakeven point moves to the right. 20. If 700 units are produced and sold, the company is making about a $5000 profit.
–5
26. f(22) 5 2 27. f(0) 5 4
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34. A cost and revenue function that has a breakeven point at (100, $5000).
Questions 28 and 29 refer to the piecewise function 4 2 2x f sxd 5 • 0 x11
for 2 3 # x , 2 for 2 # x # 3 for 3 , x # 5
35. A demand equation (downward sloping) that, with the supply function p 5 q 1 20, has an equilibrium point at (20 units, $40).
28. f(21) 5 6
36. A system of linear inequalities that has as its solution all of quadrant IV (the region where x . 0 and y , 0).
29. f(3) 5 0 II. In Problems 30–36, give an example of a function or functions with the specified properties. Express your answer using formulas, and specify the independent and dependent variables. 30. A system of two linear equations in two variables that has no solution.
III. Is each of the statements in Problems 37– 40 true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample. 37. Linear systems of equations in two variables always have exactly one pair of numbers as a solution.
31. A system of two linear equations in two variables that has an infinite number of solutions, including the pairs (2, 3) and (21, 9).
38. A linear system of equations in two variables can have exactly two pairs of numbers that are solutions to the system.
32. A system of two linear inequalities in two variables that has no solution.
39. Any solution(s) to a linear system of equations in two variables is either a point or a line.
33. A system of two linear equations in variables c and r with the ordered pair (21, 0) as a solution, where c is a function of r.
40. Solutions to a linear system of inequalities in two variables can be a region in the plane.
CHAPTER 3 REVIEW: PUTTING IT ALL TOGETHER 1. The following graph shows worldwide production and consumption of grain in millions of metric tons (MMT). It is based on data from 1998 to May 2006 and includes projections to the end of 2006. a. Estimate the maximum production of grain in this time period. In what year did it occur? What was the minimum production and in what year did it occur? b. Explain the meaning of the intersection point(s) in this context. c. In what year was the difference between consumption and production the largest? Was there a surplus or deficit? d. Create a title for the graph. 1700
y
y 5
5
x 5
–5
x 5
–5
–5
–5
System A
System B
For each system: a. Determine the number of solutions using the graph of the system. b. Construct the equations of the lines. c. Solve the system using your equations from part (b). Explain why your answer makes sense.
1600
Consumption MMT
2. The accompanying graphs show two systems of linear equations.
1500
Production
1400
1300 2006
2005
2004
2003
2002
2001
2000
1999
1998
Year
Source: Foreign Agricultural Service Circular Series, FG 05-06, May 2006, www.fas.usda.gov/grain/circular/2006/ 05-06/graintoc.htm.
3. a. Construct a system of linear equations where both of the following conditions are met: i. The coordinates of the point of intersection are (4, 10). ii. The two lines are perpendicular to each other and one of the lines has a slope of –4. b. Graph the system of equations you found in part (a). Estimate the coordinates of the point of intersection on your graph. Does your estimate confirm your answer for part (a)?
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4. New York City taxi fares are as follows: initial fare $2.00, $0.30 per 1/5 mile, and night surcharge (8 p.m.–6 a.m.) of $0.50.9 a. Create and graph the equation for Cd, the cost of a daytime taxi ride for m miles. b. Create and graph the equation for Cn, the cost of a night taxi ride for m miles. c. How do the two graphs compare?
a. Compute the average annual rate of change of gas prices for each country using data for 1990 and 2004. b. Compute the average (mean) gas price from 1990 to 2004 for each country. c. Write a paragraph comparing U.S. gas prices in this 14-year period with prices in Japan and Germany. 9. In 2006 China was the world’s second largest consumer of oil, behind the United States.
5. Graph and shade the region bounded by the following inequalities:
6. Determine the inequalities that describe the shaded region in the following graph. y 5
China’s Oil Production and Consumption, 1986–2006 Thousands of barrels per day
y,2 y $ 22x 1 3 x#5
205
8,000 7,000 6,000
Net imports
Consumption
5,000 4,000
Production
3,000 2,000 1,000
2006
2004
2002
2000
1998
1996
1994
1992
1990
1988
1986
0
Year
Source: EIA International Petroleum Monthly. Data include projections
x –5
for last four months of 2006.
5
–1
7. A musician produces and sells CDs on her website. She estimates fixed costs of $10,000, with an additional cost of $7 to produce each CD. She currently sells the CDs for $12 each. a. Create the cost equation, C, and revenue equation R, in terms of x number of CDs produced and sold. b. Find the breakeven point. c. By how much should the musician charge for each CD if she wants to break even when producing and selling 1600 CDs, assuming other costs remain the same? d. By how much would she need to reduce fixed costs if she wants to break even when producing and selling 1600 CDs, assuming other costs and prices remain the same as originally stated? 8. Data from the U.S. Department of Energy show that gasoline prices vary immensely for different countries. These prices include taxes and are in U.S. dollars/gallon of unleaded regular gasoline, standardized on the U.S. gallon. Year
Germany
Japan
United States
1990 1992 1994 1996 1998 2000 2002 2004
2.65 3.26 3.52 3.94 3.34 3.45 3.67 5.24
3.16 3.59 4.39 3.65 2.83 3.65 3.15 3.93
1.16 1.13 1.11 1.23 1.06 1.51 1.36 1.88
a. Estimate the amount of oil consumption and production in China in 1990. What is the net difference between production and consumption? What does this mean? b. Estimate the year when oil consumption equaled oil production. Explain the meaning of this condition. c. Estimate the amount of oil consumption and oil production in China in 2006. What is the net difference? What does this mean? 10. You keep track of how much gas your car uses and estimate that it gets 32 miles/gallon. a. If you start out with a full tank of 14 gallons, write a formula for how many gallons of gas G are left in the tank after you have gone M miles. b. You consider borrowing your friend’s larger car, which gets 18 miles/gallon and holds 20 gallons. Write a formula for how many gallons of gas GF are left in the tank after you have gone M miles. Generate a data table and graph the gas remaining (vertical axis) versus miles (horizontal axis) for both cars. c. Which car has the longer mileage range? d. Is there a distance at which they both have the same amount of gas left? If so, what is it? 11. a. Solve algebraically each of the following systems of linear equations. i. 4x 1 21y 5 25 3x 1 7y 5 210
9 http://www.ny.com/transportation/taxis/ gives more detailed information on NYC taxi fares.
ii. 2.5a 2 b 5 7 a b5 6
b. Create a system of linear equations for which there is no solution.
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The accompanying table is for problems 12 and 13. Calories Burned per Minute
Activity Running (moderate pace) Swimming laps
10 8
Let r and s represent the number of minutes spent running and swimming, respectively. 12. Use the table to answer the following questions. Your goal is to run and swim so that you burn at least 800 calories but no more than 1200 calories. a. Create a system of inequalities that represents the set of combinations of minutes running and minutes swimming that meet your goal. (Note: r $ 0 and s $ 0.) b. Graph the solution to the system of inequalities using s on the horizontal axis and r on the vertical axis. c. Give an example of one combination of running and swimming times that is in the solution set and one that is not in the solution set. d. How would your solution set change if your goal were to burn at least 600 calories but not more than 1000 calories? 13. Use the table to answer the following questions. Your goal is to run and swim so that you spend no more than 60 minutes exercising but burn at least 560 calories. a. Create a system of inequalities that represents the set of combinations of minutes running and minutes swimming that meet your goal. (Note: r $ 0 and s $ 0.) b. Graph the solution to the system of inequalities with s on the horizontal axis and r on the vertical axis. c. Give an example of one combination of running and swimming times that is in the solution set and one that is not in the solution set. d. How would your solution set change if you spend no more than 70 minutes exercising but still burn at least 560 calories? 14. In a 400-meter relay swim, each team has four swimmers. In sequence, the swimmers each swim 100 meters. The total time (cumulative) and rates for team A are provided in the accompanying table. Total Time, Rate Total Distance, Cumulative (seconds/meter), Cumulative Swimmer (seconds) Rounded (meters) 1 2 3 4
99 201 279 341
0.99 1.02 0.78 0.62
100 200 300 400
a. Which swimmer is the fastest? The slowest? b. The total swim time function can be written as 0.99mfor 0 # m # 100 99 1 1.02sm 2 100d for 100 # m # 200 tsmd 5 µ 201 1 0.78sm 2 200dfor 200 , m # 300 279 1 0.62sm 2 300dfor 300 , m # 400 Find and interpret the following: t(50), t(125), t(250), t(400).
15. The accompanying table lists the monthly charge, the number of minutes allowed, and the charge per additional minute for three wireless phone plans. Cell Phone Plan
Monthly Charge
Number of Minutes
Overtime Charge/Minute
A B C
$39.99 $59.99 $79.99
450 900 1350
$0.45 $0.40 $0.35
Let m be the number of minutes per month, 0 # m # 2500. a. Construct a piecewise function for the cost A(m) for plan A. b. Construct a piecewise function for the cost B(m) for plan B. c. Construct a piecewise function for the cost C(m) for plan C. d. Complete the following table to determine the best plan for each of the estimated number of minutes per month. Number of Minutes/Month
Cost Plan A
Plan B
Plan C
500 800 1000 16. It is often said that 1 year of a dog’s life is equivalent to 7 years of a human’s life. A more accurate veterinarian’s estimate is that for the first 2 years of a dog’s life, each dog year is equivalent to 10.5 years of a human’s life, and after 2 years each dog year is equivalent to 4 years of a human’s life. a. Write formulas to describe human-equivalent years as a function of dog years for the two methods of relating dog years to human years. Use H and D for the popular formula, and Hv and Dv for the veterinary method. b. At what dog age do the 2 methods give the same human years, and what human age is that? 17. The time series at the top of the next page shows the price per barrel of crude oil from 1861 to 2006 in both dollars actually spent (nominal) and dollars adjusted for inflation (real 2006 dollars). Write a 60-second summary about crude oil prices. 18. Suppose a flat tax amounts to 10% of income. Suppose a graduated tax is a fixed $1000 for any income # $20,000 plus 20% of any income over $20,000. a. Construct functions for the flat tax and the graduated tax. b. Construct a small table of values: Income
Flat Tax
Graduated Tax
$0 $10,000 $20,000 $30,000 $40,000 c. Graph both tax plans and estimate any point(s) at which the two plans would be equal.
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207
Crude Oil Prices, 1861–2006 $100 Nominal Real (in 2006 Dollars)
Price per barrel
$80
$60
$40
$20 $0 2011
2001
1991
1981
1971
1961
1951
1941
1931
1921
1911
1901
1891
1881
1871
1861
Source: http://en.wikipedia.org/wiki/Image:Oil_Prices_1861_2006.jpg Graph created by Michael Ströck, 2006. Released under the GFDL.
d. Use the function definitions to calculate any point(s) at which the two plans would be equal. e. For what levels of income would the flat tax be more than the graduated tax? For what levels of income would the graduated tax be more than the flat tax. f. Are there any conditions in which an individual might have negative income under either of the above plans, that is, the amount of taxes would exceed an individual’s income? This is not as strange as it sounds. For example, many states impose a minimum corporate tax on a company, even if it is a small, one-person operation with no income in that year. 19. You check around for the best deal on your prescription medicine. At your local pharmacy it costs $4.39 a bottle; by mail-order catalog it costs $3.85 a bottle, but there is a flat shipping charge of $4.00 for any size order; by a source found on the Internet it costs $3.99 a bottle and shipping costs $1.00 for each bottle, but for orders of ten or more bottles it costs $3.79 a bottle, and handling is $2.50 per order plus shipping costs of $1.00 for each bottle. a. Find a formula to express each of these costs if N is the number of bottles purchased and Cp ,Cc, and CI are the respective costs for ordering from the pharmacy, catalog, and Internet. b. Graph the costs for purchases up to thirty bottles at a time. Which is the cheapest source if you buy fewer than ten bottles at a time? If you buy more than ten bottles at a time? Explain. 20. Older toilets use about 7 gallons per flush. Since using this much pure water to transport human waste is especially undesirable in areas of the country where water is scarce, new toilets now must meet a water conservation standard and are designed to use 1.6 gallons or less per flush. a. An old toilet that leaks about 2 cups of water an hour from the tank to the bowl and uses 7 gallons/flush is replaced with a new toilet, which does not leak and uses 1.6 gallons/flush. There are 16 cups in a gallon. For each toilet write an equation for daily water loss, W (gallons), as a function of number of flushes per day, F.
b. Graph the equations in part (a) on the same plot with F on the horizontal axis, where 0 # F # 25. From your graph, estimate the amount of water used by each system for 20 flushes a day, and check your estimates with your equations. What is the net difference? What does this mean? c. If a family flushes the toilet an average of 10 times a day, how much water do they save every day by replacing their leaky old toilet with a new water saver toilet? How much water would be saved over a year? d. In England during World War II the citizens were asked to flush their toilets only once every five times the toilet was used, in order to save water. A pencil was hung near the toilet, and each user made a vertical mark on the wall, the fifth user made a horizontal line through the last four marks and then flushed the toilet. If this method were used today for the family with the old leaky toilet, would the amount saved be greater than the amount saved with the new toilet if the new toilet is flushed after every use? 21. Regular aerobic exercise at a target heart rate is recommended for maintaining health. Someone starting an exercise program might begin at an intensity level of 60% of the target rate and work up to 70%; athletes need to work at 85% or higher. The American College of Sports Medicine method to compute target heart rate, H (in beats per minute), is based on maximum heart rate, Hmax, age in years, A, and exercise intensity level, I, where Hmax 5 220 2 A and
H 5 I ? Hmax
thus
H 5 I ? (220 2 A)
a. Write a formula for beginners’ target heart rate Hb if the intensity level is 60%, that is I 5 0.60. b. Write a formula for intermediate target heart rate Hi if the intensity level is 70%, that is I 5 0.70. c. Write a formula for athletes’ target heart rate Ha if the intensity level is 85%, that is I 5 0.85.
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d. Construct a graph showing all three heart rate levels, Hb, Hi, Ha, and the maximum rate, Hmax, for ages 15 to 75 years. Put age on the horizontal axis. Mark the zone in which athletes should work. e. Compute the target heart rate for a 20-year-old to work at all three levels. What is Hmax for a 20-year-old?
f. If a 65-year-old is working at a heart rate of 134 beats/minute, what is her intensity level? Has she exceeded the maximum heart rate for her age? If her 45-year-old son is working at the same heart rate, what is his intensity level? If her 25-year-old granddaughter is also working at the same heart rate, what is her intensity level?
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E X P L O R AT I O N 3 . 1 Flat vs. Graduated Income Tax: Who Benefits?
Objectives • compare the effects of different tax plans on individuals in different income brackets • interpret intersection points
Material/Equipment • spreadsheet or graphing calculator (optional). If using a graphing calculator, see examples on graphing piecewise functions in the Graphing Calculator Manual. • graph paper
Related Readings “How a Flat Tax Would Work, for You and for Them,” The New York Times, Jan. 21, 1996. “Flat Tax Goes from ‘Snake Oil’ to G.O.P. Tonic,” The New York Times, Nov. 14, 1999. “A Taxation Policy to Make John Stuart Mill Weep,” The New York Times, April 18, 2004. “Your Real Tax Rate,” www.msnmoney.com, Feb. 21, 2007. Procedure A variety of tax plans were debated in all recent elections for president. (See related readings.) One plan recommended a flat tax of 19% on income after exemptions and deductions. In this exploration we examine who benefits from this flat tax as opposed to the current graduated tax plan. The questions we explore are For what income will taxes be equal under both plans? Who will benefit under the graduated income tax plan compared with the flat tax plan? Who will pay more taxes under the graduated plan compared with the flat tax plan? In a Small Group or with a Partner The accompanying table gives the 2006 federal graduated tax rates on income after deductions for single people. 1. Construct a function for flat taxes of 19%, where income, i, is income after deductions. 2. Construct a piecewise linear function for the graduated federal tax for single people in 2006, where income, i, is income after deductions. 3. Graph your two functions on the same grid. Estimate from your graph any intersection points for the two functions. 4. Use your equations to calculate more accurate values for the points of intersection. 5. (Extra credit.) Use your results to make changes in the tax plans. Decide on a different income for which taxes will be equal under both plans. You can use what you know about the distribution of income in the United States from the FAM1000 data to make your decision. Alter one or both of the original functions such that both tax plans will generate the same tax given the income you have chosen.
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2006 Tax Rate Schedule for Single Persons Schedule X—Use if your filing status is Single If the amount on Form 1040, line 40, is: Over— $0 $7,550 $30,650 $74,200 $154,800 $336,550
But not over—
Enter on Form 1040, line 41
of the amount over—
$7,550 $30,650 $74,200 $154,800 $336,550 No limit
10% $7551 15% $4,2201 25% $15,107.501 28% $37,675.501 33% $97,653.001 35%
$0 $7,550 $30,650 $74,200 $154,800 $336,550
Source: www.irs.gov.
Analysis • Interpret your findings. Assume deductions are treated the same under both the flat tax plan and the graduated tax plan. What do the intersection points tell you about the differences between the tax plans? • What information would be useful in deciding on the merits of each of the plans? • What if the amount of deductions that most people can take under the flat tax is less than the graduated tax plan? How will your analysis be affected?
Exploration-Linked Homework Reporting Your Results Take a stance for or against a flat federal income tax. Using supportive quantitative evidence, write a 60-second summary for a voters’ pamphlet advocating your position. Present your arguments to the class.
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THE LAWS OF EXPONENTS AND LOGARITHMS: MEASURING THE UNIVERSE OVERVIEW Most of the examples we’ve studied so far have come from the social sciences. In order to delve into the physical and life sciences, we need to compactly describe and compare the extremes in deep time and deep space. In this chapter, we introduce the tools that scientists use to represent very large and very small quantities. After reading this chapter, you should be able to • write expressions in scientific notation • convert between English and metric units • simplify expressions using the rules of exponents • compare numbers of widely differing sizes • calculate logarithms base 10 and plot numbers on a logarithmic scale
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4.1 The Numbers of Science: Measuring Time and Space On a daily basis we encounter quantities measured in tenths, tens, hundreds, or perhaps thousands. Finance or politics may bring us news of “1.3 billion people living in China” or “a federal debt of over $7 trillion.” In the physical sciences the range of numbers encountered is much larger. Scientific notation was developed to provide a way to write numbers compactly and to compare the sizes found in our universe, from the largest object we know—the observable universe—to the tiniest—the minuscule quarks oscillating inside the nucleus of an atom. We use examples from deep space and deep time to demonstrate powers of 10 and the use of scientific notation.
Powers of 10 and the Metric System The international scientific community and most of the rest of the world use the metric system, a system of measurements based on the meter (which is about 39.37 inches, a little over 3 feet). In daily life Americans have resisted converting to the metric system and still use the English system of inches, feet, and yards. Table 4.1 shows the conversions for three standard metric units of length: the meter, the kilometer, and the centimeter. For a more complete conversion table, see the inside back cover. Conversions from Metric to English for Some Standard Units Metric Unit
Abbreviation
In Meters
Equivalent in English Units
meter
m
1m
3.28 ft
kilometer
km
1000 m
0.62 mile
centimeter
cm
0.01 m
0.39 in
Informal Conversion The width of a twin bed, a little more than a yard A casual 12-minute walk, a little over half a mile The length of a black ant, a little under half an inch
Table 4.1
For an appreciation of the size of things in the universe, we highly recommend the video by Charles and Ray Eames and related book by Philip and Phylis Morrison titled Powers of Ten: About the Relative Size of Things.
Deep space The Observable Universe. Current measurements with the most advanced scientific instruments generate a best guess for the radius of the observable universe at about 100,000,000,000,000,000,000,000,000 meters, or “one hundred trillion trillion meters.” Obviously, we need a more convenient way to read, write, and express this number. To avoid writing a large number of zeros, exponents can be used as a shorthand: 1026 can be written as a 1 with twenty-six zeros after it. 1026 means: 10 ? 10 ? 10 ? c ? 10, the product of twenty-six 10s. 1026 is read as “10 to the twenty-sixth” or “10 to the twenty-sixth power.” So the estimated size of the radius of the observable universe is 1026 meters. The sizes of other relatively large objects are listed in Table 4.2.1 The Relative Sizes of Large Objects in the Universe Object Milky Way Our solar system Our sun Earth
Radius (in meters) 1,000,000,000,000,000,000,000 5 1021 1,000,000,000,000 5 1012 1,000,000,000 5 109 10,000,000 5 107
Table 4.2 1
The rough estimates for the sizes of objects in the universe in this section are taken from Timothy Ferris, Coming of Age in the Milky Way (New York: Doubleday, 1988).
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Us. Human beings are roughly in the middle of the scale of measurable objects in the universe. Human heights, including children’s, vary from about one-third of a meter to 2 meters. In the wide scale of objects in the universe, a rough estimate for human height is 1 meter. To continue the system of writing all sizes using powers of 10, we need a way to express 1 as a power of 10. Since 103 5 1000, 102 5 100, and 101 5 10, a logical way to continue would be to say that 100 5 1. Since reducing a power of 10 by 1 is equivalent to dividing by 10, the following calculations give justification for defining 100 as equal to 1. 102 5
103 s10ds10ds10d 5 5 100 10 10
101 5
s10ds10d 102 5 5 10 10 10
100 5
101 10 5 51 10 10
By using negative exponents, we can continue to use powers of 10 to represent numbers less than 1. For consistency, reducing the power by 1 should remain equivalent to dividing by 10. So, continuing the pattern established above, we define 1021 5 1/10, 1022 5 1/102, and so on. For any positive integer, n, we define 102n 5
1 10 n
DNA Molecules. A DNA strand provides genetic information for a human being. It is made up of a chain of building blocks called nucleotides. The chain is tightly coiled into a double helix, but stretched out it would measure about 0.01 meter in length. How does this DNA length translate to a power of 10? The number 0.01, or one-hundredth, equals 1/102. We can write 1/102 as 1022. So a DNA strand, uncoiled and measured lengthwise, is approximately 1022 meters, or one centimeter. Table 4.3 shows the sizes of some objects relative to the size of human beings. The Relative Sizes of Small Objects in the Universe Object
Radius (in meters)
Human beings DNA molecules
1 5 10 10 5 1 5 101 2 5 0.01 5 100
100 1022
Living cells
1 0.000 01 5 100,000 5 101 5 5
1025
Atoms
1 0.000 000 000 1 5 10,000,000,000 5 10110 5
10210
Table 4.3
The following box gives the definition for various powers of 10. Powers of 10 When n is a positive integer: 10n 5 10 ? 10 ? 10 ? c ? 10 which can be written as 1 followed by n zeros. (11111)11111* n factors
100 5 1 102n 5
1 10n
which can be written as a decimal point followed by n21 zeros and a 1.
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Multiplying by 10 n is equivalent to moving the decimal point to the right n places. Multiplying by 102n is equivalent to dividing by 10 n, or moving the decimal point to the left n places.
The metric language By international agreement, standard prefixes specify the power of 10 that is attached to a specific unit of measure. They indicate the number of times the basic unit has been multiplied or divided by 10. Usually these prefixes are attached to metric units of measure, but they are occasionally used with the English system. Table 4.4 gives prefixes and their abbreviations for certain powers of 10. A more complete table is on the inside back cover. Prefixes for Powers of 10 piconanomicromillicenti-
p n m m c
10212 1029 1026 1023 1022
(unit) kilomegagigatera-
k M G T
100 103 106 109 1012
Table 4.4
EXAMPLE
1
SOLUTION
Indicate the number of meters in each unit of measure: cm, mm, Gm. 1 cm 5 1 centimeter 5 1022 m 5 1 mm 5 1 millimeter 5 1023 m 5
1 1 m 5 0.01 meter 2 m 5 10 100 1 1 m5 m 5 0.001 meter 103 1000
1 Gm 5 1 gigameter 5 109 m 5 1,000,000,000 meters
EXAMPLE
2
Translate the following underlined expressions. A standard CD holds about 700 megabytes of information. Translation: 700 ? 106 bytes or 700,000,000 bytes A calculator takes about one millisecond to add or multiply two 10-digit numbers. Translation: 1 ? 1023 second or 0.001 second In Tokyo on January 11, 1999, the NEC company announced that it had developed a picosecond pulse emission, optical communications laser. Translation: 1 ? 10212 second or 0.000 000 000 001 second It takes a New York City cab driver one nanosecond to beep his horn when the light changes from red to green. Translation: 1 ? 1029 second or 0.000 000 001 second
Scientific Notation In the previous examples we estimated the sizes of objects to the nearest power of 10 without worrying about more precise measurements. For example, we used a gross estimate of 107 meters for the measure of the radius of Earth. A more accurate measure
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is 6,368,000 meters. This number can be written more compactly using scientific notation as 6.368 ? 106 meters The number 6.368 is called the coefficient. The absolute value of the coefficient must always lie between 1 and 10. The power of 10 tells us how many places to shift the decimal point of the coefficient in order to get back to standard decimal form. Here, we would multiply 6.368 times 10 6, which means we would move the decimal place six places to the right, to get 6,368,000 meters. Any nonzero number, positive or negative, can be written in scientific notation, that is, written as the product of a coefficient N multiplied by 10 to some power, where 1 # k N k , 10. Thus 2 million, 2,000,000, and 2 ? 106 are all equivalent representations of the same number. The one you choose depends on the context. In the following examples, you’ll learn how to write numbers in scientific notation. Later we’ll use scientific notation to simplify operations with very large and very small numbers.
Scientific Notation A number is in scientific notation if it is in the form N ? 10n
where
N is called the coefficient and 1 # k N k , 10 n is an integer
EXAMPLE
3
SOLUTION
The distance to Andromeda, our nearest neighboring galaxy, is 15,000,000,000,000,000,000,000 meters. Express this number in scientific notation. The coefficient needs to be a number between 1 and 10. We start by identifying the first nonzero digit and then placing a decimal point right after it to create the coefficient of 1.5. The original number written in scientific notation will be of the form 1.5 ? 10? What power of 10 will convert this expression back to the original number? The original number is larger than 1.5, so the exponent will be positive. If we move the decimal place 22 places to the right, we will get back 1.5,000,000,000,000,000,000,000. This is equivalent to multiplying 1.5 by 1022. So, in scientific notation, 15,000,000,000,000,000,000,000 is written as 1.5 ? 1022
EXAMPLE
4
The radius of a hydrogen atom is 0.000 000 000 052 9 meter across. Express this number in scientific notation.
SOLUTION
The coefficient is 5.29. The original number written in scientific notation will be of the form 5.29 ? 10? What power of 10 will convert this expression back to the original number? The original number is smaller than 5.29, so the exponent will be negative. If we move the
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decimal place 11 places to the left, we will get back 0.000 000 000 05.2 9. This is equivalent to dividing 5.29 by 1011 or multiplying it by 10211: 0.000 000 000 052 9 5
5.29 1011
5 5.29 ¢
1 ≤ 1011
5 5.29 ? 10211 This number is now in scientific notation.2 EXAMPLE
5
SOLUTION
Express 20.000 000 000 052 9 in scientific notation. In this case the coefficient, 25.29, is negative. Notice that the absolute value of the coefficient, k 25.29 k , is equal to 5.29, which is between 1 and 10. In scientific notation, 20.000 000 000 052 9 is written as 25.29 ? 10211 Converting from Standard Decimal Form to Scientific Notation Place a decimal point to the right of the first nonzero digit, creating the coefficient N, where 1 # k N k , 10. Determine n, the power of 10 needed to convert the coefficient back to the original number. Write in the form N ? 10 n, where the exponent n is an integer. Examples: 346,800,000 5 3.468 ? 1080.000 008 4 5 8.4 ? 1026
The poem “Imagine” offers a creative look at the Big Bang.
Deep time The Big Bang. In 1929 the American astronomer Edwin Hubble published an astounding paper claiming that the universe is expanding. Most astronomers and cosmologists now agree with his once-controversial theory and believe that approximately 13.7 billion years ago the universe began an explosive expansion from an infinitesimally small point. This event is referred to as the “Big Bang,” and the universe has been expanding ever since it occurred.3 Scientific notation can be used to record the progress of the universe since the Big Bang Theory, as shown in Table 4.5. The Tale of the Universe in Scientific Notation Object
Age (in years)
Universe Earth Human life
13.7 billion 5 13,700,000,000 5 1.37 ? 1010 4.6 billion 5 4,600,000,000 5 4.6 ? 109 100 thousand 5 100,000 5 1.0 ? 105
Table 4.5
2
Most calculators and computers automatically translate a number into scientific notation when it is too large or small to fit into the display. The notation is often slightly modified by using the letter E (short for “exponent”) to replace the expression “times 10 to some power.” So 3.0 ? 1026 may appear as 3.0 E126. The number after the E tells how many places, and the sign (1 or 2) indicates in which direction to move the decimal point of the coefficient. 3 Depending on its total mass and energy, the universe will either expand forever or collapse back upon itself. However, cosmologists are unable to estimate the total mass or total energy of the universe, since they are in the embarrassing position of not being able to find about 90% of either. Scientists call this missing mass dark matter and missing energy dark energy, which describes not only their invisibility but also the scientists’ own mystification.
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Carl Sagan’s video Cosmos and book Dragons of Eden condense the life of the universe into one calendar year.
217
Table 4.5 tells us that humans, Homo sapiens sapiens, first walked on Earth about 100,000 or 1.0 ? 105 years ago. In the life of the universe, this is almost nothing. If all of time, from the Big Bang to today, were scaled down into a single year, with the Big Bang on January 1, our early human ancestors would not appear until less than 4 minutes before midnight on December 31, New Year’s Eve.
Algebra Aerobics 4.1 1. Express as a power of 10: a. 10,000,000,000 b. 0.000 000 000 000 01 c. 100,000 d. 0.000 01 2. Express in standard notation (without exponents): a. 1028 c. 1024 13 b. 10 d. 107 3. Express as a power of 10 and then in standard notation: a. A nanosecond in terms of seconds b. A kilometer in terms of meters c. A gigabyte in terms of bytes 4. Rewrite each measurement in meters, first using a power of 10 and then using standard notation: a. 7 cm b. 9 mm c. 5 km 5. Avogadro’s number is 6.02 ? 1023. A mole of any substance is defined to be Avogadro’s number of particles of that substance. Express this number in standard notation.
6. The distance between Earth and its moon is 384,000,000 meters. Express this in scientific notation. 7. An angstrom (denoted by Å), a unit commonly used to measure the size of atoms, is 0.000 000 01 cm. Express its size using scientific notation. 8. The width of a DNA double helix is approximately 2 nanometers, or 2 ? 1029 meter. Express the width in standard notation. 9. Express in standard notation: a. 27.05 ? 108 c. 5.32 ? 106 25 b. 24.03 ? 10 d. 1.021 ? 1027 10. Express in scientific notation: a. 243,000,000 c. 5,830 b. 20.000 008 3 d. 0.000 000 024 1 11. Express as a power of 10: 1 a. 100,000 b.
1 1,000,000,000
Exercises for Section 4.1 1. Write each expression as a power of 10. a. 10 ? 10 ? 10 ? 10 ? 10 ? 10 1 b. 10 ? 10 ? 10 ? 10 ? 10 c. one billion d. one-thousandth e. 10,000,000,000,000 f. 0.000 000 01
5. Computer storage is often measured in gigabytes and terabytes. Write these units as powers of 10. 6. Express each of the following using powers of 10. 1 a. 10,000,000,000,000 d. 10 ? 10 ? 10 ? 10 b. 0.000 000 000 001 e. one million c. 10 ? 10 ? 10 ? 10 f. one-millionth
2. Express in standard decimal notation (without exponents): a. 1027 c. 2108 e. 1023 7 25 b. 10 d. 210 f. 105
7. Write each of the following in scientific notation: a. 0.000 29 d. 0.000 000 000 01 g. 20.0049 b. 654.456 e. 0.000 002 45 c. 720,000 f. 21,980,000
3. Express each in meters, using powers of 10. (See inside back cover.) a. 10 cm c. 3 terameters b. 4 km d. 6 nanometers
8. Why are the following expressions not in scientific notation? Rewrite each in scientific notation. a. 25 ? 104 c. 0.012 ? 1022 23 b. 0.56 ? 10 d. 2425.03 ? 102
4. Express each unit using a metric prefix. (See inside back cover.) a. 1023 seconds b. 103 grams c. 102 meters
9. Write each of the following in standard decimal form: a. 7.23 ? 105 d. 1.5 ? 106 24 b. 5.26 ? 10 e. 1.88 ? 1024 23 c. 1.0 ? 10 f. 6.78 ? 107
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10. Express each in scientific notation. (Refer to the chart in Exploration 4.1.) a. The age of the observable universe b. The size of the first living organism on Earth c. The size of Earth d. The age of Pangaea e. The size of the first cells with a nucleus 11. Determine if the expressions are true or false. If false, change the right-hand side to make the expression true. a. 0.00 756 5 7.56 ? 1022 b. 3.432 ? 105 5 343,200 c. 49 megawatts 5 4.9 ? 106 watts d. 1,596,000,000 5 1.5 ? 109 e. 5 megapixels 5 5.0 ? 106 pixels f. 6 picoseconds 5 6.0 ? 1012 seconds
18. a. Generate a small table of values and plot the function y 5 k x k for 25 # x # 5. b. On the same graph, plot the function y 5 k x 2 2 k . 19. The accompanying amusing graph shows a roughly linear relationship between the “scientifically” calculated age of Earth and the year the calculation was published. For instance, in about 1935 Ellsworth calculated that Earth was about 2 billion years old. The age is plotted on the horizontal axis and the year the calculation was published on the vertical axis. The triangle on the horizontal coordinate represents the presently accepted age of Earth. a. Who calculated that Earth was less than 1 billion years old? Give the coordinates of the points that give this information. b. In about what year did scientists start putting the age of Earth at over a billion years? Give the coordinates that represent this point. c. On your graph sketch an approximation of a best-fit line for these points. Use two points on the line to calculate the slope of the line. d. Interpret the slope of that line in terms of the year of calculation and the estimated age of Earth.
12. Express each quantity in scientific notation. a. The mass of an electron is about 0.000 000 000 000 000 000 000 000 001 67 gram. b. One cubic inch is approximately 0.000 016 cubic meter. c. The radius of a virus is 0.000 000 05 meter. 13. Evaluate: a. k 9 k
b. k 29 k
c. k 21000 k
d. 2 k 21000 k
14. Determine the value of each expression. a. k 25 2 3 k c. k 2 2 6 k b. k 6 2 2 k d. 22 k 2113 k 1 k 25 k
16. What values for x would make the following true? a. k x k 5 7 c. k x 2 2 k 5 7 e. k 2 2 x k 5 7 b. k x 2 1 k 5 5 d. k 2x k , 0 f. k 2x k 5 8 17. Substitute the value x 5 5 into the statement. Then replace the ? with the sign (., ,, or 5) that would make the statement true. Then repeat for x 5 25. a. k x 2 1 k ? 5 c. k x 2 1 k ? 0 e. k 2x 2 1 k ? 11 b. 2 k 32x k ? 10 d. k 2x k ? 4 f. k 2x k ? 6
Tilton & Steiger Polkanov & Gerling
1960 Patterson Tilton & Ingrham
1950 Year of calculation
15. Determine which statements are true. a. k a 2 b k 5 k b 2 a k b. k 27a k 5 7a c. 2 k 2114 k 5 2 k 21 k 12 k 4 k d. k 22p k 5 k 22 k ? k p k
A Graph of Calculations for the Age of Earth 1970
Holmes 1940 Ellsworth 1930 Clarke 1920 1910 1900
Barrell
Sollas Wolcott 1
2
3
4
5
6
Age of Earth (in aeons = billions = 109 years)
Source: American Scientist, Research Triangle Park, NC. Copyright © 1980.
4.2 Positive Integer Exponents
?
SOMETHING TO THINK ABOUT
What can we say about the value of (21)n when n is an even integer? When n is an odd integer?
No matter what the base, whether it is 10 or any other number, repeated multiplication leads to exponentiation. For example, 3 ? 3 ? 3 ? 3 5 34 Here 4 is the exponent of 3, and 3 is called the base. In general, if a is a real number and n is a positive integer, then we define an as the product of n factors of a.
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Definition of an In the expression an, the number a is called the base and n is called the exponent or power. If n is a positive integer, then a n 5 (a-++)+ ?a?ac a -+*
(the product of n factors of a)
n factors
Exponent Rules In this section we’ll see how the rules for manipulating expressions with exponents make sense if we remember what the exponent tells us to do to the base. First we focus on cases where the exponents are positive integers. Later, we extend these rules to cases where the exponents can be any rational numbers, such as negative integers or fractions. In later courses you will extend the rules to all real numbers.
Rules for Exponents 1. a n ? a m 5 a sn1md an 2. m 5 a sn2md a 2 0 a 3. sa m d n 5 a sm?nd
4. sabd n 5 a nb n a n an 5. a b 5 n b 2 0 b b
We show below how Rules 1, 3, and 5 make sense and leave Rules 2 and 4 for you to justify in the exercises. Rule 1. To justify this rule, think about the total number of times a is a factor when an is multiplied by am: a n ? a m 5 a ? a ? a c a ? a ? a c a 5 a ? a ? a c a 5 a sn1md (+-+)-++* n factors
Rule 3. power:
(++)++* m factors
(-++)+--+* n1m factors
First think about how many times am is a factor when we raise it to the nth sa m d n 5 (+ a m -?+) a m -c am ++* n factors of am n terms
sm1m1c1md
Use Rule 1:
5a
Represent adding m n times as m ? n
5 a sm?nd
Rule 5. Remember that the exponent n in the expression (a/b)n applies to the whole expression within the parentheses: a n a a a a b 5 a b ? a b ca b b b b b ( +++)+ ++* -
n factors of a
5
a ? a ca ? b cb (b ++)++* n factors of b
5
an bn
--
n factors of a/b
(++)++*
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(+-+)+-+*
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EXAMPLE
1
Simplify and write as an expression with exponents: 73 ? 72 5 7312 5 75
(x5 ) 3 5 x5?3 5 x15
w3 ? w5 5 w315 5 w8
(112 ) 4 5 112?4 5 118
10 8 5 10823 5 10 5 10 3 EXAMPLE
2
z8 5 z823 5 z5 z3
Simplify: (3a) 4 5 34a4 5 81a4 (25x) 3 5 (25) 3x3 5 2125x3 2 3 23 8 a b 5 35 3 3 27
EXAMPLE
3
Simplify: a
22a 3 (22a) 3 (22) 3a3 28a3 b 5 5 5 3b (3b) 3 33b3 27b3
25(x3 ) 2 25x6 5 (2y2 ) 3 8y6 EXAMPLE
4
Using scientific notation to simplify calculations Deneb is 1600 light years from Earth. How far is Earth from Deneb when measured in miles?
SOLUTION
The distance that light travels in 1 year, called a light year, is approximately 5.88 trillion miles. Since
1 light year 5 5,880,000,000,000 miles
then the distance from Earth to Deneb is 1600 light years 5 (1600) ? (5,880,000,000,000 miles) 5 (1.6 ? 103 ) ? (5.88 ? 1012 miles) 5 (1.6 ? 5.88) ? (103 ? 1012 ) miles < 9.4 ? 103112 miles < 9.4 ? 1015 miles Using ratios to compare sizes of objects In comparing two objects of about the same size, it is common to subtract one size from the other and say, for instance, that one person is 6 inches taller than another. This method of comparison is not effective for objects of vastly different sizes. To say that the difference between the estimated radius of our solar system (1 terameter, or 1,000,000,000,000 meters) and the average size of a human (about 100 or 1 meter) is 1,000,000,000,000 2 1 5 999,999,999,999 meters is not particularly useful. In fact, since our measurement of the solar system certainly isn’t accurate to within 1 meter, this difference is meaningless. As shown in the following example, a more useful method for comparing objects of wildly different sizes is to calculate the ratio of the two sizes. EXAMPLE
5
The ratio of two quantities In April 2007, the U.S. federal government reported that the estimated gross federal debt was $8.87 trillion and the estimated U.S. population was 301 million. What was the approximate federal debt per person?
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8.87 ? 1012 dollars federal debt 5 U.S. population 3.01 ? 108 people
SOLUTION
5 ¢
8.87 dollars 1012 dollars ≤ ? ¢ 8≤ < 2.95 ? 104 3.01 10 people people
So the federal debt amounted to about $2.95 ? 104 or $29,500 per person.
EXAMPLE
6
How many times larger is the sun than Earth?
SOLUTION
1
The radius of the sun is approximately 109 meters and the radius of Earth is about 107 meters. One way to answer the question “How many times larger is the sun than Earth?” is to form the ratio of the two radii: 109 m radius of the sun 5 7 radius of Earth 10 m 5
109 m 5 10 927 5 102 107 m
The units cancel, so 102 is unitless. The radius of the sun is approximately 102, or 100, times larger than the radius of Earth. SOLUTION
2
Another way to answer the question is to compare the volumes of the two objects. The sun and Earth are both roughly spherical. The formula for the volume V of a sphere with radius r is V 5 43pr 3. The radius of the sun is approximately 109 meters and the radius of Earth is about 107 meters. The ratio of the two volumes is volume of the sun (4/3)p(109 ) 3 m3 5 volume of Earth (4/3)p(107 ) 3 m3 5
(109 ) 3 (Note: 43 p and m3 cancel.) (107 ) 3
1027 1021 5 106 5
So while the radius of the sun is 100 times larger than the radius of Earth, the volume of the sun is approximately 106 5 1,000,000, or 1 million, times larger than the volume of Earth!
Common Errors The first question to ask in evaluating expressions with exponents is: To what does the exponent apply? Consider the following expressions: 1.
–a n = –(an) but –an ≠ (–a)n (unless n is odd) For example, in the expression 224, the exponent 4 applies only to 2, not to 22. The order of operations says to compute the power first, before applying the negation sign. So 224 5 2(24) 5 216. If we want to raise 22 to the fourth power, we write (22)4 5 (22)(22)(22)(22) 5 16. In the expression (23b)2, everything inside the parentheses is squared. So (23b)2 5 (23b)(23b) 5 9b2. But in the expression 23b2, the exponent 2 applies only to the base b. In the case where n is an odd integer, then (2a)n will equal 2(a)n. For example, (22)3 5 (22)(22)(22) 5 28 5 223.
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2.
abn = a(bn) and –abn = –a(bn) but abn ≠ (ab)n Remember, the exponent applies only to the variable to which it is attached. In the expressions abn and 2abn, only b is raised to the nth power. For example, 2 ? 53 5 2 ? 125 5 250 22 ? 53 5 22 ? 125 5 2250
but
s2 ? 5d 3 5 s10d 3 5 1000
but
s22 ? 5d 3 5 s210d 3 5 21000
You can use parentheses () to indicate when more than one variable is raised to the nth power. 3.
(ab)n = anbn but For example,
(a + b)n ≠ an + bn (if n ≠ 1)
s2 ? 5d 3 5 23 ? 53 s10d 3 5 8 ? 125
but
1000 5 1000 4.
an • am = an+m but For example,
343 2 133
an + am ≠ an+m
102 ? 103 5 105
but
100 ? 1000 5 100,000
?
SOMETHING TO THINK ABOUT
What are some other exceptions to the generalizations made about common errors?
s2 1 5d 3 2 23 1 53 s7d 3 2 8 1 125
102 1 103 2 105 100 1 1000 2 100,000
Common Errors Involving Exponents In general, 2a n 2 s2ad n a n 1 a m 2 a n1m n n ab 2 sabd sa 1 bd n 2 a n 1 b n
Algebra Aerobics 4.2a 1. Simplify where possible, leaving the answer in a form with exponents: a. 105 ? 107 d. 55 ? 67 g. 34 1 7 ? 34 6 14 3 3 b. 8 ? 8 e. 7 1 7 h. 23 1 24 c. z 5 ? z 4 f. 5 ? 56 i. 25 1 52 2. Simplify (if possible), leaving the answer in exponent form: 1015 35 5 23 ? 34 a. c. 4 e. 6 g. 7 10 3 5 2 ? 32 86 5 34 6 b. 4 d. 7 f. h. 4 8 6 3 2 3. Write each number as a power of 10, then perform the indicated operation. Write your final answer as a power of 10. 1,000,000 a. 100,000 ? 1,000,000 e. 0.001 b. 1,000 ? 0.000 001
f.
c. 0.000 000 000 01 ? 0.000 01
g.
d.
1,000,000,000 10,000
0.000 01 0.0001 0.000 001 10,000
4. Simplify: 4 5 a. s10 d
4 d. s2xd
2 3 g. s23x d
2 3 b. s7 d
4 3 e. s2a d
4 5 c. sx d
3 f. s22ad
3 2 4 h. ssx d d 2 3 i. s25y d
c. 252
e. s23yz 2 d 4
5. Simplify: 22x 3 a. a b 4y
b. s25d 2 d. 23syz 2 d 4 f. s23yz 2 d 3 6. A compact disk or CD has a storage capacity of about 737 megabytes (7.37 ? 108 bytes) . If a hard drive has a capacity of 40 gigabytes (4.0 ? 1010 bytes) , how many CD’s would it take to equal the storage capacity of the hard drive? 7. Write as a single number with no exponents: a. s3 1 5d 3 c. 3 ? 52 b. 33 1 53
d. 23 ? 52
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223
Estimating Answers By rounding off numbers and using scientific notation and the rules for exponents, we can often make quick estimates of answers to complicated calculations. In this age of calculators and computers we need to be able to roughly estimate the size of an answer, to make sure our calculations with technology make sense.
EXAMPLE
7
Estimate the value of s382,152d ? s490,572,261d s32,091d ? s1942d Express your answer in both scientific and standard notation.
SOLUTION
Round each number: s382,152d ? s490,572,261d s400,000d ? s500,000,000d < s32,091d ? s1942d s30,000d ? s2000d s4 ? 105 d ? s5 ? 108 d s3 ? 104 d ? s2 ? 103 d
rewrite in scientific notation
<
group the coefficients and the powers of 10
simplify each expression
<
we get in scientific notation or in standard notation
< 3.33 ? 106 < 3,330,000
4?5 105 ? 108 b ? a 4 b 3?2 10 ? 103
20 1013 ? 6 107
Using a calculator on the original problem, we get a more precise answer of 3,008,200.
EXAMPLE
8
As of 2007 the world population was approximately 6.605 billion people. There are roughly 57.9 million square miles of land on Earth, of which about 22% are favorable for agriculture. Estimate how many people per square mile of farmable land there are as of 2007. size of world population 6.605 billion people 5 amount of farmable land 22% of 57.9 million square miles
SOLUTION
rewrite as powers of 10
5
6.605 ? 109 people (0.22) ? (57.9) ? 106 mile2
round each number
<
6.6 ? 109 people (0.2) ? 60 ? 106 mile2
simplify
<
we get in scientific notation
6.6 ? 109 people 12 ? 106 mile2 < 0.55 ? 103 people/mile2
or in standard notation
< 550 people/mile2
So there are roughly 550 people/mile2 of farmable land in the world. Using a calculator and the original numbers, we get a more accurate answer of 519 people/mile2 of farmable land, which is close to our estimate.
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Algebra Aerobics 4.2b 1. Estimate the value of: a. s0.000 297 6d ? s43,990,000d 453,897 ? 2,390,702 b. 0.004 38 0.000 000 319 c. 162,000 d. 28,000,000 ? 7,629 e. 0.000 021 ? 391,000,000 2. Evaluate the following without the aid of a calculator: 2.0 ? 105 a. s3.0 ? 103 d s4.0 ? 102 d c. 5.0 ? 103 2 2 s5.0 ? 10 d b. d. s4.0 ? 102 d 3 ? s2.0 ? 103 d 2 2.5 ? 103
3. The radius of Jupiter, the largest of the planets in our solar system, is approximately 7.14 ? 104 km. (If r is the radius of a sphere, the sphere’s surface area equals 4pr 2 and its volume equals 43pr 3.) Assuming Jupiter is roughly spherical, a. Estimate the surface area of Jupiter. b. Estimate the volume of Jupiter. (Express your answers in scientific notation.) 4. Only about three-sevenths of the land favorable for agriculture is actually being farmed. Using the facts in Example 8, estimate the number of people per square mile of farmable land that is being used. Should your estimate be larger or smaller than the ratio of people to farmable land? Explain. (Round your answer to the nearest integer.)
Exercises for Section 4.2 1. Simplify, when possible, writing your answer as an expression with exponents: z7 a. 104 ? 103 d. x 5 ? x 10 g. 2 j. 45 ? s42 d 3 z b. 104 1 103
e. sx 5 d 10
h. 2560
c. 103 1 103
f. 47 1 52
i.
2. Simplify: a. s21d 4 b. 2s1d 4
c. sa 4 d 3
e. s2a 4 d 3
g. s10a 2b 3 d 3
d. 2s2a 2 d 3
f. s22a 4 d 3
h. s2abd 223a 2b 2
3. Simplify: a. s22ad 4 b. 22sad
35 ? 32 38
e. s2x 4 d 5 f. s24x 3 d 2 1 x 3 s2x 3 d
4
c. s2x 5 d 3
g. s50a 10 d 2 h. s3abd 3 1 ab
2 3
d. s22ab d
4. Simplify and write each variable as an expression with positive exponents: 3 2 25a 3 4 10a 3 2 22x 3 3 b b a. 2a b b. a 2 b c. a d. a 5 a 5b 3y 2 5. Simplify and write each variable as an expression with positive exponents: 5 2 a. 2a b 8
3x 3 3 b. a 2 b 5y
210x 5 4 b c. a 2b 2
2x 5 3 d. a 2 b x
6. Evaluate and express your answer in standard decimal form: a. 224 1 22 e. 103 1 23 b. 22 1 s24d 3
2
f. 2 ? 10 1 10 1 10 3
3
c. 2 ? 32 2 3s22d 2
g. 2 ? 103 1 s210d 3
d. 2104 1 105
h. s1000d 0
2
7. Convert each number into scientific notation then perform the indicated operation. Leave your answer in scientific notation. a. 2,000,000 ? 4000 b. 1.4 million 4 7000 c. 50 billion ? 60 trillion d. 2500 billion 4 500 thousand 8. Convert each number into scientific notation and then perform the operation without a calculator. a. 60,000,000,000 1 40,000,000,000 (20,000) 6 b. (400) 3 c. (2,000,000) ? (40,000) 9. Simplify each expression using the properties of exponents. 22x5y5 3 a. (x5y)(x6 )(x2y3 ) c. a 2 2 b e. s3x 2y 5 d 4 xy 3x3y 2 5x6y3 b b. 2 2 d. sx 2 d 5 ? s2y 2 d 4 f. a xy 5xy 10. Each of the following simplifications contains an error made by students on a test. Find the error and correct the simplification so that the expression becomes true. a. [sx 2 d 3]5 5 [x 5]5 5 x 25 b.
7x2y6 7x2y6 5 7x4y8 2 5 (xy) x2y2
c. a
4x3y5 3 2x2y 3 2 6 3 b 5 3x y 4b 5 a 6xy 3
d. s1.1 ? 106 d ? s1.1 ? 104 d 5 1.1 ? 106 4 ? 106 5 0.5 ? 103 5 5.0 ? 104 8 ? 103 f. 6 ? 103 1 7 ? 105 5 13 ? 108 e.
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11. Express your answer as a power of 10 and in standard decimal form. (Refer to table on inside back cover.) a. How many times larger is a gigabyte of memory than a megabyte? b. How many times farther is a kilometer than a dekameter? c. How many times heavier is a kilogram than a milligram? d. How many times longer is a microsecond than a nanosecond? 12. In 2006 the People’s Republic of China was estimated to have about 1,314,000,000 people, and Monaco about 33,000. Monaco has an area of 0.75 miles2, and China has an area of 3,705,000 miles2. a. Express the populations and geographic areas in scientific notation. b. By calculating a ratio, determine how much larger China’s population is than Monaco’s. c. What is the population density (the number of people per square mile) for each country? d. Write a paragraph comparing and contrasting the population size and density for these two nations. 13. a. In 2006 Japan had a population of approximately 127.5 million people and a total land area of about 152.5 thousand square miles. What was the population density (the number of people per square mile)? b. In 2006 the United States had a population of approximately 300 million people and a total land area of about 3620 thousand square miles. What was the population density of the United States? c. Compare the population densities of Japan and the United States. 14. The distance that light travels in 1 year (a light year) is 5.88 ? 1012 miles. If a star is 2.4 ? 108 light years from Earth, what is this distance in miles? 15. An average of 1.5 ? 104 Coca-Cola beverages were consumed every second worldwide in 2005. There are 8.64 ? 104 seconds in a day. What was the daily consumption of CocaCola in 2005? (Source: World of Coca-Cola® Atlanta) 16. Change each number into scientific notation, then perform the indicated calculation without a calculator. a. A $600,000 lottery jackpot is divided among 300 people. What are the winnings per person? b. A total of 2500 megawatts are used over 500 hours. What is the rate in watts per hour? c. If there were 6 million births in 30 years, what is the birth rate per year? 17. a. For any nonzero real number a, what can we say about the sign of the expression s2ad n when n is an even integer? What can we say about the sign of s2ad n when n is an odd integer? b. What is the sign of the resulting number if a is a positive number? If a is a negative number? Explain your answer. 18. Round off the numbers and then estimate the value of each of the following expressions without using a calculator. Show
225
your work, writing your answers in scientific notation. If available, use a calculator to verify your answers. a. s2,968,001,000d ? s189,000d b. s0.000 079d ? s31,140,284,788d 4,083,693 ? 49,312 c. 213 ? 1945 19. Simplify each expression using two different methods, and then compare your answers. Method I: Simplify inside the parentheses first, and then apply the exponent rule outside the parentheses. Method II: Apply the exponent rule outside the parentheses, and then simplify. a. a
m 2n 3 2 b mn
b. a
2a 2b 3 4 b ab 2
20. Verify that sa 2 d 3 5 sa 3 d 2 using the rules of exponents. 16a 12 2a 3 4 21. Verify that a 2 b 5 using the rules of exponents. 5b 625b 8 22. An article in the journal Nature (October 2000) analyzed samples of the ballast water from ships arriving in the Chesapeake Bay from foreign ports. It reported that ballast was an important factor in the global distribution of microorganisms. One gallon of ballast water contained on average 3 billion bacteria, including some that cause cholera. The scientists estimated that about 2.5 billion gallons of ballast water are discharged into the Chesapeake Bay each year. Estimate the number of bacteria per year discharged in ballast water into the Chesapeake Bay. Write your answer in scientific notation. 23. Justify the following rule for exponents. If a and b are any nonzero real numbers and n is an integer $ 0, then sabd n 5 a nb n 24. Justify the following rule for exponents. Consider the case of n $ m and assume m and n are integers . 0. If a is any nonzero real number, then an 5 a sn2md am 25. In 2006 the United Kingdom generated approximately 81 terawatt-hours of nuclear energy for a population of about 60.6 million on 94,525 miles2. In the same year the United States generated approximately 780 terawatt-hours of nuclear energy for a population of about 300 million on 3,675,031 miles2. A terawatt is 1012 watts. a. How many terawatt-hours is the United Kingdom generating per person? How many terawatt-hours is it generating per square mile? Express each in scientific notation. b. How many terawatt-hours is the United States generating per person? How many terawatt-hours are we generating per square mile? Express each in scientific notation. c. How much nuclear energy is being generated in the United Kingdom per square mile relative to the United States? d. Write a brief statement comparing the relative magnitude of generation of nuclear power per person in the United Kingdom and the United States.
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26. Hubble’s Law states that galaxies are receding from one another at velocities directly proportional to the distances separating them. The accompanying graph illustrates that Hubble’s Law holds true across the known universe. The plot includes ten major clusters of galaxies. The boxed area at the lower left represents the galaxies observed by Hubble when he discovered the law. The easiest way to understand this graph is to think of Earth as being at the center of the universe (at 0 distance) and not moving (at 0 velocity). In other words, imagine Earth at the origin of the graph (a favorite fantasy of humans). Think of the horizontal axis as measuring the distance of the galaxy from Earth, and the vertical as measuring the velocity at which a galaxy cluster is moving away from Earth (the recession velocity). Then answer the following questions. a. Identify the coordinates of two data points that lie on the regression line drawn on the graph. b. Use the coordinates of the points in part (a) to calculate the slope of the line. That slope is called the Hubble constant. c. What does the slope mean in terms of distance from Earth and recession velocity? d. Construct an equation for our line in the form y 5 mx 1 b. Show your work.
Distance from Earth vs. Recession Velocity 70 Hydra Recession velocity (thousands of kilometers per second)
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60
50 .. Bootes 40
30
Corona Borealis 20 Ursa Major 10 Perseus Virgo 0
1 2 3 Distance from Earth (billions of light years)
Source: T. Ferris, Coming of Age in the Milky Way (New York: William Morrow, 1988). Copyright © by Timothy Ferris. By permission of William Morrow & Company, Inc.
4.3 Negative Integer Exponents The definitions for raising any base to the zero power or to a negative power follow a logic that is similar to the one used to define 100 5 1 and 102n 5 1 n. 10
Zero and Negative Exponents If a is nonzero and n is a positive integer, then a0 5 1 1 a 2n 5 n a
It is important to note that a 1 5 a, so a 21 5 a11 5 a1 . In the following examples, we show how to apply the five rules for exponents when the exponents are negative integers or zero. EXAMPLE
1
SOLUTION
Simplify x 2 ? x 25. Using Rule 1 for exponents, x 2 ? x 25 5 x 21s25d 5 x 23 or we can simplify by first writing x 25 as x15 and then use Rule 2 for exponents: x 2 ? x 25 5 x 2 ?
1 x2 225 5 x 23 5 5 5 5 x x x
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EXAMPLE
2
Simplify. Express your answer with positive exponents. 10 2 s25d 2 a. c. 6 10 s25d 6 b.
SOLUTION
EXAMPLE
3
62 627
d.
b.
62 5 622s27d 5 6 9 627
c.
s25d 2 1 1 1 226 5 s25d 24 5 6 5 s25d 4 5 4 4 5 4 s25d s25d s21d s5d 5
d.
x 22 1 2224 5 x 26 5 6 4 5 x x x
Simplify: b. sw 2 d 27
Using Rule 3 for exponents, a. s1328 d 3 5 13s28d3 5 13224
EXAMPLE
4
x 22 x4
Using Rule 2 for exponents, 10 2 1 a. 5 10226 5 1024 5 10 6 10 4
a. s1328 d 3 SOLUTION
227
b. sw 2 d 27 5 w 2s27d 5 w 214
Simplify: v 22 sw 5 d 2 sv 21 d 4 w 23
SOLUTION
Apply Rule 3 twice:
v22 (w5 ) 2 v22 w10 5 (v21 ) 4w23 v24 w23 5 v222(24)w102(23)
Apply Rule 2 twice:
5 v2w13
Evaluating Aa B 2n b The rule for applying negative powers is the same whether a is an integer or a fraction: a 2n 5 1/a n
where a 2 0
For example, 1 21 1 1 2 a b 5 514 a b 51? a b 52 2 s1/2d 1 2 1
In general, if a and b are nonzero, then a 2n 1 a n b n b n bn a b 5 5 1 4 a b 5 1 ? a b 5 a b 5 b sa/bd n b a a an
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EXAMPLE
5
Simplify: 1 211 1 22 a. a b ? a b 2 2
SOLUTION
a 3 a 25 b. a b ? a b b b
a. Using Rule 1 for exponents and the definition of a 2n, 1 211 1 22 1 2111(22) a b ? a b 5 a b 2 2 2 1 213 2 13 5 a b 5 a b 5 213 5 8192 2 1 b. Using Rules 1 and 5 for exponents and the definition of a2n, a 3 a 25 a 31(25) a b ? a b 5 a b b b b a 22 b 2 b2 5 a b 5 a b 5 2 b a a
Algebra Aerobics 4.3 1. Simplify (if possible). Express with a single positive exponent, if possible. 73 a. 105 ? 1027 e. 3 7 116 b. f. a 22 ? a 23 1124 c.
325 324
g. 34 ? 33
55 h. s22 ? 3ds26 ds24 ? 3d 67 2. A typical TV signal, traveling at the speed of light, takes 3.3 ? 1026 seconds to travel 1 kilometer. Estimate how long it would take the signal to travel across the United States (a distance of approximately 4300 kilometers). d.
3. Distribute and simplify: a. x 22 sx 5 1 x 26 d b. 2a 2 sb 2 2 3ab 1 5a 2 d 4. Simplify: a. s104 d 25
8 22 d. a b x
b. s722 d 23
e. s2x 22 d 21
c. s2a 3 d 22
f. 2sx 22 d 21
5. Simplify: t 23t 0 a. 24 3 st d b.
v 23w 7 sv d w 210 22 3
3 24 b 2y 2 3 h. s2y 2 d 24 g. a
c.
728x 21y 2 725xy 3
d.
as5b 21c3 d 2 5ab 2c26
Exercises for Section 4.3 1. Simplify and express your answer using positive exponents. Check your answers by applying the rules for exponents and doing the calculations. a. 103 # 1022 c. (1023 ) 2 b.
1022 103
d.
103 1022
2. Simplify and express your answer with positive exponents: a. sx 23 d ? sx 4 d d. sn 22 d 23 b. sx 23 d ? sx 22 d
e. (2n22 ) 23
c. sx 2 d 23
f. n 24 sn 5 2 n 2 d 1 n 23 sn 2 n 4 d
3. Simplify where possible. Express your answer with positive exponents. 23x 4 x 22y a 22bc25 a. 5 8 c. e. 3 2x xy sab 2 d 23c b.
x 4y 7 x 3y 25
d.
sx 1 yd 4 sx 1 yd 27
4. Simplify where possible. Express your answer with positive exponents. a. s3 ? 38 d 22 d. 2x 23 1 3xsx 24 d 3 24 12 b. x ? x ? x e. 1025 1 522 1 1010 6 6 7 24 c. 2 1 2 1 2 1 2
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5. Evaluate and write the result using scientific notation: 3.25 ? 108 a. s2.3 ? 104 ds2.0 ? 106 d d. 6.29 ? 1015 25 8 b. s3.7 ? 10 ds1.1 ? 10 d e. s6.2 ? 1052 d 3 23 8.19 ? 10 c. f. s5.1 ? 10211 d 2 5.37 ? 1012 6. Write each of the following in scientific notation: 1 a. 725 ? 1023 c. e. 2725 ? 10223 725 ? 1023 b. 725 ? 10223 d. 2725 ? 1023 7. Change each number to scientific notation, then simplify using rules of exponents. Show your work, recording your final answer in scientific notation. 8000 a. 10% of 0.000 01 d. 0.000 8 0.006 4 0.000 05 b. e. 50,000 8000 3 c. f. 5,000,000 ? 40,000 0.006 8. Use scientific notation and the rules of exponents to perform the indicated operation without a calculator. Show your work, recording your answer in decimal form. 20 10,000,000 a. d. 200,000 25,000 0.006 b. e. 0.06 ? 600 60,000 c. 200 ? 0.000 007 5 f. 10% of 0.000 05 9. For each equation determine the value of x that makes it true. 1 5 0.000 1 a. 10 x 5 0.000 001 c. 10 x 1 b. 10 x 5 d. 102x 5 100,000 1,000,000 10. For each equation determine the value of x that makes it true. 1 a. 6.3 ? 10 x 5 0.000 63 d. x3 5 1000 b. 1023 5 x e. 423 ? 225 5 2x 1 c. 5x 5 f. 921 ? 2722 5 3x 125 11. Simplify the following expressions by using properties of exponents. Write your final answers with only positive exponents. s22x 3y 21 d 23 3x 2y 25 21 a. c. a 3 4 b 2 22 0 sx y d 5x y b.
s22x 3y 21 d 22 sx 2y 22 d 21
d. c A3x 21z 4B
22
R
23
12. Each of the following simplifications is false. In each case identify the error and correct it. a. x 22x 25 5 x 10 x 2x 21 x 221x 2y 22 5 22 5 5 3 b. 21 5 x y 2y y 2y
229
1 2 1 b 5 2 3x 9x 1 1 5 1 x y
c. s3x 21 d 2 5 a d. sx 1 yd 21
13. A TV signal traveling at the speed of light takes about 8 ? 1025 second to travel 15 miles. How long would it take the signal to travel a distance of 3000 miles? 14. Round off the numbers and then estimate the values of the following expressions without a calculator. Show your work, writing your answers in scientific notation. If available, use a calculator to verify your answers. a. s0.000 359d ? s0.000 002 47d b.
0.000 007 31 ? 82,560 1,891,000
15. Simplify and express your answer with positive exponents. x 22 2 y 21 a. b. s5x 22y 23 d 22 sxy 2 d 21 16. (Requires a calculator that can evaluate powers.) Calculators and spreadsheets use slightly different formats for scientific notation. For example, if you type in Avogadro’s number either as 602,000,000,000,000,000,000,000 or as 6.02 ? 10 23, the calculator or spreadsheet will display 6.02 E 23, where E stands for “exponent” or power of 10. Perform the following calculations using technology, then write the answer in standard scientific notation rounded to three places. 9 50 7 a. a b d. 15 5 6 b. 235
e. s5d 210 s2d 10
1 7 c. a b 3
f. s24d 5 a
1 b s16d 12
17. Describe at least three different methods for entering 5.23 ? 1023 into a calculator or spreadsheet. 18. Using rules of exponents, show that
95 5 331. 2727
19. Using rules of exponents, show that
1 5 xn. x2n
20. Write an expression that displays the calculation(s) necessary to answer the question. Then use scientific notation and exponent rules to determine the answer. a. Find the number of nickels in $500.00. b. The circumference of Earth is about 40.2 million meters. Find the radius of Earth, in kilometers, using the formula C 5 2pr. 21. According to the National Confectioners Association, in 2006 there were 35 million pounds (or 9 billion kernels) of candy corn made for Halloween. How many kernels are in a pound? 22. The robot spacecraft NEAR (Near Earth Asteroid Rendezvous) is on a four-year mission through the inner solar system to study asteroids. In February 2001 the spacecraft
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landed on Eros, a Manhattan-sized asteroid 160 million miles from Earth. a. If radio messages travel at the speed of light, how long would it take for a message sent back from the NEAR spacecraft to reach the scientists? b. The near-Earth asteroid Cruithne is now known to be a companion, and an unusual one, of Earth. This asteroid shares Earth’s orbit, its motion “choreographed” in such a
way as to remain stable and avoid colliding with our planet. At its closest approach Cruithne gets to within 0.1 astronomical unit of Earth (about 15 million kilometers). The asteroid in 2004 was about 0.3 astronomical unit (45 million kilometers) from Earth. If the NEAR spacecraft was in orbit around Cruithne at that time, how long would a radio signal transmitted from Earth take to reach the spacecraft?
4.4 Converting Units Problems in science constantly require converting back and forth between different units of measure. To do so, we need to be comfortable with the laws of exponents and the basic metric and English units (see Table 4.1 or a more complete table on the inside back cover). The following unit conversion examples describe a strategy based on conversion factors.
Exploration 4.1 will help you understand the relative ages and sizes of objects in our universe and give you practice in scientific notation and unit conversion.
Converting Units within the Metric System EXAMPLE
1
Conversion Factors Light travels at a speed of approximately 3.00 ? 105 kilometers per second (km/sec). Describe the speed of light in meters per second (m/sec).
SOLUTION
The prefix kilo means thousand. One kilometer (km) is equal to 1000 or 103 meters (m): 1 km 5 103 m
(1)
Dividing both sides of Equation (1) by 1 km, we can rewrite it as 15
103 m 1 km
If instead we divide both sides of Equation (1) by 103 m, we get 1 km 51 103 m 3
m 1 km The ratios 10 1 km and 103 m are called conversion factors, because we can use them to convert between kilometers and meters. What is the right conversion factor? If units in km sec are multiplied by units in meters per kilometer, we have
km m ? sec km and the result is in meters per second. So multiplying the speed of light in km/sec by a conversion factor in m/km will give us the correct units of m/sec. A conversion factor always equals 1. So we will not change the value of the original quantity by multiplying it by a conversion factor. In this case, we use the 3 m conversion factor of 10 1 km . km 103 m ? sec 1 km 5 3 5 3.00 ? 10 ? 10 m/sec
3.00 ? 105 km/sec 5 3.00 ? 105
5 3.00 ? 108 m/sec Hence light travels at approximately 3.00 ? 108 m/sec.
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4.4 Converting Units
EXAMPLE
2
SOLUTION
231
Check your answer in Example 1 by converting 3.00 ? 108 m/sec back to km/sec. Here we use the same strategy, but now we need to use the other conversion factor. Multiplying 3.00 ? 108 m/sec by (1 km)/(103 m) gives us 3.00 ? 108
m 1 km 108 km ? 3 5 3.00 ? 3 sec 10 m 10 sec 5 3.00 ? 105 km/sec
which was the original value given for the speed of light.
Converting between the Metric and English Systems EXAMPLE
3
You’re touring Canada, and you see a sign that says it is 130 kilometers to Toronto. How many miles is it to Toronto?
SOLUTION
The crucial question is, “What conversion factor should be used?” From Table 4.1 we know that 1 km < 0.62 mile This equation can be rewritten in two ways: 1<
0.62 mile 1 km or 1 < 1 km 0.62 mile
It produces two possible conversion factors:
?
SOMETHING TO THINK ABOUT
0.62 mile 1 km and 1 km 0.62 mile
Why is the conversion factor 1 km/ 0.62 miles not helpful in solving this problem?
Which one will convert kilometers to miles? We need one with kilometers in the mile denominator and miles in the numerator, namely 0.62 1 km , so that the km will cancel when we multiply by 130 km: 130 km ?
0.62 mile 5 80.6 miles 1 km
So it is a little over 80 miles to Toronto.
Using Multiple Conversion Factors 4
Light travels at 3.00 ? 105 km/sec. How many kilometers does light travel in one year?
SOLUTION
Here our strategy is to use more than one conversion factor to convert from seconds to years. Use your calculator to perform the following calculations:
EXAMPLE
3.00 ? 105
km 60 sec 60 min 24 hr 365 days ? ? ? ? 5 94,608,000 ? 105 km/year sec 1 min 1 hr 1 day 1 year < 9.46 ? 107 ? 105 km/year 5 9.46 ? 1012 km/year
So a light year, the distance light travels in one year, is approximately equal to 9.46 ? 1012 kilometers.
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Algebra Aerobics 4.4 On the back inside cover of the text there are tables containing metric prefixes and conversion facts. Round your answers to two decimal places. 1. Coca-Cola Classic comes in a 2-liter container. How many fluid ounces is that? (Note: There are 32 ounces in a quart.) 2. A child’s height is 120 cm. How tall is she in inches? 3. Convert to the desired unit: a. 12 inches 5 __________ cm b. 100 yards 5 __________ meters c. 20 kilograms 5 __________ pounds d. $40,000 per year 5 $__________ per hour (assume a 40-hour work week for 52 weeks) e. 24 hr/day 5 __________ sec/day f. 1 gallon 5 __________ ml g. 1 mph 5 __________ ft/sec 4. The distance between the sun and the moon is 3.84 ? 108 meters. Express this in kilometers. 5. The mean distance from our sun to Jupiter is 7.8 ? 108 kilometers. Express this distance in meters. 6. A light year is about 5.88 ? 1012 miles. Verify that 9.46 ? 1012 kilometers < 5.88 ? 1012 miles. 7. 1 angstrom (Å) 5 1028 cm. Express 1 angstrom in meters. 8. If a road sign says the distance to Quebec is 218 km, what is the distance in miles? 9. The distance from Earth to the sun is about 93,000,000 miles. There are 5280 feet in a mile, and a dollar bill is approximately 6 inches long. Estimate how many dollar bills would have to be placed end to end to reach from Earth to the sun. 10. Fill in the missing parts of the following conversion.
11. Anthrax spores, which were inhaled by postal workers, causing severe illness and death, are no larger than 5 microns in diameter. How much larger than a spore is the tip of a pencil that is 1 millimeter in diameter? (Note: A micron is the same as a micrometer, mm.) 12. Use the conversion factor of 1 light year 5 9.46 ? 1012 kilometers or 5.88 ? 1012 miles to determine the following. a. Alpha Centauri, the nearest star to our sun, is 4.3 light years away. What is the distance in kilometers? How many miles away is it? b. The radius of the Milky Way is 108 light years. How many meters is that? c. Deneb is a star 1600 light years from Earth. How far is that in feet? 13. If 1 angstrom, Å, 5 10210 meter, determine the following values. a. The radius of a hydrogen atom is 0.5 angstrom. How many meters is the radius? b. The radius of a cell is 105 angstroms. How many meters is the cell’s radius? c. A radius of a proton is 0.00001 angstrom. Express the proton’s radius in meter. 14. The Harvard Bridge, which connects Cambridge to Boston along Massachusetts Avenue, is literally marked off in units called Smoots. A Smoot is equal to about 5.6 feet, the height of an M.I.T. fraternity pledge named Oliver Smoot. The bridge is approximately 364 Smoots long. How long is the bridge in feet? Show all units when doing your conversion.
2560 mi 2560 mi 1.6 km ? 5 ? 5 4.2 hrs 4.2 hrs ? ? ? ? km ? 5 5 ? ? 60 min ? min
Exercises for Section 4.4 Use the conversion table on the back cover of the text for problems in this section. 1. Change the following English units to the metric units indicated. a. 50 miles to kilometers d. 12 inches to centimeters b. 3 feet to meters e. 60 feet to meters c. 5 pounds to kilograms f. 4 quarts to liters
2. Change the following metric units to the English units indicated. a. 25 kilometers to miles d. 50 grams to ounces b. 700 meters to yards e. 10 kilograms to pounds c. 250 centimeters to inches f. 10,000 milliliters to quarts
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3. For the following questions, make an estimate and then check your estimate using the conversion table on the inside back cover: a. One foot is how many centimeters? b. One foot is what part of a meter? 4. A football field is 100 yards long. How many meters is this? What part of a kilometer is this? 5. How many droplets of water are in a river that is 100 km long, 250 m wide, and 25 m deep? Assume a droplet is 1 milliliter. (Note: one liter 5 one cubic decimeter and 10 decimeters 5 l meter.) 6. a. A roll of aluminum foil claims to be 50 sq ft or 4.65 m2. Show the conversion factors that would verify that these two measurements are equivalent. b. One cm3 of aluminum weighs 2.7 grams. If a sheet of aluminum foil is 0.003 8 cm thick, find the weight of the roll of aluminum foil in grams. 7. If a falling object accelerates at the rate of 9.8 meters per second every second, how many feet per second does it accelerate each second? 8. Convert the following to feet and express your answers in scientific notation. a. The radius of the solar system is approximately 1012 meters. b. The radius of a proton is approximately 10215 meter. 9. The speed of light is approximately 1.86 ? 105 miles/sec. a. Write this number in decimal form and express your answer in words. b. Convert the speed of light into meters per year. Show your work.
233
c. Speed is distance divided by time. Find the orbital speed of Earth in miles per hour. 13. A barrel of U.S. oil is 42 gallons. A barrel of British oil is 163.655 liters. Which barrel is larger and by how much? 14. A barrel of wheat is 3.2812 bushels (U.S. dry) or 4.0833 cubic feet. a. How many cubic feet are in a bushel of wheat? b. How many cubic inches are in a barrel? c. How many cubic centimeters are in a bushel? 15. In the United States, land is measured in acres and one acre is 43,560 sq ft. a. If you buy a one-acre lot that is in the shape of a square, what would be the length of each side in feet? b. A newspaper advertisement states that all lots in a new housing development will be a minimum of one and a half acres. Assuming the lot is rectangular and has 150 ft of frontage, how deep will the minimal-size lot be? If the new home owner wants to fence in the lot, how many yards of fencing would be needed? c. The metric unit for measuring land is the square hectometer. (A hectometer is a length of 100 meters.) Find the size of a one-acre lot if it were measured in square hectometers. d. A hectare is 100 acres. How many one-acre lots can fit in a square mile? How many hectares is that? 16. Estimate the number of heartbeats in a lifetime. Explain your method. 17. A nanosecond is 1029 second. Modern computers can perform on the order of one operation every nanosecond. Approximately how many feet does an electrical signal moving at the speed of light travel in a computer in 1 nanosecond?
10. The average distance from Earth to the sun is about 150,000,000 km, and the average distance from the planet Venus to the sun is about 108,000,000 km. a. Express these distances in scientific notation. b. Divide the distance from Venus to the sun by the distance from Earth to the sun and express your answer in scientific notation. c. The distance from Earth to the sun is called 1 astronomical unit (1 A.U.) How many astronomical units is Venus from the sun? d. Pluto is 5,900,000,000 km from the sun. How many astronomical units is it from the sun?
18. Since light takes time to travel, everything we see is from the past. When you look in the mirror, you see yourself not as you are, but as you were nanoseconds ago. a. Suppose you look up tonight at the bright star Deneb. Deneb is 1600 light years away. When you look at Deneb, how old is the image you are seeing? b. Even more disconcerting is the fact that what we see as simultaneous events do not necessarily occur simultaneously. Consider the two stars Betelgeuse and Rigel in the constellation Orion. Betelgeuse is 300 and Rigel 500 light years away. How many years apart were the images generated that we see simultaneously?
11. The distance from Earth to the sun is approximately 150 million kilometers. If the speed of light is 3.00 ? 105 km/sec, how long does it take light from the sun to reach Earth? If a solar flare occurs right now, how long would it take for us to see it?
19. The world population in 2005 was approximately 6.45 billion people. During that year the Coca-Cola company claimed that 15,000 of their beverages were consumed every second. What was the worldwide consumption of their beverages per year per person in 2005?
12. Earth travels in an approximately circular orbit around the sun. The average radius of Earth’s orbit around the sun is 9.3 ? 107 miles. Earth takes one year, or 365 days, to complete one orbit. a. Use the formula for the circumference of the circle to determine the distance the Earth travels in one year. b. How many hours are in one year?
20. A homeowner would like to spread shredded bark (mulch) over her flowerbeds. She has three flowerbeds measuring 25 ft by 3 ft, 15 ft by 4 ft, and 30 ft by 1.5 ft. The recommended depth for the mulch is 4 inches, and the shredded bark costs $27.00 per one cubic yard. How much will it cost to cover all of the flowerbeds with shredded bark? (Note: You cannot buy a portion of a cubic yard of mulch.)
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21. A circular swimming pool is 18 ft in diameter and 4 ft deep. a. Determine the volume of the pool in gallons if one gallon is 231 cubic inches. b. The pool’s filter pump can circulate 2500 gal per hour. How many hours do you need to run the filter in order to filter the number of gallons contained in the pool? c. One pound of chlorine shock treatment can treat 10,000 gal. How much of the shock treatment should you use?
24. Computer technology refers to the storage capacity for information with its own special units. Each minuscule electrical switch is called a “bit” and can be off or on. As the information capacity of computers has increased, the industry has developed some much larger units based on the bit:
22. An angstrom, Å, is a metric unit of length equal to one ten billionth of a meter. It is useful in specifying wavelengths of electromagnetic radiation (e.g., visible light, ultraviolet light, X-rays, and gamma rays). a. The visible-light spectrum extends from approximately 3900 angstroms (violet light) to 7700 angstroms (red light) Write this range in centimeters using scientific notation. b. Some gamma rays have wavelengths of 0.0001 angstrom. Write this number in centimeters using scientific notation. c. The nanometer (nm) is 10 times larger than the angstrom, so 1 nm is equal to how many meters?
1 kilobyte 5 210 bytes, or 1024 bytes (a kilobyte is sometimes abbreviated Kbyte)
23. The National Institutes of Health guidelines suggest that adults over 20 should have a body mass index, or BMI, under 25. This index is created according to the formula BMI 5
weight in kilograms sheight in metersd 2
a. Given that 1 kilogram 5 2.2 pounds, and 1 meter 5 39.37 inches, calculate the body mass index of President George W. Bush, who is 6 feet tall and in 2003 weighed 194 pounds. According to the guidelines, how would you describe his weight? b. Most Americans don’t use the metric system. So in order to make the BMI easier to use, convert the formula to an equivalent one using weight in pounds and height in inches. Check your new formula by using Bush’s weight and height, and confirm that you get the same BMI. c. The following excerpt from the article “America Fattens Up” (The New York Times, October 20, 1996) describes a very complicated process for determining your BMI: To estimate your body mass index you first need to convert your weight into kilograms by multiplying your weight in pounds by 0.45. Next, find your height in inches. Multiply this number by 0.254 to get meters. Multiply that number by itself and then divide the result into your weight in kilograms. Too complicated? Internet users can get an exact calculation at http://141.106.68.17/bsa.acgl. Can you do a better job of describing the process? d. A letter to the editor from Brent Kigner, of Oneonta, N.Y., in response to the New York Times article says: Math intimidates partly because it is often made unnecessarily daunting. Your article “American Fattens Up” convolutes the procedure for calculating the Body Mass Index so much that you suggest readers retreat to the Internet. In fact, the formula is simple: Multiply your weight in pounds by 703, then divide by the square of your height in inches. If the result is above 25, you weigh too much. Is Brent Kigner right?
1 byte 5 8 bits 1 kilobit 5 210 bits, or 1024 bits (a kilobit is sometimes abbreviated Kbit)
1 megabit 5 220 bits, or 1,048,576 bits 1 megabyte 5 220 bytes, or 1,048,576 bytes 1 gigabyte 5 230 bytes, or 1,073,741,824 bytes a. How many kilobytes are there in a megabyte? Express your answer as a power of 2 and in scientific notation. b. How many bits are there in a gigabyte? Express your answer as a power of 2 and in scientific notation. 25. The accompanying excerpt is from an article about Planck’s length, which at 10235 meter is believed to be the smallest length or size anything can be in the universe (from G. Johnson, “How Is the Universe Built? Grain by Grain,” in the science section of the Dec. 7, 1999, New York Times, p. D1). Read the accompanying excerpt and then answer the following questions. a. How many kilometers is Planck’s length? b. How many miles is Planck’s length? c. If light travels at 3 ? 108 m/sec, how long will it take light to cross a distance equivalent to Planck’s length? Slightly smaller than what Americans quaintly insist on calling half an inch, a centimeter (one-hundredth of a meter) is easy enough to see. Divide this small length into 10 equal slices and you are looking, or probably squinting, at a millimeter (one-thousandth, or 10 to the minus 3 meters). By the time you divide one of these tiny units into a thousand minuscule micrometers, you have far exceeded the limits of the finest bifocals. But in the mind’s eye, let the cutting continue, chopping the micrometer into a thousand nanometers and the nanometers into a thousand picometers, and those in steps of a thousandfold into femtometers, attometers, zeptameters, and yoctometers. At this point, 10 to the minus 24 meters, about one-billionth the radius of a proton, the roster of Greek names runs out. But go ahead and keep dividing, again and again until you reach a length only one hundred-billionth as large as that tiny amount: 10 to the minus 35 meters. . . . You have finally hit rock bottom: a span called the Planck length, the shortest anything can get. According to recent developments in the quest to devise the “theory of everything,” space is not an infinitely divisible continuum. It is not smooth but granular, and the Planck length gives the size of the smallest possible grains. The time it takes for a light beam to zip across this ridiculously tiny distance . . . is called Planck time, the shortest possible tick of an imaginary clock.
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4.5 Fractional Exponents So far we have derived rules for operating with expressions of the form a n, where n is any integer. These rules can be extended to expressions of the form a m/n, where the exponent is a fraction. We need first to consider what an expression such as a m/n means. The expression m/n can also be written as m ? s1/nd or s1/nd ? m. If the laws of exponents are consistent, then a m/n 5 sa m d 1/n 5 sa 1/n d m For example, if we apply Rule 3 for exponents to the expression sa 1/3 d 2, then the following is true: sa 1/3 d 2 5 a s1/3d2 5 a 2/3
Square Roots: Expressions of the Form a1/2 The expression a 1/2 is called the principal square root (or just the square root) of a and is often written as "a. The symbol " is called a radical. The principal square root of a is the nonnegative number b such that b 2 5 a. Both the square of 22 and the square of 2 are equal to 4, but the notation "4 is defined as only the positive root. If both 22 and 2 are to be considered, we write 6 "4, which means “plus or minus the square root of 4.” So, !4 5 2 and 6 !4 5 2 and 22. When you solve x2 5 4, the solution is 2 and 22. In the real numbers, "a is not defined when a is negative. For example, "24 is undefined, since there is no real number b such that b 2 5 24.
The Square Root For a $ 0, a 1/2 5 a 0.5 5 "a where "a is the nonnegative number b such that b2 5 a. For example, 251/2 5 "25 5 5, since 52 5 25.
Estimating Square Roots. A number is called a perfect square if its square root is an integer. For example, 25 and 36 are both perfect squares since 25 5 52 and 36 5 62, so "25 5 5 and "36 5 6. If we don’t know the square root of some number x and don’t have a calculator handy, we can estimate the square root by bracketing it between two perfect squares, a and b, for which we do know the square roots. If 0 # a , x , b, then "a , "x , "b. For example, to estimate "10, we know
9 , 10 , 16
so
"9 , "10 , "16
and
3 , "10 , 4
where 9 and 16 are perfect squares
Therefore "10 lies somewhere between 3 and 4, probably closer to 3 because 10 is closer to 9 than to 16. According to a calculator, "10 < 3.16.
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EXAMPLE
1
SOLUTION
Estimate "27. We know therefore and
25 , 27 , 36 "25 , "27 , "36 5 , "27 , 6
So "27 lies somewhere between 5 and 6. Would you expect "27 to be closer to 5 or to 6? Check your answer with a calculator. Using a calculator Many calculators and spreadsheet programs have a square root function, often labeled " or perhaps “SQRT.” You can also calculate square roots by raising a number to the 1 2 or 0.5 power using the ^ key, as in 4 ^ 0.5. Try using a calculator to find "4 and "9. In any but the simplest cases where the square root is immediately obvious, you will probably use the calculator. For example, use your calculator to find 81/2 5 "8 < 2.8284 Double-check the answer by verifying that s2.8284d 2 < 8. EXAMPLE
2
Calculating square roots The function S 5 "30d describes the relationship between S, the speed of a car in miles per hour, and d, the distance in feet a car skids after applying the brakes on a dry tar road. Use a calculator to estimate the speed of a car that: a. Leaves 40-foot-long skid marks on a dry tar road. b. Leaves 150-foot-long skid marks.
SOLUTION
a. If d 5 40 feet, then S 5 "30 ? 40 5 "1200 < 35, so the car was traveling at about 35 miles per hour. b. If d 5 150 feet, then S 5 "30 ? 150 5 "4500 < 67, so the car was traveling at almost 70 miles per hour.
EXAMPLE
3
SOLUTION
Assuming that the surface area of Earth is approximately 200 million square miles, estimate the radius of Earth. Step 1. Find the formula for the radius of a sphere. If we assume that Earth is roughly spherical, we can solve for the radius r in the formula for the surface of a sphere, S 5 4pr 2. We get r5
S Å 4p
5
1 Å4
?
S Åp
5
1 S 2Å p
Step 2. Estimate the radius of Earth. Given that Earth’s surface area is approximately 200,000,000 square miles, we can use our derived formula to estimate Earth’s radius. Substituting for S, we get 1 200,000,000 miles2 2Ç p 1 < "63,661,977 miles2 2 1 < ? 7979 miles < 3989 miles 2
r5
So Earth has a radius of about 4000 miles.
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nth Roots: Expressions of the Form a1/n n The term a 1/n denotes the nth root of a, often written as " a. For a $ 0, the nth root of a is the nonnegative number whose nth power is a. 3 81/3 5 " 852
since 23 5 8 (we call 2 the third or cube root of 8)
4 161/4 5 " 16 5 2
since 24 5 16 (we call 2 the fourth root of 16)
n For a , 0, if n is odd, " a is the negative number whose nth power is a. Note that n if n is even, then "a is not a real number when a , 0. 3 (28) 1/3 5 " 28 5 22
since (22) 3 5 28.
3 (227) 1/3 5 " 227 5 23
since (23) 3 5 227
4 (216) 1/4 5 " 216 is not a real number
The nth Root If a is a real number and n is a positive integer, n a 1/n 5 " a,
the nth root of a
For a $ 0, n " a is the nonnegative number b such that bn 5 a.
For a , 0, If n is odd, "a is the negative number b such that bn 5 a. n If n is even, " a is not a real number. n
If the nth root exists, you can find its value on a calculator. For example, to determine a fifth root, raise the number to the 15 or the 0.2 power. So 5 31251/5 5 " 3125 5 5
Double-check your answer by verifying that 55 5 3125.
EXAMPLE
4
SOLUTION
EXAMPLE
5
SOLUTION
Simplify: a. 6251/4 a. b. c. d.
b. s2625d 1/4
c. 1251/3
d. s2125d 1/3
6251/4 5 5 since 54 5 625 s2625d 1/4 does not have a real-number solution 1251/3 5 5 since 53 5 125 s2125d 1/3 5 25 since s25d 3 5 2125
a. The volume of a sphere is given by the equation V 5 43pr 3. Rewrite the formula, solving for the radius as a function of the volume. b. If the volume of a sphere is 370 cubic inches, what is its radius? What common object might have that radius? c. What are the dimensions of a cube that contains this volume? a. Given: multiply both sides by 3
V 5 43pr3 3V 5 4pr3
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3V 5 r3 4p
divide by 4p
r5
take the cube root and switch sides
3
3V
Ç4p
b. Substituting 370 for V and 3.14 for p in our derived formula in part (a), we have r<
3
3 ? 370 < 4.45 inches ? 3.14
Ç4
A regulation-size soccer ball is basically a sphere with a radius of about 4.45 inches. c. A cube has the same length on all three sides. If s is the side length, then the volume 3 of the desired cube is s 3 5 370 cubic inches. So s 5 " 370 < 7.18 inches. A cube of length 7.18 inches on each side would give a volume equivalent to a sphere with a radius of 4.45 inches.
Rules for Radicals The following rules can help you compute with radicals. They represent extensions of the rules for integer exponents. In the following table we assume that m and n are positive n n integers and that " a and " b exist.
Rules for Radicals Example n n n 1. " a? " b5 " ab
"a 5 n " b n
2.
a b 2 0 Åb n
n n 3. s "ad n 5 "a n 5 aa . 0
EXAMPLE
6
SOLUTION
"3 ? "2 5 "6 4 " 125 4 " 25
5
4
Å
125 4 5 "5 25
s "7d 2 5 "72 5 7
Simplifying radicals Simplify the following radical expressions. Assume all variables are nonnegative real numbers. 3 a. "625x 4 b. 3 "48 2 5 "27 a. Factor 625
3 3 4 " 625x4 5 " 5 ? x4
rewrite using perfect cube factors
3 3 5 " 5 ? 5 ? x3 ? x
use Rule 1 for radicals
3 3 3 3 5 " 5x ? " 5x
extract the perfect cubes (Rule 3), leaving the remaining factors under the radical
3 5 5x ? " 5x
b. Find the largest perfect square factors 3 "48 2 5 "27 5 3 " 16 ? 3 2 5 "9 # 3 extract the perfect squares (Rule 3)
5 3 ? 4 "3 2 5 ? 3 "3
multiply
5 12 "3 2 15 "3
use distributive law
5 s12 2 15d "3 5 23 "3
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Algebra Aerobics 4.5a 1. Evaluate each of the following without a calculator. a. 811/2 c. 361/2 e. s236d 1/2 b. 1441/2 d. 2491/2 2. Assume that all variables represent nonnegative quantities. Then simplify and rewrite the following without radical signs. (Use fractional exponents if necessary.) 49 a. "9x c. "36x 2 e. Ç x2 4a 9y 2 x2 b. d. f. Ç 25 Ç 25x 4 Ç 169 3. Use the formula in Example 2 in this section to estimate the following: a. The speed of a car that leaves 60-foot-long skid marks on a dry tar road. b. The speed of a car that leaves 200-foot-long skid marks on a dry tar road. 4. Without a calculator, find two consecutive integers between which the given square root lies. a. "29
c. "117
b. "92
d. "79
You may want to do Exploration 4.2 on Kepler’s laws of planetary motion after reading this section.
e. "39
5. Evaluate each of the following without a calculator: 21/3
a. 271/3
c. 821/3
e. 2721/3
8 R g. Q27
b. 161/4 6. Evaluate: 3 227 a. "
d. 321/5
f. 2521/2
1 R h. Q16
c. s21000d 1/3
1/2
e. s28d 1/3
b. s210,000d 1/4 d. 2161/4 f. "2500 7. Estimate the radius of a spherical balloon with a volume of 2 cubic feet. 8. Simplify if possible. a. "9 1 16
3 c. "2125
b. 2 "49 d. "45 2 3 "125 9. Change each radical expression into exponent form, then simplify. Assume all variables are nonnegative. a. "36
3 b. "27x 6
4 c. "81a 4b 12
10. Solve for the indicated variable. Assume all variables represent nonnegative quantities. a. V 5 pr 2h for r b. V 5 13pr 2h for r c. V 5 s3
d. c2 5 a2 1 b2for a for x e. S 5 6x2
for s
Fractional Powers: Expressions of the Form am/n In the beginning of this section, we saw that we can write am/n either as (am)1/n or (a1/n)m. Writing it as (am)1/n means that we would first raise the base, a, to the mth power and then take the nth root of that. Writing it as (a1/n)m implies first finding the nth root of a and then raising that to the mth power. For example,
using a calculator Equivalently,
23/2 5 (23 ) 1/2 5 (8) 1/2 < 2.8284 23/2 5 (21/2 ) 3 < (1.414) 3 < 2.8284
We could, of course, use a calculator to compute 23/2 (or 21.5) directly by raising 2 to the 3 2 or 1.5 power. If a $ 0 and m and n are integers sn 2 0d, then using radical notation, m
n am/n 5 Q " aR
or equivalently
n m 5 " a
Exponents expressed as ratios of the form m/n are called rational exponents. The set of laws for simplifying expressions with integer exponents also holds for real exponents, which includes rational exponents.
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EXAMPLE
7
3 Find the product of s "5d ? s " 5d, leaving the answer in exponent form.
3 Q "5R ? Q " 5R 5 51/2 ? 51/3 5 5s1/2d1s1/3d 5 55/6
SOLUTION
EXAMPLE
8
According to McMahon and Bonner in On Size and Life,4 common nails range from 1 to 6 inches in length. The weight varies even more, from 11 to 647 nails per pound. Longer nails are relatively thinner than shorter ones. A good approximation of the relationship between length and diameter is given by the equation d 5 0.07L2/3 where d 5 diameter and L 5 length, both in inches. Estimate the diameters of nails that are 1, 3, and 6 inches long. When L 5 1 inch, the diameter d 5 0.07 ? s1d 2/3 5 0.07 ? 1 5 0.07 inches.
SOLUTION
When L 5 3 inches, then d 5 0.07 ? s3d 2/3 < 0.07 ? 2.08 < 0.15 inches. When L 5 6 inches, then d 5 0.07 ? s6d 2/3 < 0.07 ? 3.30 < 0.23 inches.
Summary of Zero, Negative, and Fractional Exponents If m and n are integers and a 2 0, then a0 5 1 1 a 2n 5 n a n a 1/n 5 " a
m
n n a m/n 5 "a m 5 Q "aR
a.0
Algebra Aerobics 4.5b Assume all variables represent positive quantities. 1. Find the product expressed in exponent form: 3 3 4 3 a. "2 ? " 2 c. "3 ? " 9 e. " x ? "x 4 4 3 3 b. "5 ? " 5 d. " x? " x f. " xy 2 ? "xy 2. Find the quotient by representing the expression in exponent form. Leave the answer in positive exponent form. 4 3 "2 2 " 5 "x " xy 2 b. 4 c. 3 d. 4 3 e. 3 "2 "2 "5 "x "xy 3. McMahon and Bonner give the relationship between chest circumference and body weight of adult primates as
a.
c 5 17.1w 3/8
4
where w 5 weight in kilograms and c 5 chest circumference in centimeters. Estimate the chest circumference of a: a. 0.25-kg tamarin b. 25-kg baboon 4. Simplify each expression by removing all possible factors from the radical. a. "20x 2
3 c. " 16x 3y 4
4 " 32x 4y 6 4 "81x 8y 5 5. Change each radical expression into a form with fractional exponents, then simplify.
b. "75a 3
d.
a. "4a 2b 6
3 c. " 8.0 ? 1029
4 b. " 16x 4y 6
d. "8a 24
T. A. McMahon and J. Tyler Bonner, On Size and Life (New York: Scientific American Library, Scientific American Books, 1983).
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241
Exercises for Section 4.5 1. Evaluate without a calculator: a. 1001/2 c. 10021/2 d. 210021/2
b. 21001/2
13. Use ,, ., or 5 to make each statement true. ? "3 1 7 a. "3 1 "7 '
e. 210001/3 f. (21000) 1/3
2. Evaluate without a calculator: a. "10,000
c. 6251/2
1 1/2 e. a b 9
b. "225
d. 1001/2
f. a
? 5 b. "32 1 22 ' 2 2 ? c. "5 2 4 ' 2 14. Fill in the missing forms in the table.
625 1/2 b 100
Radical Form a. "64 5 4 b.
3. Assume that all variables represent positive quantities and simplify. a.
a 2b 4 Ç c6
b. #36x 4y
c.
49x Ç y6
d.
c. 10 "32 2 6 "18
b. 3 "27 2 2 "75
3 3 d. 2 "16 1 4 "54
4 4 3 c. Q " 81xR 5 27 ? " x
x4y2 Ç 100z6
b.
x2
d. "64x 4y 5
Ç 4x4y6
6. Estimate the radius, r, of a circular region with an area, A, of 35 ft 2 (where A 5 pr 2 d. 7. Evaluate each expression without using a calculator. 4 a. "36 ? 106 c. "625 ? 10 20 d. "1.0 ? 1024
3 b. "8 ? 109
8. Calculate the following: a. 41/2 c. 271/3 b. 24
d. 227
1/2
1/3
e. 82/3
g. 161/4
f. 28
3/4
2/3
h. 16
9. Calculate: a. a
1 1/2 b 100
b. 2521/2
c. a
9 21/2 b 16
d. a
1 1/3 b 1000
10. Estimate the length of a side, s, of a cube with volume, V, of 6 cm3 (where V 5 s 3 d. 11. Evaluate when x 5 2: a. s2xd 2 c. x 1/2 b. 2x
2
d. s2xd
e. x 3/2 1/2
f. x 0
12. Determine if the following statements are true or false. 4 a. "s3x 2 d 4 5 3x 2 b. "(x 1 1) 5 (x 1 1)Q "x 1 1R 3
c.
3
4
9
Ç25
"45 5 3
5
d. "15 2 "3 5 "12
d.
s2243d 1/5 5 23
e.
165/4 5 32
15. Evaluate: a. 272/3
5. Simplify by removing all possible factors for each radical. Assume all variable quantities are positive. a. "125a c. "8x 3y 2
2s144d 1/2 5 212
3
4. Simplify each expression by removing all possible factors from the radical, then combining any like terms. a. 2 "50 1 12 "8
Rational Exponent Form
3
b. 1623/4
c. 2523/2
d. 8123/4
16. Without using a calculator, find two consecutive integers such that one is smaller and one is larger than each of the following (for example, 3 , "11 , 4). Show your reasoning. a. "13 b. "22 c. "40 17. Estimate the radius of a spherical balloon that has a volume of 4 ft 3. 18. Constellation. Reduce each of the following expressions to the form u a ? m b; then plot the exponents as points with coordinates (a, b) on graph paper. Do you recognize the constellation? su 2 d 2 ? m u 23/2 ? u 27/2 ? m 1 ? sm 3 d 3 a. 2 e. 24 u ?m sumd 2 b.
u 29/5 ? m 3 sumu 2 d 1 ? m 21
f.
1 u 12 ? m 29
c.
u 2 ? u 24 u ? sm 22 d 3
g.
smud 0 ? su 10 d 21 ? m 1/4 sm 23 ? u 21/3 d 3
d.
sum 2 d 3 ? u 2 sumd 4
3
19. An equilateral triangle has sides of length 8 cm. a. Find the height of the triangle. (Hint: Use the Pythagorean theorem on the inside back cover.) 1 b. Find the area A of the triangle if A 5 bh. 2
8 cm
8 cm
h
3
4 cm
4 cm
b = 8 cm
20. An Egyptian pyramid consists of a square base and four triangular sides. A model of a pyramid is constructed using
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four equilateral triangles each with a side length of 30 inches. Find the surface area of the pyramid model, including the base. (Note: Surface area is the sum of the areas of the four triangular sides and the rectangular base. The previous exercise gives the formula for finding the area of a triangle.)
4 ft off the ground. The base of the ramp is 48 ft from the porch. How long is the ramp? (Hint: Use the Pythagorean theorem on the inside back cover.)
30
4
30
48
21. The time it takes for one complete swing of a pendulum is called the period of its motion. The period T (in seconds) of a L swinging pendulum is found using the formula T 5 2p , Å 32 where L is the length of the pendulum in feet and 32 is the acceleration of gravity in feet per second.2 a. Find the period of a pendulum whose length is 2 ft 8 in. b. How long would a pendulum have to be to have a period of 2 seconds? 22. (Requires the use of a calculator that can evaluate powers.) A wheelchair ramp is constructed at the end of a porch, which is
23. (Requires the use of a calculator that can evaluate powers.) The breaking strength S (in pounds) of a three-strand manila rope is a function of its diameter, D (in inches). The relationship can be described by the equation S 5 1700D 1.9. Calculate the breaking strength when D equals: a. 1.5 in
b. 2.0 in
24. (Requires the use of a calculator that can evaluate powers.) If a rope is wound around a wooden pole, the number of pounds of frictional force, F, between the pole and the rope is a function of the number of turns, N, according to the equation F 5 14 ? 100.70N. What is the frictional force when the number of turns is: a. 0.5 b. 1 c. 3
4.6 Orders of Magnitude Comparing Numbers of Widely Differing Sizes We have seen that a useful method of comparing two objects of widely different sizes is to calculate the ratio rather than the difference of the sizes. The ratio can be estimated by computing orders of magnitude, the number of times we would have to multiply or divide by 10 to convert one size to the other. Each factor of 10 represents one order of magnitude. For example, the radius of the observable universe is approximately 10 26 meters and the radius of our solar system is approximately 10 12 meters. To compare the radius of the observable universe to the radius of our solar system, calculate the ratio radius of the universe 10 26 meters < 12 radius of our solar system 10 meters < 10 26212 < 1014
Orders of Magnitude The radius of the universe is roughly 1014 times larger than the radius of the solar system; that is, we would have to multiply the radius of our solar system by 10 fourteen times in order to obtain the radius of the universe. Since each factor of 10 is counted as a single order of magnitude, the radius of the universe is fourteen orders of magnitude larger than the radius of our solar system. Equivalently, we could say that the radius of our solar system is fourteen orders of magnitude smaller than the radius of the universe. When something is one order of magnitude larger than a reference object, it is 10 times larger. You multiply the reference size by 10 to get the other size. If the object is
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243
two orders of magnitude larger, it is 100 or 10 2 times larger, so you would multiply the reference size by 100. If it is one order of magnitude smaller, it is 10 times smaller, so you would divide the reference size by 10. Two orders of magnitude smaller means the reference size is divided by 100 or 102. EXAMPLE
1
The radius of the sun (109 meters) is how much larger than the radius of a hydrogen atom (10211 meter)? radius of sun 109 meters < 211 radius of the hydrogen atom 10 meters
SOLUTION
< 1092(211) < 1020 So the radius of the sun is 1020 times, or twenty orders of magnitude, larger than the radius of the hydrogen atom. EXAMPLE
2
Compare the length of an unwound DNA strand (1022 meter) with the size of a living cell (radius of 1025 meter). length of DNA strand 1022 meter < 25 radius of the living cell 10 meter
SOLUTION
< 10222(25) < 102215 < 103 Surprisingly enough, the average width of a living cell is approximately three orders of magnitude smaller than one of the single strands of DNA it contains, if the DNA is uncoiled and measured lengthwise.
The reading “Earthquake Magnitude Determination” describes how earthquake tremors are measured.
The Richter Scale The Richter scale, designed by the American Charles Richter in 1935, allows us to compare the magnitudes of earthquakes throughout the world. The Richter scale measures the maximum ground movement (tremors) as recorded on an instrument called a seismograph. Earthquakes vary widely in severity, so Richter designed the scale to measure order-of-magnitude differences. The scale ranges from less than 1 to over 8. Each increase of one unit on the Richter scale represents an increase of ten times, or one order of magnitude, in the maximum tremor size of the earthquake. So an increase from 2.5 to 3.5 indicates a 10-fold increase in maximum tremor size. An increase of two units from 2.5 to 4.5 indicates an increase in maximum tremor size by a factor of 102 or 100. Description of the Richter Scale Richter Scale Magnitude 2.5 3.5 4.5 6 7 8 and above
Table 4.6
Description Generally not felt, but recorded on seismographs Felt by many people locally Felt by all locally; slight local damage may occur Considerable damage in ordinary buildings; a destructive earthquake “Major” earthquake; most masonry and frame structures destroyed; ground badly cracked “Great” earthquake; a few per decade worldwide; total or almost total destruction; bridges collapse, major openings in ground, tremors visible
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Table 4.6 contains some typical values on the Richter scale along with a description of how humans near the center (called the epicenter) of an earthquake perceive its effects. There is no theoretical upper limit on the Richter scale. The U.S. Geological Survey reports that the largest measured earthquake in the United States was in Prince William Sound, Alaska, in 1964 (magnitude 9.2), and the largest in the world was in Chile in 1960 (magnitude 9.5).5
Graphing Numbers of Widely Differing Sizes: Log Scales Exploration 4.1 asks you to construct a graph using logarithmic scales on both axes.
If the sizes of various objects in our solar system are plotted on a standard linear axis, we get the uninformative picture shown in Figure 4.1. The largest value stands alone, and all the others are so small when measured in terameters that they all appear to be zero. When objects of widely different orders of magnitude are compared on a linear scale, the effect is similar to pointing out an ant in a picture of a baseball stadium.
Radius of solar system
Average radius of atoms Average height of humans Radius of Earth 0
0.2
0.4
0.6
0.8
1
Terameters
Figure 4.1 Sizes of various objects in the universe on a linear scale. (Note: One terameter 5 1012 meters.)
A more effective way of plotting sizes with different orders of magnitude is to use an axis that has powers of 10 evenly spaced along it. This is called a logarithmic or log scale. The plot of the previous data graphed on a logarithmic scale is much more informative (see Figure 4.2).
Average radius of atoms
Average height of humans
10–12 10–8 (0.000 000 000 001) (0.000 000 01)
10–4 (0.0001)
100 (1) Meters
Radius of Earth
104 (10,000)
108 (100,000,000)
Radius of solar system
1012 (1,000,000,000,000)
Figure 4.2 Sizes of various objects in the universe on an order-of-magnitude (logarithmic) scale.
Reading Log Scales Graphing sizes on a log scale can be very useful, but we need to read the scales carefully. When we use a linear scale, each move of one unit to the right is equivalent to adding one unit to the number, and each move of k units to the right is equivalent to adding k units to the number (Figure 4.3).
5
See the National Earthquake Information Center website at http://earthquake.usgs.gov/eqcenter/ historic_eqs.php.
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Start at 4 Move one unit to right (4 + 1) End up at 5
0
5
245
Start at 11 Move 3 units to right (11 + 3) End up at 14
10
15
Figure 4.3 Linear scale.
When we use a log scale (see Figure 4.4), we need to remember that one unit of length now represents a change of one order of magnitude. Moving one unit to the right is equivalent to multiplying by 10. So moving from 104 to 105 is equivalent to multiplying 104 by 10. Moving three units to the right is equivalent to multiplying the starting number by 103, or 1000. In effect, a linear scale is an “additive” scale and a logarithmic scale is a “multiplicative” scale.
Start at 104 Move one unit to right (104 • 101) End up at 105
100
105
Start at 1011 Move 3 units to right (1011 • 103) End up at 1014
1010
1015
Figure 4.4 Order-of-magnitude
(logarithmic) scale.
Algebra Aerobics 4.6 1. In 1987 Los Angeles had an earthquake that measured 5.9 on the Richter scale. In 1988 Armenia had an earthquake that measured 6.9 on the Richter scale. Compare the sizes of the two earthquakes using orders of magnitude. 2. On July 15, 2003, Little Rock, Arkansas, had an earthquake that measured 6.5 on the Richter scale. Compare the size of this earthquake to the largest ever recorded, 9.5 in Chile in 1960. 3. If my salary is $100,000 per year and you make an order of magnitude more, what is your salary? If Henry makes two orders of magnitude less money than I do, what is his salary? 4. For each of the following pairs, determine the orderof-magnitude difference: a. The radius of the sun (109 meters) and the radius of the Milky Way (1021 meters) b. The radius of a hydrogen atom (10211 meter) and the radius of a proton (10215 meter)
5. Joe wants to move from Wyoming to California, but he has been advised that houses in California cost an order of magnitude more than houses in Wyoming. a. If Joe’s house in Wyoming is worth $400,000, how much would a similar house cost in California? b. If a house in California sells for $650,000, how much would it cost in Wyoming? 6. How many orders of magnitude greater is a kilometer than a meter? Than a millimeter? 7. By rounding the number to the nearest power of 10, find the approximate location of each of the following on the logarithmic scale in Figure 4.2 on page 244. a. The radius of the sun, at approximately 1 billion meters b. The radius of a virus, at 0.000 000 7 meter c. An object whose radius is two orders of magnitude smaller than that of Earth
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Exercises for Section 4.6 1. What is the order-of-magnitude difference between the following units? (Refer to table on inside back cover.) a. A millimeter and a gigameter b. A second and a day c. A square centimeter and an acre s1 acre 5 43,560 ft 2 d d. A microfarad and a picofarad 2. Fill in the blanks to make each of the following statements true. a. Attaching the prefix “micro” to a unit __________ the size by __________ orders of magnitude. b. Attaching the prefix “kilo” to a unit __________ the size by __________ orders of magnitude. c. Scientists and engineers have designated prefix multipliers from septillionths s10224 d to septillions s10 24 d, a span of __________ orders of magnitude. 3. Compare the following numbers using orders of magnitude. a. 5.261 and 52.61 c. 5.261 ? 106 and 526.1 b. 5261 and 5.261 4. An ant is roughly 1023 meter in length and the average human roughly one meter. How many times longer is a human than an ant? 5. Refer to the chart in Exploration 4.1. a. How many orders of magnitude larger is the Milky Way than the first living organism on Earth? b. How many orders of magnitude older is the Pleiades (a cluster of stars) than the first Homo sapiens? 6. Water boils (changes from a liquid to a gas) at 373 kelvins. The temperature of the core of the sun is 20 million kelvins. By how many orders of magnitude is the sun’s core hotter than the boiling temperature of water? 7. An electron weighs about 10227 gram, and a raindrop weighs about 1023 gram. How many times heavier is a raindrop than an electron? How many times lighter is an electron than a raindrop? What is the order-of-magnitude difference? 8. On Nov. 20, 2001, The New York Times reported that FBI scientists had found a sealed plastic bag with a letter addressed to Senator Patrick Leahy that was highly contaminated with anthrax. The article said that a sample taken from the bag “showed the presence of 23,000 anthrax spores. This, the scientists said, was roughly three orders of magnitude more spores than found in samples from any of the other 600 bags of mail the bureau examined.” Estimate the number of spores found in any of the 600 other bags of mail. 9. In the December 1999 issue of the journal Science, two Harvard scientists describe a pair of “nanotweezers” they created that are capable of manipulating objects as small as one-50,000th of an inch in width. The scientists used the tweezers to grab and pull clusters of polystyrene molecules, which are of the same size as structures inside
cells. A future use of these nanotweezers may be to grab and move components of biological cells. a. Express one-50,000th of an inch in scientific notation. b. Express the size of objects the tweezers are able to manipulate in meters. c. The prefix “nano” refers to nine subdivisions by 10, or a multiple of 1029. So a nanometer would be 1029 meters. Is the name for the tweezers given by the inventors appropriate? d. If not, how many orders of magnitude larger or smaller would the tweezers’ ability to manipulate small objects have to be in order to grasp things of nanometer size? 10. Determine the order-of-magnitude difference in the sizes of the radii for: a. The solar system (1012 meters) compared with Earth (107 meters) b. Protons (10215 meter) compared with the Milky Way (1021 meters) c. Atoms (10210 meter) compared with neutrons (10215 meter) 11. To compare the sizes of different objects, we need to use the same unit of measure. a. Convert each of these to meters: i. The radius of the moon is approximately 1,922,400 yards. ii. The radius of Earth is approximately 6400 km. iii. The radius of the sun is approximately 432,000 miles. b. Determine the order-of-magnitude difference between: i. The surface areas of the moon and Earth ii. The volumes of the sun and the moon 12. The pH scale measures the hydrogen ion concentration in a liquid, which determines whether the substance is acidic or alkaline. A strong acid solution has a hydrogen ion concentration of 1021 M. One M equals 6.02 ? 1023 particles, such as atoms, ions, molecules, etc., per liter, or 1 mole per liter.6 A strong alkali solution has a hydrogen ion concentration of 10214 M. Pure water, with a concentration of 1027 M, is neutral. The pH value is the power without the minus sign, so pure water has a pH of 7, acidic substances have a pH less than 7, and alkaline substances have a pH greater than 7. a. Tap water has a pH of 5.8. Before the industrial age, rain water commonly had a pH of about 5. With the spread of modern industry, rain in the northeastern United States and parts of Europe now has a pH of about 4, and in extreme cases the pH is about 2. Lemon juice has a pH of 2.1. If acid rain with a pH of 3 is discovered in an area, how much more acidic is it than preindustrial rain?
You may recall from Algebra Aerobics 4.1 that 6.02 ? 1023 is called Avogadro’s number. A mole of a substance is defined as Avogadro’s number of particles of that substance. M is called a molar unit. 6
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b. Blood has a pH of 7.4; wine has a pH of about 3.4. By how many orders of magnitude is wine more acidic than blood? 13. Which is an additive scale? Explain why. Which is a multiplicative or logarithmic scale? Explain why. a. 10
20 30 40 50 60 70 80 90 100
b. 10
20
30
40 50
100
14. Graph the following on a power-of-10 (logarithmic) scale. (See sample log scale at the end of the exercises.) a. 1 meter d. 1000 kilometers b. 10 meters e. 10 gigameters c. 1 hectometer 15. (Refer to the chart in Exploration 4.1.) Plot on the logarithmic scale in Figure 4.2 an object whose radius is: a. Five orders of magnitude larger than the radius of the first atoms b. Twenty orders of magnitude smaller than the radius of the sun 16. a. Read the chapter entitled “The Cosmic Calendar” from Carl Sagan’s book The Dragons of Eden. b. Carl Sagan tried to give meaning to the cosmic chronology by imagining the almost 15 billion–year lifetime of the universe compressed into the span of one calendar year. To get a more personal perspective, consider your date of birth as the time at which the Big Bang took place. Map the following five cosmic events onto your own life span: i. The Big Bang iv. First Homo sapiens ii. Creation of Earth v. American Revolution iii. First life on Earth Once you have done the necessary mathematical calculations and placed your results on either a chart or a timeline, form a topic sentence and write a playful paragraph about what you were supposedly doing when these cosmic events took place. Hand in your calculations along with your writing. 17. Graph the following on a power-of-10 (logarithmic) scale. (See sample log scale at the end of the exercises.) a. 1 watt c. 100 billion kilowatts b. 10 kilowatts d. 1000 terawatts
100
101
102
103
104
105
106
107
247
18. Radio waves, sent from a broadcast station and picked up by the antenna of your radio, are a form of electromagnetic (EM) radiation, as are microwaves, X-rays, and visible, infrared, and ultraviolet light. They all travel at the speed of light. Electromagnetic radiation can be thought of as oscillations like the vibrating strings of a violin or guitar or like ocean swells that have crests and troughs. The distance between the crest or peak of one wave and the next is called the wavelength. The number of times a wave crests per minute, or per second for fast-oscillating waves, is called its frequency. Wavelength and frequency are inversely proportional: the longer the wavelength, the lower the frequency, and vice versa—the faster the oscillation, the shorter the wavelength. For radio waves and other EM, the number of oscillations per second of a wave is measured in hertz, after the German scientist who first demonstrated that electrical waves could transmit information across space. One cycle or oscillation per second equals 1 hertz (Hz). For the following exercise you may want to find an old radio or look on a stereo tuner at the AM and FM radio bands. You may see the notation kHz beside the AM band and MHz beside the FM band. AM radio waves oscillate at frequencies measured in the kilohertz range, and FM radio waves oscillate at frequencies measured in the megahertz range. Wavelength
a. The Boston FM rock station WBCN transmits at 104.1 MHz. Write its frequency in hertz using scientific notation. b. The Boston AM radio news station WBZ broadcasts at 1030 kHz. Write its frequency in hertz using scientific notation. The wavelength (Greek lambda) in meters and frequency m (Greek mu) in oscillations per second are related by the formula c l 5 where c is the speed of light in meters per second. m c. Estimate the wavelength of the WBCN FM radio transmission. d. Estimate the wavelength of the WBZ AM radio transmission. e. Compare your answers in parts (c) and (d), using orders of magnitude, with the length of a football field (approximately 100 meters).
108
Sample log Scale
109
1010
1011
1012
1013
1014
1015
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4.7 Logarithms Base 10 In Section 4.6 we used a logarithmic scale to graph numbers of widely disparate sizes. We labeled the axis with powers of 10, so it was easy to plot numbers such as 1000 5 103 or 100,000 5 105 that are integer powers of 10. But how would we plot a number such as 4,600,000,000 5 4.6 ? 10 9, the approximate age of Earth in years? To do that we need to understand logarithms.
Finding the Logarithms of Powers of 10 For handling very large or very small numbers, it is often easier to write the number using powers of 10. For example, 100,000 5 105 We say that 100,000 equals the base 10 to the fifth power But we could rephrase this as 5 is the exponent of the base 10 that is needed to produce 100,000 The more technical way to say this is 5 is the logarithm base 10 of 100,000 In symbols we write 5 5 log 10 100,000 So the expressions 100,000 5 105
and
5 5 log 10 100,000
are two ways of saying the same thing. The key point to remember is that a logarithm is an exponent. Definition of Logarithm The logarithm base 10 of x is the exponent of 10 needed to produce x: log 10 x 5 c
means
10c 5 x
So to find the logarithm of a number, write it as 10 to some power. The power is the logarithm of the original number. EXAMPLE
1
SOLUTION
Find log (1,000,000,000) without using a calculator. Since then
1,000,000,000 5 109 log10 1,000,000,000 5 9
and we say that the logarithm base 10 of 1,000,000,000 is 9. The logarithm of a number tells us the exponent of the number when written as a power of 10. Here the logarithm is 9, so that means that when we write 1,000,000,000 as a power of 10, the exponent is 9. EXAMPLE
2
SOLUTION
Find log 1 without using a calculator. Since then
1 5 100 log10 1 5 0
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249
and we say that the logarithm base 10 of 1 is 0. Since logarithms represent exponents, this says that when we write 1 as a power of 10, the exponent is 0.
EXAMPLE
3
SOLUTION
How do we calculate the logarithm base 10 of decimals such as 0.000 01? Since then
0.000 01 5 1025 log10 0.000 01 5 25
and we say that the logarithm base 10 of 0.000 01 is 25.
In the previous example, we found that the log (short for “logarithm”) of a number can be negative. This makes sense if we think of logarithms as exponents, since exponents can be any real number. But we cannot take the log of a negative number or zero; that is, log10 x is not defined when x # 0. Why? If log 10 x 5 c , where x # 0, then 10c 5 x (a number # 0 ). But 10 to any power will never produce a number that is negative or zero, so log 10 x is not defined if x # 0.
log10 x is not defined when x # 0.
Table 4.7 gives a sample set of values for x and their associated logarithms base 10. To find the logarithm base 10 of x, we write x as a power of 10, and the logarithm is just the exponent. Most scientific calculators and spreadsheet programs have a LOG function that calculates logarithms base 10. Try using technology to double-check some of the numbers in Table 4.7. Logarithms of Powers of 10 x 0.0001 0.001 0.01 0.1 1 10 100 1000 10,000
Exponential Notation 24
10 1023 1022 1021 100 101 102 103 104
log 10 x 24 23 22 21 0 1 2 3 4
Table 4.7
Logarithms base 10 are used frequently in our base 10 number system and are called common logarithms. We write log 10 x as log x.
Common Logarithms Logarithms base 10 are called common logarithms. log 10 x is written as log x.
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Algebra Aerobics 4.7a 1. Without using a calculator, find the logarithm base 10 of: a. 10,000,000 c. 10,000 e. 1000 g. 1 b. 0.000 000 1 d. 0.0001 f. 0.001 2. Rewrite the following expressions in an equivalent form using powers of 10: a. log 100,000 5 5 c. log 10 5 1 b. log 0.000 000 01 5 28 d. log 0.01 5 22 3. Evaluate without using a calculator. Find a number if its log is:
a. 3 c. 6 e. 22 b. 21 d. 0 4. Find c and then rewrite as a logarithm: a. 10 c 5 1000 d. 10 c 5 0.000 01 b. 10 c 5 0.001 e. 10 c 5 1,000,000 c c. 10 5 100,000 f. 10 c 5 0.000 001 5. Find the value of x that makes the statement true. a. 10 x23 5 102 c. log (x 2 2) 5 1 2x21 4 b. 10 d. log 5x 5 21 5 10
Finding the Logarithm of Any Positive Number Scientific calculators have a LOG function that will calculate the log of any positive number. However, it’s easy to make errors typing in numbers, so it’s important not to rely solely on technology-generated answers. To verify that the calculated number is the right order of magnitude, you should estimate the answer without using technology.
EXAMPLE
4
SOLUTION
Estimating, then using technology to calculate logs a. Estimate the size of log (2000) and log (0.07). b. Use a calculator to find the logarithm of 2000 and 0.07. a. i. If we place 2000 between the two closest integer powers of 10, we have 1000 , 2000 , 10,000 Rewriting 1000 and 10,000 as powers of 10 gives 103 , 2000 , 104 Taking the log of each term preserves the inequality, so we would expect 3 , log 2000 , 4 ii. If we place 0.07 between the two closest integer powers of 10, we have 0.01 , 0.07 , 0.10 Rewriting 0.01 and 0.10 as powers of 10 gives 1022 , 0.07 , 1021 Taking the log of each term preserves the inequality, so we would expect 22 , log 0.07 , 21 b. Using a calculator, we have i. log 2000 < 3.301 ii. log s0.07d < 21.155. So our estimates were correct.
EXAMPLE
5
Calculating logs of very large or small numbers Find the logarithm of a. 3.7 trillion b. A Planck length of 0.000 000 000 000 000 000 000 000 000 000 000 016 meter
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SOLUTION
251
Our strategy in each case is to • Write the number in scientific notation. • Then convert the number into a single power of 10. The resulting exponent is the desired log. a. In scientific notation 3.7 trillion is 3.7 ? 1012. To convert the entire expression into a single power of 10, we need to first convert the coefficient 3.7 to a power of 10. Using a calculator, we have log 3.7 < 0.568 so 3.7 < 100.568 If we substitute for 3.7, 3.7 ? 1012 < 100.568 ? 1012 and use rules for exponents, 5 100.568112 we have 5 1012.568 So the exponent 12.568 is the desired logarithm. b. In scientific notation a Planck length is 1.6 ? 10235 meter. We need to convert 1.6 to a power of 10. Using a calculator, we have log s1.6d < 0.204, so 1.6 5 100.204. If we substitute for 1.6 use rules for exponents and subtract, we get
1.6 ? 10235 < 100.204 ? 10235 5 100.204235 5 10234.796
So the exponent 234.796 is the desired log. So far we have dealt with finding the logarithm of a given number. Logarithms can, of course, occur in expressions involving variables. EXAMPLE
6
SOLUTION
EXAMPLE
7
SOLUTION
Finding the number given the log Rewrite the following expressions using exponents, and then solve for x without using a calculator. a. log x 5 3 b. log x 5 0 c. log x 5 22 a. If log x 5 3, then 103 5 x, so x 5 1000. b. If log x 5 0, then 100 5 x, so x 5 1. c. If log x 5 22, then 1022 5 x, so x 5 1/102 5 1/100 5 0.01. Rewrite the following expressions using logarithms and then solve for x using a calculator. Round off to three decimal places. a. 10 x 5 11 b. 10 x 5 0.5 c. 10 x 5 0 a. If 10 x 5 11, then log 11 5 x. Using a calculator gives x < 1.041. b. If 10 x 5 0.5, then log 0.5 5 x. Using a calculator gives x < 20.301. c. There is no power of 10 that equals 0. Hence there is no solution for x.
Plotting Numbers on a Logarithmic Scale We are finally prepared to answer the question posed at the very beginning of this section: How can we plot on a logarithmic (or order-of-magnitude) scale a number such as 4.6 million years, the estimated age of Earth? A Strategy for Plotting Numbers on a Logarithmic Scale • Write the number in scientific notation. • Then convert the number into a single power of 10. • Use the exponent to help plot the number.
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In scientific notation the age of Earth equals 4.6 billion 5 4,600,000,000 5 4.6 ? 109 years. To plot this number on a logarithmic scale, we need to convert 4.6 into a power of 10. Using a calculator, we have log 4.6 < 0.663 , so 4.6 < 100.663. If we 4.6 ? 109 < 100.663 ? 109
substitute for 4.6 and use rules of exponents
5 100.66319
we have
5 109.663
The power of 10 seems reasonable since 109 , 4.6 ? 109 , 1010. Having converted our original number 4.6 ? 109 into 109.663, we can plot it on an order-of-magnitude graph between 109 and 1010 (see Figure 4.5). Earth 109.663 109
109.1
109.2
109.3
109.4
109.5
109.6
1,000,000,000
109.7
109.8
109.9
4,600,000,000
1010 10,000,000,000
Years
Figure 4.5 Age of Earth plotted on an order-of-magnitude (or logarithmic) scale.
EXAMPLE
8
Plot the numbers 100, 200, 300, 400, 500, 600, 700, 800, 900, and 1000 on a logarithmic scale.
SOLUTION
Using our log plotting strategy, we first convert each number to a single power of 10. We have 100 200 300 400 500
5 102 < 102.301 < 102.477 < 102.602 < 102.699
600 < 102.778 700 < 102.845 800 < 102.903 900 < 102.954 1000 5 103
We can now use the exponents of each power of 10 to plot the numbers directly onto a logarithmic scale (see Figure 4.6). 100 10
2
200 10
2.1
2.2
10
10
2.3
300 10
2.4
400 2.5
10
10
2.6
500 10
600
2.7
10
700 2.8
800 900 1000 102.9
103
Figure 4.6 A logarithmic plot of the numbers 100, 200, 300, . . . , 1000.
Note that on the logarithmic scale in Figure 4.6 the point halfway between 102 and 10 is at 102.5 5 316. 3
When are numbers evenly spaced on a logarithmic scale? On a linear (additive) scale the numbers 100, 200, . . . , 1000 would be evenly spaced, since you add a constant amount to move from one number to the next. On a logarithmic (multiplicative) scale, numbers that are evenly spaced are generated by multiplying by a constant amount to get from one number to the next. For example, the integer powers of 10 are all evenly spaced on a log plot since you multiply each number by 10 to get the next number in the sequence. Similarly, the numbers 100, 200, 400, and 800 are evenly spaced in Figure 4.6 since you multiply by the constant 2 to get from one number in the sequence to the next. The sequence 100, 200, 300, . . . , 1000 is not evenly spaced, since there is not a constant factor that you could multiply one number by to get to the next. In this last sequence, since the multiplication factor needed to move from one number to the next decreases as the numbers approach 1000, the distance between points decreases.
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253
Labeling using only the exponent Instead of labeling the axis using powers of 10, we can label it using just the exponents of 10 as in Figure 4.7. Remember that exponents are logarithms, which is why we call the scale logarithmic. Surtsey 101.602
x 101 log x
Pangaea 108.301
years
Earth 109.663
102
103
104
105
106
107
108
109
2
3
4
5
6
7
8
9
1
log (years)
1.602
8.301
1010 10 9.663
Figure 4.7 The age of Surtsey, Pangaea, and Earth plotted using an order-of-magnitude or
logarithmic scale.
EXAMPLE
9
Add on to the logarithmic plot in Figure 4.7 the following numbers: a. 200 million, the number of years since all of Earth’s continents collided to form one giant land mass called Pangaea b. 40, the number of years since the volcanic island of Surtsey, Earth’s newest land mass, emerged near Iceland
SOLUTION
a. In scientific notation 200 million 5 200,000,000 5 2.0 ? 108. Using a calculator, we have log 2.0 < 0.301, so 2.0 < 100.301. Hence 2.0 ? 108 < 100.301 ? 108 5 108.301 is the age of Pangaea in a form easily plotted on a logarithmic scale. b. In scientific notation 40 5 4.0 ? 101. Using a calculator, we have log 4.0 < 0.602, so 4.0 < 100.602. Therefore 4.0 ? 101 < 100.602 ? 101 5 101.602 is the age of Surtsey in a form easily plotted on a log scale. The two numbers are plotted in Figure 4.7.
Algebra Aerobics 4.7b Check your estimate with a calculator. Round the value of x to the nearest integer. a. log x 5 4.125 b. log x 5 5.125 c. log x 5 2.125 5. Rewrite the following equations using logs instead of exponents. Estimate a solution for x and check your estimate with a calculator. Round the value of x to three decimal places. a. 10 x 5 250 c. 10 x 5 0.075 b. 10 x 5 250,000 d. 10 x 5 0.000 075 6. Write each number as a power of 10 and then plot them all on the logarithmic scale below. a. 57 c. 25,000 b. 182 d. 7,200,000,000
Most of these problems require a calculator that can evaluate logs. 1. Use a calculator to estimate each of the following: a. log 3 b. log 6 c. log 6.37 2. Use the answers from Problem 1 to estimate values for: a. log 3,000,000 b. log 0.006 Then use a calculator to check your answers. 3. Write each of the following as a power of 10: a. 0.000 000 7 m (the radius of a virus) b. 780,000,000 km (the mean distance from our sun to Jupiter) c. 0.0042 d. 5,400,000,000 4. Rewrite the following equations using exponents instead of logarithms. Estimate the solution for x. 100
101
102
103
104
105
106
107
108
109
1010
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Exercises for Section 4.7 Many of the problems in this section require the use of a calculator that can evaluate logs. 1. Rewrite in an equivalent form using logarithms: a. 104 5 10,000 c. 100 5 1 22 b. 10 5 0.01 d. 1025 5 0.000 01
11. Solve for x. (Hint: Rewrite each expression so that you can use a calculator to solve for x.) a. log x 5 0.82 c. log x 5 0.33 x b. 10 5 0.012 d. 10 x 5 0.25
2. Use your calculator to evaluate to two decimal places: a. 100.4 c. 100.6 e. 100.8 0.5 0.7 b. 10 d. 10 f. 100.9 3. Express the number 375 in the form 10 x. 4. Estimate the value of each of the following: a. log 4000 b. log 5,000,000 c. log 0.0008 5. Rewrite the following statements using logs: a. 102 5 100 b. 107 5 10,000,000 c. 1023 5 0.001 Rewrite the following statements using exponents: d. log 10 5 1 e. log 10,000 5 4 f. log 0.0001 5 24 6. Evaluate the following without a calculator. a. Find the following values: i. log 100 ii. log 1000 iii. log 10,000,000 What is happening to the values of log x as x gets larger? b. Find the following values: i. log 0.1 ii. log 0.001 iii. log 0.000 01 What is happening to the values of log x as x gets closer to 0? c. What is log 0? d. What is log(210)? What do you know about log x when x is any negative number? 7. Rewrite the following equations using exponents instead of logs. Estimate a solution for x and then check your estimate with a calculator. Round the value of x to the nearest integer. a. log x 5 1.255 c. log x 5 4.23 b. log x 5 3.51 d. log x 5 7.65 8. Rewrite the following equations using exponents instead of logs. Estimate a solution for x and then check your estimate with a calculator. Round the value of x to the nearest integer. a. log x 5 1.079 c. log x 5 2.1 b. log x 5 0.699 d. log x 5 3.1 9. Rewrite the following equations using logs instead of exponents. Estimate a solution for x and then check your estimate with a calculator. Round the value of x to three decimal places. a. 10 x 5 12,500 c. 10 x 5 597 x b. 10 5 3,526,000 d. 10 x 5 756,821
100
101
102
103
104
105
106
10. Rewrite the following equations using logs instead of exponents. Estimate a solution for x and then check your estimate with a calculator. Round the value of x to three decimal places. a. 10 x 5 153 c. 10 x 5 0.125 x b. 10 5 153,000 d. 10 x 5 0.001 25
107
12. Without using a calculator, show how you can solve for x. a. 10 x22 5 100 c. 102x23 5 1000 b. log(x 2 4) 5 1 d. log(6 2 x) 5 22 13. Without using a calculator show how you can solve for x. a. 10 x25 5 1000 c. 10 3x21 5 0.0001 b. log(2x 1 10) 5 2 d. log(500 2 25x) 5 3 14. Find the value of x that makes the equation true. a. log x 5 22 b. log x 5 23 c. log x 5 24 15. Without using a calculator, for each number in the form log x, find some integers a and b such that a , log x , b. Justify your answer. Then verify your answers with a calculator. a. log 11 b. log 12,000 c. log 0.125 16. Use a calculator to determine the following logs. Double-check each answer by writing down the equivalent expression using exponents, and then verify this equivalence using a calculator. a. log 15 b. log 15,000 c. log 1.5 17. On a logarithmic scale, what would correspond to moving over to the right: 1 a. 0.001 unit b. unit c. 2 units d. 10 units 2 18. The difference in the noise levels of two sounds is measured in I2 decibels, where decibels 5 10 log a b and I1 and I2 are the I1 intensities of the two sounds. Compare noise levels when I1 5 10215 watts/cm2 and I2 5 1028 watts/cm2. 19. The concentration of hydrogen ions in a water solution typically ranges from 10 M to 10215 M. (One M equals 6.02 ? 1023 particles, such as atoms, ions, molecules, etc., per liter or 1 mole per liter.) Because of this wide range, chemists use a logarithmic scale, called the pH scale, to measure the concentration (see Exercise 12 of Section 4.6). The formal definition of pH is pH 5 2log[H 1], where [H 1] denotes the concentration of hydrogen ions. Chemists use the symbol H 1 for hydrogen ions, and the brackets [ ] mean “the concentration of.” a. Pure water at 25ºC has a hydrogen ion concentration of 1027 M. What is the pH?
108
Sample log Scale
109
1010
1011
1012
1013
1014
1015
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b. In orange juice, [H 1] < 1.4 ? 1023 M. What is the pH? c. Household ammonia has a pH of about 11.5. What is its [H 1]? d. Does a higher pH indicate a lower or a higher concentration of hydrogen ions? e. A solution with a pH . 7 is called basic, one with a pH 5 7 is called neutral, and one with a pH , 7 is called acidic. Identify pure water, orange juice, and household ammonia as either acidic, neutral, or basic. Then plot their positions on the accompanying scale, which shows both the pH and the hydrogen ion concentration. pH 0
1
1M [H+]
10–1
2
3
4
5
10–2 10–3 10–4
6
10–5
7
10–6 10–7
8 10–8
9
10
11
12
10–9 10–10 10–11 10–12
20. a. Place the number 50 on the additive scale below. 0
100
b. Place the number 50 on the multiplicative scale below. 10 101
100 101.1
101.2
101.3
101.4
101.5
101.6
101.7
101.8
101.9
102
21. The coordinate system below uses multiplicative or log scales on both axes. Position the point whose coordinates are (708, 25).
255
102
101
100 100
101
102
103
22. Change each number to a power of 10, then plot the numbers on a power-of-10 scale. (See sample log scale on page 254.) a. 125 b. 372 c. 694 d. 840 23. Compare the times listed below by plotting them on the same order-of-magnitude scale. (Hint: Start by converting all the times to seconds.) a. The time of one heartbeat (1 second) b. Time to walk from one class to another (10 minutes) c. Time to drive across the country (7 days) d. One year (365 days) e. Time for light to travel to the center of the Milky Way (38,000 years) f. Time for light to travel to Andromeda, the nearest large galaxy (2.2 million years)
C H A P T E R S U M M A RY Powers of 10
If a is nonzero real and n is a positive integer, then
If n is a positive integer, we define 10n 5 (11111)11111* 10 ? 10 ? 10 ? c ? 10 n factors
10 5 1 1 102n 5 n 10 0
ca an 5 (11 a ? a1) ? a111* n factors
a 51 0
a2n 5
1 an
If m and n are positive integers and the base, a, is restricted to values for which the power is defined, then
Scientific Notation A number is in scientific notation if it is in the form N ? 10
n
where N is called the coefficient, 1 # k N k , 10, and n is an integer. Example: In scientific notation 67,000,000 is written as 6.7 ? 107 and 20.000 000 000 008 1 is written as 28.1 ? 10212. Powers of a In the expression a n, a is called the base and n is called the exponent or power.
a1/2 5 "a n a1/n 5 " a
am/n 5 (am ) 1/n 5 (a1/n ) m n m n 5 " a 5 (" a) m
Rules of Exponents If a and b are nonzero, then 1. am ? an 5 a(m1n) an 2. m 5 a sn2md a 3. sa m d n 5 a sm?nd
4. sabd n 5 a nb n a n an 5. a b 5 n b b
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Orders of Magnitude We use orders of magnitude when we compare objects of widely different sizes. Each ƒactor of 10 is counted as a single order of magnitude. Example: The radius of the universe is 1014 times or fourteen orders of magnitude larger than the radius of the solar system. And vice versa: The radius of the solar system is fourteen orders of magnitude smaller than the radius of the universe.
Logarithms
Example: log 10 6,370,000 < 6.804 means that 106.804 < 6,370,000. Logarithms base 10 are called common logarithms. We usually write log 10 x as log x. When x # 0, log x is not defined.
Plotting Numbers on a Logarithmic Scale Logarithmic or powers-of-10 scales are used to graph objects of widely differing sizes. We can plot a number on a log scale by converting the number to a power of 10.
The logarithm base 10 of x is the exponent of 10 needed to produce x. So log 10 x 5 c
means
10 c 5 x
We say that c is the logarithm base 10 of x. Earth 109.663 109
109.1
109.2
109.3
109.4
109.5
109.6
1,000,000,000
109.7
109.8
109.9
4,600,000,000
1010 10,000,000,000
Years
Age of Earth plotted on a logarithmic scale.
C H E C K Y O U R U N D E R S TA N D I N G I. Are the statements in Problems 1–26 true or false? Give an explanation for your answer. 1. A distance of 10 miles is longer than a distance of 10 kilometers. 2. There are 39 centimeters in 1 inch. 15
3. 10
is 10 followed by fifteen zeros.
4. 10 5 0. 0
5.
1 5 10m. 102m
13. To convert a distance D in kilometers to miles, you 1 km could multiply D by . 0.62 mile 14. The units of 300
km 1 hr 1 min 103 m are ? ? ? hr 60 min 60 sec 1 km
meters per second. 15. 29 , 2 "75 , 28. 16. 81/2 5 80.5. 17. log 10 0.0001 5 1024. 18. log 10 1821 is not defined since 1821 is not a power of 10.
6. 20.000 005 62 5 25.62 ? 1026.
19. s81d 1/2 5 69 becauses9d 2 5 81 and (29)2 5 81.
7. 15 ? 10 is correct scientific notation for the number 150,000.
20. log 0 5 1.
8. The age of the universe s1.37 ? 1010 yearsd is about three times the age of Earth s4.6 ? 109 yearsd.
22. 24 , log 0.00015 , 23.
9. In July 2004, the population of the world (about 6,377,642,000) was approximately three orders of magnitude larger than the population of the United States (about 293,028,000).
24. If P . 0, log P 5 Q means that 10P 5 Q.
4
10. 282 5 s28d 2. 5 23 215 . 11. a b 5 3 29 12. 102 1 101 1 105 5 1021115 5 108.
21. log(23) 5 2log(3). 23. log 0.143 < 20.845 means that 1020.845 < 0.143. 25. The following figure illustrates the number 7,500,000 plotted correctly on a logarithmic scale. x 106 log x
6
107
108
7
8
7,500,000
26. If 10 x 5 36, then x < 1.556.
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II. In Problems 27–32, give examples with the specified properties. 27. Populations of two cities A and B, where the population of city A is two orders of magnitude larger than that of city B. 28. A number x such that log x lies between 8 and 9.
257
33. If one quantity is four orders of magnitude larger than a second quantity, it is four times as large as the second quantity. 34. k c k 5 c for any real number c. 35. log x is defined only for numbers x . 0.
29. A number x such that log x is a negative number.
36. Raising a number to the 13 power is the same as taking the cube root of that number.
30. A positive number b such that "b . b.
37. b m ? b m 5 b m
31. A non-zero number b such that b m 5 b n for any numbers m and n.
38. sb p d q 5 b p1q
32. A number b such that k b k 5 2b.
40. s2bd q 5 b q
2
III. Are the statements in Problems 33–42 true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample.
39. sb 1 cd m 5 b m 1 cm 41. b m ? cn 5 sb ? cd m1n n 42. If n is odd, "b can be positive, negative, or zero depending on the value of b.
CHAPTER 4 REVIEW: PUTTING IT ALL TOGETHER 1. Evaluate each of the following without a calculator. a. 4.2 ? 103
b. (25)23
c. 242
5
d. 10021/2
e.
3 32
2. Use the rules of exponents to simplify the following. Express your answer with positive exponents. c. (22xy2)3 e. (x23y)(x2y21/2) a. x4x3 2 4 10x y b. d. (x21/2)2 5xy3 3. For what integer values of x will the following statement be true? (210) x 5 210x 4. a. Show with an example why the following is not a true statement for all values of x: x3 ⫹ x5 5 x8. b. For what value of x is the above statement true? 5. Elephant seals can weigh as much as 5000 lb (for males) and 2000 lb (for females). On land, these seals can travel short distances quite quickly, as much as 20 feet in 3 seconds. How many miles per hour is this? 6. An NFL regulation playing field for football is 120 yd (110 m) long including the end zones, and 53 yd 1 ft (48.8 m) wide. An acre is 4840 square yards, and 1 yard ⫽ 3 feet. a. Which is larger, a football field (including the end zones) or an acre? By how much? b. If you bought a house on a square lot that measured half an acre, what would the dimensions of the lot be in feet? 7. In 2006, Tiger Woods was the highest paid athlete in the world (taking into account on and off the field earnings), making $11.9 million in salary and $100 million in endorsements for a total of $111.9 million. By what order of magnitude is his salary greater than that of a minimum-wage worker in the
same state making $6.40/hr working 40 hours/week for 52 weeks/year? How many years would the minimum-wage worker have to work to earn what Tiger Woods made in 1 year? 8. The Yangtze River (China) is 6380 km long. The Colorado River is 1400 miles long. a. Which river is longer? b. Compare the lengths of these rivers using orders of magnitude. 9. Use the accompanying table to answer the following questions. Country Russia Chile Canada South Africa Norway Monaco
Area 17,075,200 km2 290,125 mi2 3,830,840 mi2 1,184,825 km2 323,895 km2 0.5 mi2
a. Which country has the largest area? The smallest? b. Using scientific notation, arrange the countries from largest area to smallest area. c. What is the order-of-magnitude difference between the country with the largest area and the country with the smallest area? 10. The Energy Information Administration of the U.S. Department of Energy estimates that in 2010 the world energy use will be 470.8 quadrillion Btu (British thermal units), where 1 Btu ⫽ 0.000 293 1 kWh (kilowatt-hours.) a. Express 470.8 quadrillion Btu and 0.000 293 1 kWh in scientific notation. b. How many kilowatt-hours are there in 470.8 quadrillion Btu? Give your answer in scientific notation.
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11. Is the following statement true or false? “An increase in one order of magnitude is the same as an increase of 100%.” If true, explain why. If false, revise the statement to make it true. 12. On October 15, 2006, the San Francisco Chronicle published the accompanying graph and table derived from U.S. Bureau of the Census data on the growing size of the U.S. population. Compare the changes from 1915 to 2006 in two different categories, using at least one rate-of-change calculation and one order-of-magnitude comparison. Show your work.
13. The U.S. Census Bureau estimates that a baby is born somewhere in the country every 7 seconds, a new immigrant arrives every 31 seconds, and someone dies every 13 seconds, for a net average gain of one resident every 10 seconds. a. On average, how many babies are born in the country per day? b. On average, how many new immigrants arrive per day? c. On average, how many people die per day? 14. Temperature can affect the speed of sound. The speed of sound, S (in feet/second), at an air temperature of T (in degrees Celsius) is S5
1087(273 1 T) 0.5 16.52
A New Milestone for U.S. Population
a. Express T in terms of S. b. The speed of sound is often given as 1120 feet/second. At what temperature in degrees Celsius would that be? At what temperature in degrees Fahrenheit would that be? (Recall that degrees Fahrenheit ⫽ 1.8(degrees Celsius) ⫹ 32.)
The number of residents has tripled since 1915––and much has changed. U.S. population in hundreds of millions
300 1967: 200,000,000
250
15. The radius of Earth is about 6.3 ? 106 m and its mass is approximately 5.97 ? 1024 kg. Find its density in kg/m3 (density 5 mass/volume).
200
150
2006: 300,000,000
1915: 100,000,000
16. Objects that are less dense than water will float; those that are more dense than water will sink. The density of water is 1.0 g/cm3. A brick has a mass of 2268 g and a volume of 1230 cm3. Show that the brick will sink in water (recall density 5 mass/volume).
100
50 2010
2000
1990
1980
1970
1960
1950
1940
1930
1920
1910
1900
17. An adult patient weighs 130 lb. The prescription for a drug is 5 mg per kg of the patient’s weight per day. This drug comes in 100-mg tablets. What daily dosage should be prescribed? 18. On March 2, 2007, the Boston Globe reported the following:
1915 Woodrow Wilson
1967 Lyndon B. Johnson
2006 George W. Bush
$3,200
$24,600
$290,600
Cost of gallon of regular gas
25¢
33¢
$2.25
Cost of a firstclass stamp
2¢
5¢
39¢
Average household size
4.5 people
3.3 people
2.6 people
Number of people age 65 and older
4.5 million
19.1 million
36.8 million
President Price of new home
Most popular baby names for boys and girls
John and Mary
Michael and Lisa
Jacob and Emily
An exabyte is 1 quintillion bytes. In 2006 alone, the human race generated 161 exabytes of digital information. So? Well, that’s about 3 million times the information in all the books ever written or the equivalent of 12 stacks of books, each extending more than 93 million miles from Earth to the sun. a. Use scientific notation to represent the amount of digital information generated in 2006. (One quintillion is 1 followed by eighteen zeros.) b. Compare the amount of digital information generated in 2006 with the amount of information in all the books ever written, using orders of magnitude. c. Estimate how many miles of books are needed to hold the equivalent information in 161 exabytes. Express your answer in scientific notation. 19. Find the logarithm of each of the following numbers: a. 1 b. 1 billion c. 0.000 001 20. Estimate the following by placing the log between the two closest integer powers of 10. a. log 3000 b. log 150,000 c. log 0.05
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Item
Value
Mass-energy of electron
0.000 000 000 000 051 J
The kinetic energy of a flying mosquito
0.000 000 160 2 J
An average person swinging a baseball bat
Value in Scientific Notation
80 J
Energy received from the sun at Earth’s orbit on one square meter in one second
1,360 J
Energy released by one gram of TNT
4,184 J
Energy released by metabolism of one gram of fat Approximate annual power usage of a standard clothes dryer
259
38,000 J 320,000,000 J
Source: http://en.wikipedia.org.
21. One way of defining the energy unit joule (J) is the amount of the energy required to lift a small apple weighing 102 grams one meter above Earth’s surface. The accompanying table lists the estimated energy in joules for different situations. Use the accompanying table to answer the following questions. a. Write each value in scientific notation. b. A year’s use of a clothes dryer requires how many times the energy of swinging a baseball bat once? c. Metabolizing one gram of fat releases how many times the kinetic energy of a flying mosquito? 22. Rewrite each number as a power of 10, then create a logarithmic scale and estimate the location of the number on that scale. (Hint: log 2 ⫽ 0.301.) a. 10 b. 100 c. 200 d. 20,000
23. (Requires a scientific calculator.) Some drugs are prescribed in dosages that depend on a patient’s BSA, or body surface area, an indicator of metabolic mass. One formula for calculating BSA is BSA ⫽ 71.84W 0.425H0.725, where BSA is measured in square centimeters, W is weight in kilograms, and H is height in centimeters. A patient weighs 180 lb and is 6 feet tall. His dosage of a particular drug is 15 mg/m2/day (that is, 15 mg per square meter of body surface area per day). What is his daily dosage in mg? (Source: DuBois & DuBois, 1916, from http://en.wikipedia.org/wiki/Body_surface_area# Calculation.)
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E X P L O R AT I O N 4 . 1 The Scale and the Tale of the Universe Objective • gain an understanding of the relative sizes and relative ages of objects in the universe using scientific notation and unit conversions Materials/Equipment • tape, pins, paper, and string to generate a large wall graph (optional) • enclosed worksheet and conversion table on inside back cover
Related Readings/Videos Powers of Ten and “The Cosmic Calendar” from The Dragons of Eden Videos: Powers of Ten and The Cosmic Calendar in the PBS series Cosmos Related Software “E1: Tale and Scale of the Universe” in Exponential & Log Functions Procedure Work in small groups. Each group should work on a separate subset of objects on the accompanying worksheet. 1. Convert the ages and sizes of objects so they can be compared. You can refer to the conversion table that shows equivalences between English and metric units (see inside back cover). In addition, 1 light year < 9.46 ? 1012 km. 2. Generate on the blackboard or on the wall (with string) a blank graph whose axes are marked off in orders of magnitude (integer powers of 10), with the units on the vertical axis representing age of object, ranging from 100 to 1011 years, and the units on the horizontal axis representing size of object, ranging from 10212 to 1027 meters. 3. Each small group should plot the approximate coordinates of their selected objects (size in meters, age in years) on the graph. You might want to draw and label a small picture of your object to plot on your graph. Discussion/Analysis • Scan the plotted objects from left to right, looking only at relative sizes. Now scan the plotted objects from top to bottom, considering only relative ages. Does your graph make sense in terms of what you know about the relative sizes and ages of these objects? • Describe the scale and the tale of the universe.
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In Scientific Notation Object
Age (in years)
Size (of radius)
Observable universe
13.7 billion
1026 meters
Surtsey (Earth’s newest land mass)
40 years
0.5 mile
Pleiades (a galactic cluster)
100 million
32.6 light years
First living organisms on Earth
4.6 billion
0.000 05 meter
Pangaea (Earth’s prehistoric supercontinent)
200 million
4500 miles
First Homo sapiens sapiens
100 thousand
100 centimeters
First Tyrannosaurus rex
200 million
20 feet
Eukaryotes (first cells with nuclei)
2 billion
0.000 05 meter
Earth
4.6 billion
6400 kilometers
Milky Way galaxy
14 billion
50,000 light years
First atoms
13.7 billion
0.000 000 0001 meter
Our sun
5 billion
1 gigameter
Our solar system
5 billion
1 terameter
Age (in years)
Size (in meters)
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E X P L O R AT I O N 4 . 2 Patterns in the Positions and Motions of the Planets
Objective • explore patterns in the positions and motions of the planets and discover Kepler’s Law Introduction and Procedure Four hundred years ago, before Newton’s laws of mechanics, Johannes Kepler discovered a law that relates the periods of planets with their average distances from the sun. (A period of a planet is the time it takes the planet to complete one orbit of the sun.) Kepler’s strong belief that the solar system was governed by harmonious laws drove him to try to discover hidden patterns and correlations among the positions and motions of the planets. He used the trial-and-error method and continued his search for years. At the time of his work, Kepler did not know the distance from the sun to each planet in terms of measures of distance such as the kilometer. But he was able to determine the distance from each planet to the sun in terms of the distance from Earth to the sun, now called the astronomical unit, or A.U. for short. One A.U. is the distance from Earth to the sun. The first column in the table below gives the average distance from the sun to each of the planets in astronomical units. Patterns in the Positions and Motions of the Planets: Kepler’s Discovery Fill in the following table and look for the relationship that Kepler found.
Kepler’s Third Law: The First Planet Table (Inner Planetary System)
Planet
Average Distance from Sun (A.U.)*
Cube of the Distance (A.U.3)
Orbital Period (years)
Mercury
0.3870
0.2408
Venus
0.7232
0.6151
Earth
1.0000
1.0000
Mars
1.5233
1.8807
Jupiter
5.2025
11.8619
Saturn
9.5387
29.4557
Square of the Orbital Period (years2)
*1 A.U. < 149.6 ? 106 km; 1 year < 365.26 days. Source: Data from S. Parker and J. Pasachoff, Encyclopedia of Astronomy, 2nd ed. (New York: McGraw-Hill, 1993), Table 1, Elements of Planetary Orbits. Copyright © 1993 by McGraw-Hill, Inc. Reprinted with permission.
262
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The planets Uranus and Neptune were discovered after Kepler made his discovery. Check to see whether the relationship you found above holds true for these two planets.
The Second Planet Table (Outer Planetary System)
Planet
Average Distance from Sun (A.U.)
Uranus
19.1911
84.0086
Neptune
30.0601
164.7839
Cube of the Distance (A.U.3)
Orbital Period (years)
Square of the Orbital Period (years2)
Source: Data from S. Parker and J. Pasachoff, Encyclopedia of Astronomy, 2nd ed. (New York: McGraw-Hill, 1993), Table 1, Elements of Planetary Orbits. Copyright © 1993 by McGraw-Hill, Inc. Reprinted with permission. Note: Pluto is no longer classified as a planet. It is now called a “dwarf planet.”
Summary • Express your results in words. • Construct an equation showing the relationship between distance from the sun and orbital period. Solve the equation for distance from the sun. Then solve the equation for orbital period. • Do your conclusions hold for all of the planets?
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GROWTH AND DECAY: AN INTRODUCTION TO EXPONENTIAL FUNCTIONS OVERVIEW Exponential and linear functions are used to describe quantities that change over time. Exponential functions represent quantities that are multiplied by a constant factor during each time period. Linear functions represent quantities to which a fixed amount is added (or subtracted) during each time period. Exponential functions can model such diverse phenomena as bacteria growth, radioactive decay, compound interest rates, inflation, musical pitch, and family trees. After reading this chapter you should be able to • recognize the properties of exponential functions and their graphs • understand the differences between exponential and linear growth • model growth and decay phenomena with exponential functions • represent exponential functions using percentages, factors, or rates • use semi-log plots to determine if data can be modeled by an exponential function
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5.1 Exponential Growth The Growth of E. coli Bacteria
Growth of E. coli Bacteria Time Periods (of 20 minutes each)
Number of E. coli Bacteria
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
100 200 400 800 1,600 3,200 6,400 12,800 25,600 51,200 102,400 204,800 409,600 819,200 1,638,400 3,276,800 6,553,600 13,107,200 26,214,400 52,428,800 104,857,600 209,715,200 419,430,400 838,860,800 1,677,721,600
120,000
100,000
No. of bacteria
The program “E2: Exponential Growth & Decay” in Exponential & Log Functions offers a dramatic visualization of growth and decay.
Measuring and predicting growth is of concern to population biologists, ecologists, demographers, economists, and politicians alike. The growth of bacteria provides a simple model that scientists can generalize to describe the growth of other phenomena such as cells, countries, or capital. Bacteria are very tiny, single-celled organisms that are by far the most numerous organisms on Earth. One of the most frequently studied bacteria is E. coli, a rod-shaped bacterium approximately 1026 meter (or 1 micrometer) long that inhabits the intestinal tracts of humans and other mammals.1 The cells of E. coli reproduce by a process called fission: The cell splits in half, forming two “daughter cells.” Under ideal conditions E. coli divide every 20 minutes. If we start with an initial population of 100 E. coli bacteria that doubles every 20-minute time period, we generate the data in Table 5.1. The initial 100 bacteria double to become 200 bacteria at the end of the first time period, double again to become 400 at the end of the second time period, and so on. At the end of the twenty-fourth time period (at 24 ? 20 minutes 5 480 minutes, or 8 hours), the initial 100 bacteria in our model have grown to over 1.6 billion bacteria! Because the numbers become astronomically large so quickly, we run into the problems we saw in Chapter 4 when graphing numbers of widely different sizes. Figure 5.1 shows a graph of the data in Table 5.1 for only the first ten time periods. We can see from the graph that the relationship between number of bacteria and time is not linear. The number of bacteria seems to be increasing more and more rapidly over time.
80,000
60,000
40,000
20,000
0
Table 5.1
2
4 6 8 No. of 20-minute time periods
10
Figure 5.1 Growth of E. coli bacteria.
A mathematical model for E. coli growth Table 5.1 shows us that the initial number of 100 bacteria repeatedly doubles. If we record in a third column (Table 5.2) the number of times we multiply 2 times the original value of 100, we begin to see a pattern emerge.
1
Most types of E. coli are beneficial to humans, aiding in digestion. A few types are lethal. You may have read about deaths resulting from people eating certain deadly strains of E. coli bacteria in undercooked hamburgers or tainted spinach. The explosive nature of exponential growth shows how a few dangerous bacteria can multiply rapidly to become a deadly quantity in a very short time.
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Pattern in E. coli Growth Number of Time Periods
Number of E. coli Bacteria
0 1 2 3 4 5 6 7 8 9 10
100 200 400 800 1,600 3,200 6,400 12,800 25,600 51,200 102,400
Generalized Expression 100 100 ? 2 100 ? 2 ? 2 100 ? 2 ? 2 ? 2 100 ? 2 ? 2 ? 2 ? 2 100 ? 2 ? 2 ? 2 ? 2 ? 2 100 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 100 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 100 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 100 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 100 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2 ? 2
5 5 5 5 5 5 5 5 5 5 5
100 100 100 100 100 100 100 100 100 100 100
? ? ? ? ? ? ? ? ? ? ?
20 21 22 23 24 25 26 27 28 29 210
Table 5.2
Remembering that 20 equals 1 by definition, we can describe the relationship by number of E. coli bacteria 5 100 ? 2number of time periods If we let N 5 number of bacteria and t 5 number of time periods, we can write the equation more compactly as N 5 100 ? 2t Since each value of t determines one and only one value for N, the equation represents N as a function of t. The number 100 is the initial bacteria population. This function is called exponential since the input or independent variable, t, occurs in the exponent of the base 2. The base 2 is the growth factor, or the multiple by which the population grows during each time period. The bacteria double every time period, which means that the amount of increase equals the previous amount. Thus it increases at a rate of 100% during each time period. If E. coli grew unchecked at this pace, the offspring from one cell would cover Earth with a layer a foot deep in less than 36 hours!
The General Exponential Growth Function The E. coli growth equation N 5 100 ? 2t is in the form output 5 sinitial quantityd ? sgrowth factord input Such an equation, with the input in the exponent and a growth factor .1, describes an exponential growth function.
Exponential Growth Function An exponential growth function can be represented by an equation of the form y 5 Ca xsa . 1, C . 0d
where
a, the base, is the growth factor, the amount by which y is multiplied when x increases by 1. C is the initial value or y-intercept.
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EXAMPLE
1
Constructing an exponential function If you start with 200 cells that triple during every time period t, an equation to model its growth is N 5 200 ? 3t
EXAMPLE
2
Interpreting an exponential equation Interpret the numbers in the following equation: Q 5 s4 ? 106 d ? 2.5T
SOLUTION
EXAMPLE
4 ? 106, or 4,000,000, represents an initial population that grows by a factor of 2.5 each time period, T. This means the initial population is multiplied by 2.5 each time period, T.
3
Take a piece of ordinary paper, about 0.1 mm in thickness. Now start folding it in half, then in half again, and so on. Assume you could continue indefinitely. a. How would you model the growth in thickness of the folded paper? b. How many folds would it take to reach: i. The height of a human? ii. The height of the Matterhorn, a famous mountain in Switzerland? iii. The sun? iv. The edge of the known universe?
SOLUTION
a. Each time we fold the paper, we double the thickness. If N 5 number of folds, the initial thickness 5 0.1 mm, and T 5 thickness of the folded paper (in mm), then T 5 s0.1d ? 2N b. If you start plugging in values for N (this is where technology is useful), it takes 14 folds to reach the height of an average person (well, a short person). At 26 folds, the paper is higher than the Matterhorn. At 51 folds, the paper would reach the sun, and at 54 folds the edge of our solar system. It would take only 84 folds to reach the limits of the Milky Way. A mere 100 folds takes you to the edge of the known universe!
Looking at Real Growth Data for E. coli Bacteria Figure 5.2 shows a plot of real E. coli growth over 24 time periods (8 hours). The growth appears to be exponential for the first 12 time periods (4 hours). We can use technology to generate a best-fit exponential function for that section of the curve. The function is N 5 s1.37 ? 107 d ? 1.5t where N 5 number of cells per milliliter and t 5 the number of 20-minute periods. So the initial quantity is 1.37 ? 107 (or 13.7 million) cells. More important, the growth factor is 1.5. Every 20 minutes the number of cells is multiplied by 1.5 (a 50% increase), as opposed to the doubling (a 100% increase) in our first, idealized example. The bacterial growth rate in the laboratory is half that of the idealized data. But even this rate can’t be sustained for long. Conditions are rarely ideal; bacteria die, nutrients are used up, space to grow is limited. In the real world, growth that starts out as exponential must eventually slow down (see Figure 5.2). The curve flattens out as the number of bacteria reaches its maximum size, called the carrying capacity. The overall shape is called sigmoid. The arithmetic of exponentials leads to the inevitable conclusion that in the long term—for bacteria, mosquitoes, or humans—the rate of growth for populations must approach zero.
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DATA
ECOLI
Number of bacteria (in millions)
2000
269
Carrying capacity
1600
1200
800
400
0
5
10 15 Time periods
20
25
Figure 5.2 Sigmoid curve of real E. coli growth over 24 time periods and best-fit exponential for 12 time periods
Algebra Aerobics 5.1 A calculator that can evaluate powers is recommended for Problems 3 and 4. 1. Identify the initial value and the growth factor in each of the following exponential growth functions: a. y 5 350 ? 5x c. P 5 s7 ? 103 d ? 4t t b. Q 5 25,000 ? 1.5 d. N 5 5000 ? 1.025t 2. Write an equation for an exponential growth function where: a. The initial population is 3000 and the growth factor is 3. b. The initial population is 4 ? 107 and the growth factor is 1.3. c. The initial population is 75 and the population quadruples during each time period. d. The initial amount is $30,000 and the growth factor is 1.12.
3. The population growth of a small country is described by the function P 5 28 ? 1.065t , where t is in years and P is in millions. a. Determine P when t 5 0. What does that quantity represent? b. Determine P when t 5 10, 20, and 30. c. Estimate the value of t that would result in a doubling of the population to 56,000,000. If available, use technology to check your estimate. 4. Fill in a table of values for the amount A 5 80 ? 1.06t for values of the time period t of 0, 1, 10, 15, and 20, and then complete the following statements. a. The initial amount is ___. b. During the first time period, the amount grew from 80 to ___. c. Based on this table, the amount doubles between the values t 5 ___ and t 5 ___. d. Estimate the number of time periods it will take the amount to double.
Exercises for Section 5.1 1. The following exponential functions represent population growth. Identify the initial population and the growth factor. a. Q 5 275 ? 3T d. A 5 25s1.18d t t b. P 5 15,000 ? 1.04 e. Pstd 5 8000s2.718d t 8 x c. y 5 s6 ? 10 d ? 5 f. ƒsxd 5 4 ? 105 s2.5d x
2. Write the equation of each exponential growth function in the form y 5 Cax where: a. The initial population is 350 and the growth factor is 43 . b. The initial population is 5 ? 109 and the growth factor is 1.25.
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c. The initial population is 150 and the population triples during each time period. d. The initial population of 2 quadruples every time period.
y
y 10
10
5
5
3. Fill in the missing parts of the table.
Initial Value C
Growth Factor a
1600
1.05 3.25
Exponential Function y 5 Ca x y 5 6.2 ? 105 s2.07d x y 5 1400s __ d x
–5
x
0
5 –5
5
–3
–3
Graph A
Graph C
y
4. The populations of four towns for time t, in years, are given by:
x
0
y
10
10
5
5
P1 (t) 5 12,000(1.05) t P2 (t) 5 6000(1.07) t P3 (t) 5 100,000(1.01) t P4 (t) 5 1000(1.9) t a. Which town has the largest initial population? b. Which town has the largest growth factor? c. At the end of 10 years, which town would have the largest population? 5. Find C and a such that the function ƒsxd 5 Ca x satisfies the given conditions. a. ƒ(0) 5 6 and for each unit increase in x, the output is multiplied by 1.2. b. ƒ(0) 5 10 and for each unit increase in x, the output is multiplied by 2.5. 6. In the United States during the decade of the 1990s, live births to unmarried mothers, B, grew according to the exponential model B 5 1.165 ? 10 6(1.013)t, where t is the number of years after 1990. a. What does the model give as the number of live births to unwed mothers in 1990? b. What was the growth factor? c. What does the model predict for the number of live births to unwed mothers in 1995? In 2000? 7. A cancer patient’s white blood cell count grew exponentially after she had completed chemotherapy treatments. The equation C 5 63s1.17d d describes C, her white blood cell count per milliliter, d days after the treatment was completed. a. What is the white blood cell count growth factor? b. What was the initial white blood cell count? c. Create a table of values that shows the white blood cell counts from day 0 to day 10 after the chemotherapy. d. From the table of values, approximate when the number of white blood cells doubled. 8. Match each function with its graph. a. y 5 2s1.5d x c. y 5 2s3d x x b. y 5 3s1.5d d. y 5 4s2d x
–5
0
x 5 –5
0
–3
–3
Graph B
Graph D
x 5
9. National health care expenditures in 2005 were approximately $2016 billion and are expected to increase by a factor of 1.076 per year. In 5 years what would be the predicted expenditures? 10. The per-capita consumption of bottled water was 8 gallons in 1990 and has been increasing yearly by a factor of 1.088. What was the per capita consumption of bottled water 12 years later? 11. A tuberculosis culture increases by a factor of 1.185 each hour. a. If the initial concentration is 5 ? 10 3 cells/ml, construct an exponential function to describe its growth over time. b. What will the concentration be after 8 hours? 12. An ancient king of Persia was said to have been so grateful to one of his subjects that he allowed the subject to select his own reward. The clever subject asked for a grain of rice on the first square of a chessboard, two grains on the second square, four on the next, and so on. a. Construct a function that describes the number of grains of rice, G, as a function of the square, n, on the chessboard. (Note: There are 64 squares.) b. Construct a table recording the numbers of grains of rice on the first ten squares. c. Sketch your function. d. How many grains of rice would the king have to provide for the 64th (and last) square?
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13. A new species of fish is introduced into a pond. The size of the fish population can be modeled by the accompanying graph of the function P(t), where t is time in months. Growth of Fish Population
271
species (wild mallard, gadwall, and blue-winged teal). Each one-day-old duckling weighed about 32 grams. Weights were tracked and are graphed on the following chart. The weight curves appear sigmoidal, each approaching a maximum sustainable (or mature) weight.
90 80
Weight of Duckling
70
1200 solid line: Females dotted line: Males
50 40 30 20 10 0
5 10 15 20 25 30 35 40 45 50 55 60
Weight of duckling (grams)
Population
60 1000
Mallard
Gadwall
800 600
BW teal
400 200
Months
a. What is the initial population of the fish? b. How long did it take for the initial fish population to double? c. What was the sustainable fish population (carrying capacity) of the pond? d. Estimate the number of months it took for the fish population to reach its sustainable size. 14. The Northern Wildlife Prairie Research Center in Jamestown, North Dakota, measured the weights of three duckling
0
20
40
60
80
100
Age (days)
Source: http://www.npwrc.usgs.gov.
a. Estimate the mature weight for the female ducklings in each species. b. Estimate the mature weight for the male ducklings in each species. c. Approximately how long did it take for each species of ducklings to attain the mature weight?
5.2 Linear vs. Exponential Growth Functions Linear vs. Exponential Growth Linear growth is intrinsically additive. Linear growth means that for each unit increase in the input, we must add a fixed amount (the slope or rate of change) to the value of the output. For example, in the linear function N 5 100 1 2t, each time we increase t by 1, we add 2 to the value of N. After t time periods, we will have added 2t to the initial value of N. Assuming t is a positive integer, we can write this linear function as N 5 100 1 8 2 1 2 1 c1 2 t times Exponential growth is multiplicative, which means that for each unit increase in the input, we multiply the value of the output by a fixed number (the growth factor). For example, in the exponential function N 5 100 ? 2t, each time we increase t by 1, we multiply the value of N by 2. After t time periods have elapsed, we will have multiplied the value of N by 2t. If t is a positive integer, we can write this exponential function as N 5 100 ? 8 2 ? 2 ? c? 2 t times Comparing the general equations There are both similarities and differences in the general forms of the equations for linear and exponential functions.
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If y is a linear function of x, then y 5 b 1 mx y 5 y-intercept 1 (slope) ? x y5
and if x is a positive integer
b
1 sm 1 m 1 m 1 c 1 md 8 x times
If y is an exponential function of x, then y 5 C ? ax y 5 y-intercept ? (base)x y 5 C ? sa ? a ? a ? a ? c ? ad
and if x is a positive integer
8
x times
Linear functions involve repeated additions, whereas exponential functions involve repeated products. In each case, x determines the number of repetitions. For both linear and exponential equations, the vertical intercept gives the initial or starting value.
Linear vs. Exponential Functions A linear function represents a quantity to which a constant amount is added for each unit increase in the input. An exponential function represents a quantity that is multiplied by a constant factor for each unit increase in the input.
In the long run, exponential growth will always outpace linear growth Exponential growth is more rapid as time goes on, as can be seen in Table 5.3. For both the linear function N 5 100 1 2t and the exponential function N 5 100 ? 2t, when t 5 0, then N 5 100. After ten time periods, the values for N are strikingly different: 102,400 for the exponential function versus 120 for the linear function. The initial value of 100 has been multiplied by 2 ten times in the exponential function to get 100 ? 210. In the linear function, 2 has been added to 100 ten times, to get 100 1 2 ? 10. Comparing the Additive Pattern of Linear Growth to the Multiplicative Pattern of Exponential Growth
y
Time, t Exponential
Linear
x
Figure 5.3 Graph comparing linear and exponential growth.
0 1 2 3 4 5 6 7 8 9 10
Linear Function, N 5 100 1 2t Pattern 100 100 100 100 100 100 100 100 100 100 100
1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2 2
? ? ? ? ? ? ? ? ? ? ?
0 1 2 3 4 5 6 7 8 9 10
5 5 5 5 5 5 5 5 5 5 5
100 102 104 106 108 110 112 114 116 118 120
Exponential Function, N 5 100 ? 2t Pattern 100 100 100 100 100 100 100 100 100 100 100
? ? ? ? ? ? ? ? ? ? ?
20 21 22 23 24 25 26 27 28 29 210
5 100 5 200 5 400 5 800 5 1,600 5 3,200 5 6,400 5 12,800 5 25,600 5 51,200 5 102,400
Table 5.3
If we compare any linear growth function (whose graph will be a straight line with a positive slope) to any exponential growth function (whose graph will curve upward),
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we see that sooner or later the exponential curve will permanently lie above the linear graph and continue to grow faster and faster (Figure 5.3). The exponential function eventually dominates the linear function.
Comparing the Average Rates of Change Another way to compare linear and exponential functions is to examine average rates of change. Recall that if N is a function of t, then average rate of change 5
change in N N 5 change in t t
Examine Table 5.4, which contains average rates of change for both the linear and exponential functions, where t 5 1. For all linear functions, we know that the average rate of change is constant. For the linear function N 5 100 1 2t, we can tell from the equation, Table 5.4, and Figure 5.4(a) that the (average) rate of change is constant at two units. Average rates of change for the exponential function N 5 100 ? 2t are calculated in Table 5.4 and then graphed in Figure 5.4(b). These suggest that the average rates of change of an exponential growth function grow exponentially. Comparing Average-Rate-of-Change Calculations Linear Function N 5 100 1 2t
Exponential Function
t
N
Average Rate of Change (between t 2 1 and t)
0 1 2 3 4 5 6 7 8 9 10
100 102 104 106 108 110 112 114 116 118 120
n.a. 2 2 2 2 2 2 2 2 2 2
N 5 100 ? 2t N
Average Rate of Change (between t 2 1 and t)
100 200 400 800 1,600 3,200 6,400 12,800 25,600 51,200 102,400
n.a. 100 200 400 800 1,600 3,200 6,400 12,800 25,600 51,200
Table 5.4 5
60,000 50,000
4
Average rate of change
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3
2
1
40,000 30,000 20,000 10,000
t 0
5 (a)
10
t 0
5
10
(b)
Figure 5.4 (a) Graph of the average rates of change between several pairs of points that satisfy the linear function N 5 100 1 2t. (b) Graph of the average rates of change between several pairs of points that satisfy the exponential function N 5 100 ? 2t.
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A Linear vs. an Exponential Model through Two Points A town’s population increased from 20,000 to 24,000 over a 5-year period. The town council is concerned about the rapid population growth and wants to predict future population size. The functions most commonly used to predict growth patterns over time are linear and exponential. If we let t 5 number of years and P 5 population size, then when t 5 0, P 5 20,000 and when t 5 5, P 5 24,000. This gives us two points, (0, 20000) and (5, 24000), which we will use to construct our models. Constructing a linear model: Adding a constant amount each year Assuming linear growth, our function will be of the form P 5 b 1 mt. The initial population, b, is 20,000. The average rate of change or slope, m, through the two points is m5
change in population 24,000 2 20,000 4000 5 5 5 800 people/yr change in time 520 5
So the equation is P 5 20,000 1 800t This linear model says that the original population of 20,000 is increasing by the constant amount of 800 people each year. Constructing an exponential model: Multiplying by a constant factor each year Assuming exponential growth and an initial population of 20,000, our function will be of the form P 5 20,000 ? a t where P is the population of the town and t the number of years. To complete our model, we need a value for a, the annual growth factor. Finding the Growth Factor from Two Data Points. We have enough information to find the 5-year growth factor. The inputs 0 and 5 years give us two outputs, 20,000 and 24,000 people, respectively. Exponential growth is multiplicative, so for each additional year, the initial population is multiplied by the growth factor, a. After 5 years, the initial population of 20,000 is multiplied by a5 to get 24,000. Given divide by 20,000
20,000 ? a5 5 24,000 24,000 a5 5 5 1.2 20,000
So the 5-year growth factor, a5, is 1.2. Using a calculator, we can find the annual growth factor, a. Take the fifth root of both sides
(a5 ) 1/5 5 (1.2) 1/5 a1 < 1.0371
Now we can represent our exponential function as P 5 20,000 ? s1.0371d t where t 5 number of years and P 5 population size. Our exponential model tells us the initial population of 20,000 is multiplied by 1.0371 each year. Making predictions with our models We can use the models not only to describe past behavior, but also to predict future population sizes (see Table 5.5 and Figure 5.5). The population size is the same in both models for year 0 and year 5. For year 10, the linear and exponential predictions are
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275
Town Population Year
Linear Model P 5 20,000 1 800t
Exponential Model P 5 20,000s1.0371d t
0 5 10 15 20 25
20,000 24,000 28,000 32,000 36,000 40,000
20,000 24,000 28,790 34,540 41,440 49,720
Table 5.5
60,000 Linear model Exponential model
50,000 40,000 Population
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30,000 20,000 10,000
0
5
10
15
20
25
Year
Figure 5.5 Predictions for the town population using linear
and exponential models.
fairly close: 28,000 versus approximately 28,790. As we move further beyond year 10, the exponential predictions exceed the linear by an increasingly greater amount. By year 25, the linear model predicts a population of 40,000 whereas the exponential model predicts almost 50,000. Both models should be considered as generating only crude future estimates, particularly since we constructed the models using only two data points. Clearly, the further out we try to predict, the more unreliable the estimates become.
Identifying Linear vs. Exponential Functions in a Data Table Table 5.5 shows a linear function side by side with an exponential function. Every 5 years, the linear model added 800 people five times, adding a total of 4000 to the population. Every 5 years, the exponential model multiplied the population size by the annual growth factor 1.0371 five times; that is, the population size is multiplied by 1.03715 < 1.2. The 1-year growth factor 5 1.0371 and the 5-year growth factor 5 1.03715 < 1.2. If P1 is the starting population size and P2 is the population size 5 years later, then In the linear model or equivalently
P1 1 4000 5 P2 4000 5 P2 2 P1
In the exponential model or equivalently
P1 ? 1.2 5 P2 1.2 5 P2 /P1
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Using these equations we can translate our statements about addition (for linear) and multiplication (for exponential) into statements about differences and ratios. In general, given any data table of the form (x, y) where ⌬x is constant, then the data are linear if the difference between successive y values is constant. The data are exponential if the ratio of successive y values is constant. We can test whether a data table represents a linear or exponential function by using differences or ratios, as long as the ⌬x values are constant.
EXAMPLE
1
Determine whether each function in the table is linear or exponential. x
0
10
20
30
40
ƒ(x)
5
25
45
65
85
g(x)
20
30
45
67.5
101.25
Table 5.6
SOLUTION
The consecutive x values or input in Table 5.6 are ten units apart, so ⌬x 5 10. Now we compare the output values. The difference between successive values of ƒ(x) is always 20, so ƒ(x) is a linear function. The difference between successive values of g(x) is not constant. For example, but
g(10) 2 g(0) 5 30 2 20 5 10 g(20) 2 g(10) 5 45 2 30 5 15
So g(x) cannot be linear. To determine if g(x) is exponential, we calculate the ratios of successive values of g(x). For example, 45 67.5 101.25 30 5 5 5 5 1.5 20 30 45 67.5 The ratios of consecutive y values are constant at 1.5 when ⌬x 5 10 (this ratio represents the 10-year growth factor). Hence the function is exponential.
For a table describing y as a function of x, where ⌬ x is constant: If the difference between consecutive y values is constant, the function is linear. If the ratio of consecutive y values is constant, the function is exponential.
Algebra Aerobics 5.2 1. Fill in the following table and sketch the graph of each function. t 0 1 2 3 4
N 5 10 1 3t
N 5 10 ? 3t
2. a. Create a table of values (from t 5 0 to t 5 5) for the functions ƒ(t), with a vertical intercept at 200 and a constant rate of change of 20, and g(t), with a vertical intercept at 200 and a growth factor of 1.20. b. Sketch the graphs of g and ƒ on the same grid. c. Compare the graphs. 3. Given the values in Table 5.7, determine which functions (if any) are linear and which are approximately exponential. Justify your answer.
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T
ƒ(t)
g(t)
h(t)
p(t)
r(t)
0 1 2 3 4 5
10 16 22 28 34 40
10 10 10 10 10 10
10 15 22.5 33.75 50.63 75.94
10 100 190 280 370 460
10 10.4 10.82 11.25 11.70 12.17
277
b. The initial amount, A, is $80,000 and one time period, t, later, the amount grew by $2300. c. A quantity, Q, grew from 30 mg to 32.7 mg after one time period, t. 6. Assume in Table 5.8 that Q is an exponential function of t.
Table 5.7
4. Write an equation for the linear function and the exponential function that pass through the given points. a. (0, 500) and (1, 620) b. (0, 3) and (1, 3.2) 5. In each of the following situations, assume growth is exponential, find the growth factor, and then construct an exponential function that models the situation. a. The initial population, P, is 1500 and one time period, t, later, the population grew to 3750.
t (years)
Q
0 2 4 6 8
10 20
Table 5.8
a. By what factor is Q multiplied when t increases by 2 years? This is the 2-year growth factor. b. Fill in the rest of the Q values in the table. c. What is the annual growth factor? d. Construct an equation to model the relationship between Q and t.
Exercises for Section 5.2 b. If the size of the herd increases by a factor of 1.8 each year, find the formula for the deer population, Q(t), over time. c. For each model create a table of values for the deer population for a 10-year period. d. Using the table, estimate when the two models predict the same population size.
A graphing program is recommended for Exercises 3 and 11. 1. Determine which of the following functions are linear, which are exponential, and which are neither. In each case identify the vertical intercept. a. C(t) 5 3t 1 5 c. y 5 5x2 1 3 e. P 5 7s1.25d t b. ƒsxd 5 3s5d x
t
d. Q 5 6 Q32R
f. T 5 1.25n
2. Create a linear or exponential function based on the given conditions. a. A function with an average rate of change of 3 and a vertical intercept of 4. b. A function with growth factor of 3 and vertical intercept of 4. c. A function with slope of 4/3 and initial value of 5. d. A function with initial value of 5 and growth factor of 4/3. 3. (Graphing program recommended.) A small village has an initial size of 50 people at time t 5 0, with t in years. a. If the population increases by 5 people per year, find the formula for the population size P(t). b. If the population increases by a factor of 1.05 per year, find a new formula Q(t) for the population size. c. Plot both functions on the same graph over a 30-year period. d. Estimate the coordinates of the point(s) where the graphs intersect. Interpret the meaning of the intersection point(s). 4. A herd of deer has an initial population of 10 at time t 5 0, with t in years. a. If the size of the herd increases by 8 per year, find the formula for the population of deer, P(t), over time.
5. Construct both a linear and an exponential function that go through the points (0, 6) and (1, 9). 6. Construct both a linear and an exponential function that go through the points (0, 200) and (10, 500). 7. Find the equation of the linear function and the exponential function that are sketched through two points on each of the following graphs. y 60
1000
30
500
0
2
4 Graph A
6
x
y
0
3
6
x
Graph B
8. Match the description to any appropriate graph(s) shown below. a. When x increases by 10, y increases by 5. b. When x increases by 10, y is multiplied by a factor of 5.
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40
60
30
40
20
y
20
10
x 0
40
b. Compare the average rate of change in the mute swan population for the entire Atlantic coast with that for Massachusetts. c. Construct both a linear and an exponential model for the mute swan population in Massachusetts since 1986. d. Compare the projected populations of mute swans in Massachusetts by the year 2010 as predicted by your linear and exponential models.
y
10 20 Graph A
x 0
30
y
800
5
10 15 Graph C
20
25
y
30 20
400
10
x 0
10
20
30
40
50
x 0
Graph B
10 20 Graph D
30
9. Each of the following tables contains values representing either linear or exponential functions. Find the equation for each function. a. x 22 21 0 1 2 ƒ(x) 1.12 2.8 7 17.5 43.75 b.
x g(x)
22 0.1
21 0.3
0 0.5
1 0.7
2 0.9
10. Each table has values representing either linear or exponential functions. Find the equation for each function. a. x 22 21 0 1 2 h(x) 160 180 200 220 240 b.
x j(x)
0 200
10 230
20 264.5
30 304.17
40 349.8
11. (Graphing program recommended.) Create a table of values for the following functions, then graph the functions. a. ƒ(x) 5 6 1 1.5x c. hsxd 5 1.5s6d x x b. gsxd 5 6s1.5d 12. Suppose you are given a table of values of the form (x, y) where x, the distance between two consecutive x values, is constant. Why is calculating y2 2 y1 , the distance between two consecutive y values, equivalent to calculating the average rate of change between consecutive points? 13. Mute swans were imported from Europe in the nineteenth century to grace ponds. Now there is concern that their population is growing too rapidly, edging out native species. Their population along the Atlantic coast has grown from 5800 in 1986 to 14,313 in 2002. The increase is most acute in the mid-Atlantic region, but Massachusetts has also seen a jump, with 2939 mute swans counted in 2002 as compared with 585 in 1986. a. Compare the growth factor in the mute swan population for the entire Atlantic coast with that for Massachusetts.
14. The price of a home in Medford was $100,000 in 1985 and rose to $200,000 in 2005. a. Create two models, ƒ(t) assuming linear growth and g(t) assuming exponential growth, where t 5 number of years after 1985. b. Fill in the following table representing linear growth and exponential growth for t years after 1985.
t
Linear Growth ƒ(t) 5 Price (in thousands of dollars)
Exponential Growth g(t) 5 Price (in thousands of dollars)
100
100
200
200
0 10 20 30 40 50
c. Which model do you think is more realistic? 15. The Mass Media, a student publication at the University of Massachusetts–Boston, reported on a proposed parking fee increase. The university administration recommended gradually increasing the daily parking fee on this campus to $7.00 by the year 2004, followed by an increase of 5% every year after that. Call this plan A. Several other plans were also proposed; one of them, plan B, recommended that every year after 2004 the rate be increased by 50 cents each year. a. Let t 5 0 for year 2004 and fill in the chart for parking fees under plans A and B. Years after 2004
Parking Fee under Plan A
Parking Fee under Plan B
0 1 2 3 4
$7.00
$7.00
b. Write an equation for parking fees FA as a function of t (years since 2004) for plan A and an equation FB for plan B. c. What will the daily parking fee be by the year 2025 under each plan? (Show your calculations.) d. Imagine that you are the student representative to the UMass Board of Trustees. Which plan would you recommend for adoption? Explain your reasons and support your position using quantitative arguments.
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5.3 Exponential Decay An exponential growth function is of the form y 5 Cax, where a . 1. Each time x increases by 1, the population is multiplied by a, a number greater than 1, so the population increases. But what if a were positive and less than 1, that is, 0 , a , 1? Then each time x increases by 1 and the population is multiplied by a, the population size will be reduced. We would have exponential decay.
The Decay of Iodine-131 Iodine-131 is one of the radioactive isotopes of iodine that is used in nuclear medicine for diagnosis and treatment. It can be used to test how well the thyroid gland is functioning. An exponential function can be used to describe the decay of iodine-131. After being generated by a fission reaction, iodine-131 decays into a nontoxic, stable isotope. Every 8 days the amount of iodine-131 remaining is cut in half. For example, we show in Table 5.9 and Figure 5.6 how 160 milligrams (mg) of iodine-131 decays over four time periods (or 4 ? 8 5 32 days).
160
Time Periods (8 days each) 0
Amount of Iodine-131 (mg) 160
General Expression 5
1
80
5
2
40
5
3
20
5
4
10
5
160 5 160 160 A12B 160 A12B A12B 160 A12B A12B A12B 160 A12B A12B A12B A12B
5 160 5 160 5 160 5 160
A12B 0 A12B 1 A12B 2 A12B 3 A12B 4
Table 5.9
Amount of iodine-131 (mg)
140 120 100 80 60 40 20
0
1
2 Time periods
3
4
Figure 5.6 Exponential decay of
iodine-131.
We can describe the decay of iodine-131 with the equation 1 T Q 5 160 ? a b 2 where Q 5 quantity of iodine-131 (in mg) and T 5 number of time periods (of 8 days each). The number 160 is the initial amount of iodine-131 and 12 is the decay factor.
The General Exponential Decay Function The equation for Q is in the form output 5 sinitial quantityd ? sdecay factord input This is the standard format for an exponential decay function.
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Exponential Decay Function An exponential decay function y 5 ƒ(x) can be represented by an equation of the form y 5 Cax
(0 , a , 1 and C . 0)
where
a is the decay factor, the amount by which y is multiplied when x increases by 1. C is the initial value or y-intercept.
EXAMPLE
1
SOLUTION
a. How would you model the decay of 500 mg of iodine-131? b. How many milligrams would be left after four time periods (or 32 days)? a. The initial value is 500 mg, and we know from the previous discussion that the decay T factor is 12 for each time period T (of 8 days). So the decay function is Q 5 500 ? A12B . 4 1 b. When T 5 4, then Q 5 500 ? A12B 5 500 ? A16 B 5 31.25. So after four time periods (32 days) there would be 31.25 mg of iodine-131 remaining.
2
Identify the initial value and the decay factor for the function N 5 s3 ? 10 4d ? s0.25d T. Interpret the decay factor.
SOLUTION
The initial value is 3 ? 104 (or 30,000) and the decay factor is 0.25 (or 1/4). So each time T increases by 1, the value for N is one-fourth of its previous value.
EXAMPLE
EXAMPLE
3
Which of the following exponential functions represent growth and which represent decay? Identify the growth or decay factor. t a. Pstd 5 2000s1.05d t c. Nstd 5 16 A23B b. Qstd 5 25s0.75d t d. ƒsxd 5 5s4d 2x
SOLUTION
a. P(t) represents exponential growth because 1.05 . 1; the term 1.05 is the growth factor. b. Q(t) represents exponential decay because 0 , 0.75 , 1; the term 0.75 is the decay factor. c. N(t) represents exponential decay because 0 , 23 , 1; the term 23 is the decay factor. x
d. We can rewrite the function as f sxd 5 5 A14B . Thus ƒ(x) represents exponential decay because 0 , EXAMPLE
4
1 4
, 1; the term 14 is the decay factor.
Finding an exponential decay function through two points: Measuring caffeine levels When you drink an 8-ounce cup of coffee, virtually all the caffeine is absorbed in your gut and passes through the liver and into your bloodstream, acting as a stimulant. The peak blood levels of caffeine are reached in about 30 minutes—and then they begin to fall exponentially. Five hours after the peak, your blood contains 60 milligrams of caffeine. Fifteen hours after the peak your blood contains 15 milligrams of caffeine. a. Construct a function to model the caffeine decrease in your bloodstream over time. b. How much caffeine was in the original cup? c. Generate a sketch of the caffeine levels in your bloodstream over time.
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SOLUTION
281
a. The maximum caffeine blood levels occur about 30 minutes after drinking the coffee. Since the decrease in caffeine is exponential from that point on, we can model the decline with an exponential decay function. Let t 5 number of hours after reaching maximum caffeine blood level and C 5 amount of caffeine in your blood (mg). Five hours after reaching your maximum level, your caffeine level is 60 mg. Fifteen hours after the maximum level, your caffeine level is 15 mg. Assuming C is a function of t, the two points (5, 60) and (15, 15) should satisfy that function. If we let a 5 the hourly decay factor, then we can find the 10-hour decay factor, a10, using the ratio of the two caffeine levels. a10 5
caffeine at t 5 15 hr 15 mg 5 5 0.25 caffeine at t 5 5 hr 60 mg a10 5 0.25 (a10 ) 1/10 5 (0.25) 1/10 a < 0.871, the hourly decay factor
So the 10-hour decay factor is Taking the 10th root Using a calculator
So the exponential decay function is of the form C 5 C0 ? s0.871d t where C0 5 the peak amount of caffeine and t 5 number of hours since the peak caffeine level. b. To find C0 we can substitute either of the two points (5, 60) and (15, 15) known to satisfy our function. Given Substitute (5, 60) Use a calculator Solve for C0
C 5 C0 ? (0.871) t 60 5 C0 ? (0.871) 5 60 < C0 ? 0.501 60 C0 < < 120 mg 0.501
So the peak amount of caffeine in your blood was approximately 120 mg. Since virtually all the caffeine in the coffee is passed into the bloodstream, your original cup contained about 120 mg of caffeine.2 c.
C Caffeine peak
120 Caffeine in blood (mg)
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–10
0
10
20
t
No. of hours before and after blood caffeine peak
Figure 5.7 Blood caffeine levels after drinking 8 oz of coffee.
2
Note to coffee drinkers: Consumption of 250 mg of caffeine (2 to 3 cups) is considered a moderate amount; over 800 mg of caffeine is considered excessive. Various factors can influence the rate of caffeine decay. For a pregnant woman or a woman on oral contraceptives, the decrease in caffeine slows down considerably; a little caffeine takes a long time to pass through the body. For smokers, the decrease in caffeine is speeded up, so they can actually drink more coffee without feeling its side effects.
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Translating the Growth (or Decay) Factor from n Time Units to One Time Unit If a is the growth (or decay) factor for n time units, then a1/n is the growth (or decay) factor for one time unit. Examples: If 2.3 is the 10-year growth factor, then 2.31/10 < 1.0869 is the annual growth factor. If 0.7 is the 5-month decay factor, then 0.71/5 < 0.9311 is the one-month decay factor.
Algebra Aerobics 5.3 1. Which of the following exponential functions represent growth and which represent decay? a. y 5 100 ? 3x b. ƒstd 5 75 ? A23B
d. gsrd 5 s2 ? 106 d ? s1.15d r t
e. y 5 s7 ? 109 d ? s0.20d z
c. w 5 250 ? s0.95d r f. hsxd 5 150 ? A52B
x
2. Write an equation for an exponential decay function where: a. The initial population is 2300 and the decay factor is 13. b. The initial population is 3 ? 109 and the decay factor is 0.35. c. The initial population is 375 and the population drops to one-tenth its previous size during each time period.
3. Does the exponential function y 5 12 ? s5d 2x represent growth or decay? (Hint: Rewrite the function in the standard form y 5 Ca x.) 4. Rewrite each of the following using the general form and indicate whether the function represents growth or decay. a. y 5 23s2.4d 2x b. ƒsxd 5 8000s0.5d 2x c. P 5 52,000s1.075d 2t 5. Given ƒ(x) 5 100(0.9) x and g(x) 5 100(0.7) x, a. Which function decreases more rapidly? b. By what percentage does each function decrease each time period?
Exercises for Section 5.3 A graphing program is recommended for Exercise 15. 1. Identify and interpret the decay factor for each of the following functions: a. P 5 450(0.43) t b. f (t) 5 3500(0.95) t c. y 5 21(3) 2x 2. Which of the following exponential functions represent growth and which decay? a. N 5 50 ? 2.5T d. gszd 5 s3 ? 105 d ? s0.8d z x b. y 5 264s5/2d e. ƒsTd 5 s1.5 ? 1011 d ? s0.35d T x c. R 5 745s1.001d t f. hsxd 5 2000 Q23R 3. Write an equation for an exponential decay function where: a. The initial population is 10,000 and the decay factor is 25. b. The initial population is 2.7 ? 1013 and the decay factor is 0.27.
c. The initial population is 219 and the population drops to one-tenth its previous size during each time period. 4. The accompanying tables show approximate values for the four exponential functions: ƒsxd 5 5s2x d, gsxd 5 5s0.7x d, hsxd 5 6s1.7x d , and jsxd 5 6s0.6x d . Which table is associated with each function? Function B
Function A x
y
x
y
22 21 0 1 2
16.67 10.00 6.00 3.60 2.16
22 21 0 1 2
10.2 7.1 5.0 3.5 2.5
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Function C
9. Match each function with its graph. a. y 5 100s0.8d x b. y 5 100s0.5d x
Function D
x
y
x
y
22 21 0 1 2
1.25 2.50 5.00 10.00 20.00
22 21 0 1 2
2.1 3.5 6.0 10.0 17.3
300
5. Determine which of the following functions are exponential. For each exponential function, identify the growth or decay factor and the vertical intercept. a. y 5 5sx2 d c. P 5 1000s0.999d t 2x b. y 5 100 ? 2 6. Determine which of the following functions are exponential. Identify each exponential function as representing growth or decay and find the vertical intercept. a. A 5 100s1.02t d d. y 5 100x 1 3 b. ƒsxd 5 4s3x d e. M 5 2 p c. gsxd 5 0.3s10x d f. y 5 x2 7. Fill in the missing parts of the table.
Initial Value, C
Decay Factor, a
500
0.95 0.25
Exponential Function y 5 Ca x
8. a. Complete the following table for the exponential function y 5 20s0.75d x.
x
y
0 1 2 3 4
20 15 11.25 8.4375 6.328125
y
y 150
200
100
100
50
y2 2 y1
Ratio y2 y1
b. Choose the correct word in each italicized pair to describe the function: For the exponential function y 5 20s0.75d x, the differences are constant/decreasing in magnitude and the ratios are constant/decreasing in magnitude.
x
x –2
0
2
4
6
8
10
–2
0
2
Graph A
4
6
8
10
Graph B
10. The per capita (per person) consumption of milk was 27.6 gallons in 1980 and has been steadily decreasing by an annual decay factor of 0.99. a. Form an exponential function for per capita milk consumption M(t) for year t after 1980. b. According to your function, what was the per capita consumption of milk in 2000? If available, use the Internet to check your predictions. 11. The U.S. Department of Agriculture’s data on per capita food commodity consumption for 1980 are listed in the following table. a. Using the data in the following table, construct exponential functions for each food category. Then evaluate each function for the year 2000. Assume t is the number of years since 1980.
y 5 (1.72 ? 106 )(0.75) x y 5 1600( ___ ) x
Difference
283
Beef Chicken Pork Fish
Per Capita Consumption (pounds) in 1980
Yearly Growth/Decay Factor
Exponential Function
72.1 32.7 52.1 12.4
0.994 1.024 0.996 1.010
B(t) 5 C(t) 5 P(t) 5 F(t) 5
b. Which commodities showed exponential growth? Which showed exponential decay? c. Write a 60-second summary about the consumption of meat, chicken, and fish from 1980 to 2000. 12. Find the formula for the exponential function that satisfies the given conditions. a. ƒ(0) 5 4 and ƒ(1) 5 2 b. g(0) 5 6 and g(1) 5 3 c. h(0) 5 100 and h(1) 5 75 d. k(0) 5 12 and k(1) 5 7.2 13. a. A linear function ƒ(t) 5 b 1 mt has a slope of 24 and a vertical intercept of 20. Find its equation. b. An exponential function g(t)5 Ca t has a decay factor of 1/4 and an initial value of 20. Find its equation. c. Plot both functions on the same grid.
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y
14. Which of the following functions (if any) are equivalent? Explain your answer. a. ƒsxd 5 40s0.625d x b. gsxd 5 40 A58B
x
c. hsxd 5 40 A85B
2x
1600 1200 800 400
15. (Graphing program recommended.) Which of the following functions declines more rapidly? Graph the functions on the same grid and check your answer. a. ƒsxd 5 25s5d 2x b. gsxd 5 25s0.5d x 16. Find the equation of the exponential function through the indicated points in graphs A and B.
y 1600 1200
(–1, 900)
800
x –1
0
1
2
x –2
–1
0
1
2
Graph B
17. Plutonium-238 is used in bombs and power plants but is dangerously radioactive. It decays very slowly into nonradioactive materials. If you started with 100 grams today, a year from now you would have 99.2 grams. a. Construct an exponential function to describe the decay of plutonium-238 over time. b. How much of the original 100 grams of plutonium-238 would be left after 50 years? After 500 years? 18. It takes 1.31 billion years for radioactive potassium-40 to drop to half its original size. a. Construct a function to describe the decay of potassium-40. b. Approximately what amount of the original potassium-40 would be left after 4 billion years? Justify your answer.
(1, 400)
400
–2
(1, 900)
(–1, 400)
3
Graph A
5.4 Visualizing Exponential Functions We can summarize what we’ve learned so far about exponential functions:
Exponential Functions Exponential functions can be represented by equations of the form y 5 Ca x
(a . 0, a 2 1)where
C is the initial value and a is the base Assuming C . 0, then if a . 1, the function represents growth and a is called the growth factor. 0 , a , 1, the function represents decay and a is called the decay factor.
The Effect of the Base a Exponential growth: a > 1 Given an exponential function y 5 Ca x, when a . 1 (and C . 0), the function represents growth. The graph of an exponential growth function is concave up and
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increasing. Figure 5.8 shows how the value of a affects the steepness of the graphs of the following functions: y 5 100 ? 1.2x
y 5 100 ? 1.3x
y 5 100 ? 1.4x
Each function has the same initial value of C 5 100 but different values of a, that is, 1.2, 1.3, and 1.4, respectively. When a . 1, the larger the value of a, the more rapid the growth and the more rapidly the graph rises. So as the values of x increase, the values of y not only increase, but increase at an increasing rate.
Exploration 5.1 along with “E3: y 5 Cax Sliders” in Exponential & Log Functions allows you to examine the effects of a and C on the graph of an exponential function.
400
y
y y = 100(1.4)x
y = 100(1.3)x
200
1000 x
y = 100(1.2)
?
2000
y = 1000(0.7)x
SOMETHING TO THINK ABOUT
What would the graph of the function y 5 Ca x look like if a 5 1? What kind of function would you have?
y = 1000(0.5)x y
= 1000(0.3)x
x –5
0
x
5
–1
0
3
Figure 5.8 Three exponential growth
Figure 5.9 Three exponential decay
functions.
functions.
Exponential decay: 0 < a < 1 In an exponential function y 5 Ca x, when a is positive and less than 1 (and C . 0), we have decay. The graph of an exponential decay function is concave up but decreasing. The smaller the value of a, the more rapid the decay. For example, if a 5 0.7, after each unit increase in x, the value of y would be multiplied by 0.7. So y would drop to 70% of its previous size—a loss of 30%. But if a 5 0.5, then when x increases by 1, y would be multiplied by 0.5, equivalent to dropping to 50% of its previous size—a loss of 50%. Figure 5.9 shows how the value of a affects the steepness of the graph. Each function has the same initial population of 1000 but different values of a, that is, 0.3, 0.5, and 0.7, respectively. When 0 , a , 1, the smaller the value of a, the more rapid the decay and the more rapidly the graph falls.
The Effect of the Initial Value C In the exponential function y 5 Ca x, the initial value C is the vertical intercept. When x 5 0, then y 5 Ca0 5C?1 5C Figure 5.10 compares the graphs of three exponential functions with the same base a of 1.1, but with different C values of 50, 100, and 250, respectively.
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700
y y = 250(1.1)x
?
SOMETHING TO THINK ABOUT
What happens if C 5 0? What happens if C , 0?
y = 100(1.1)x
y = 50(1.1)x x –10
0
10
Figure 5.10 Three exponential func-
tions with the same growth factor but different y-intercepts, or initial values.
In Exploration 5.1, you can examine what happens when C , 0.
EXAMPLE
Changing the value of C changes where the graph of the function will cross the vertical axis. 1
Match each of the graphs (A to D) in Figure 5.11 to one of the following equations and explain your answers. a. ƒsxd 5 1.5s2d x c. jsxd 5 5s0.6d x d. ksxd 5 5s0.8d x
b. gsxd 5 1.5s3d x
y A
B
10
CD
x –10
0
10
Figure 5.11 Graphs of four exponential functions. SOLUTION
A is the graph of ksxd 5 5s0.8d x. B is the graph of jsxd 5 5s0.6d x. Reasoning: Graphs A and B both have a vertical intercept of 5 and represent exponential decay. The steeper graph (B) must have the smaller decay factor (in this case 0.6). C is the graph of gsxd 5 1.5s3d x. D is the graph of ƒsxd 5 1.5s2d x. Reasoning: Graphs C and D both have a vertical intercept of 1.5 and both represent exponential growth. The steeper graph (C) must have the larger growth factor (in this case 3).
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Horizontal Asymptotes In the exponential decay function y 5 1000 ? ( 12 ) x, graphed in Figure 5.12, the initial population of 1000 is cut in half each time x increases by 1. So when x 5 0, 1, 2, 3, 4, 5, 6, . . . , the corresponding y values are 1000, 500, 250, 125, 62.5, 31.25, 15.625, . . . . As x gets larger and larger, the y values come closer and closer to, but never reach, zero. Thus as x approaches positive infinity sx S 1`d , y approaches zero sy S 0d . The graph of the function y 5 1000 s 12 d x is said to be asymptotic to the x-axis, or we say the x-axis is a horizontal asymptote to the graph. In general, a horizontal asymptote is a horizontal line that the graph of a function approaches for extreme values of the input or independent variable. 1500
y
1 x __ 2
() Exponential growth:
y=1000•2x
y=1000
Exponential growth: as x → –∞, y → 0
as x → +∞, y → 0
x –10
0
10
Figure 5.12 The x-axis is a horizontal asymptote for both exponential growth and decay functions of the form y 5 Ca x.
Exponential decay functions of the form y 5 Cax are asymptotic to the x-axis. Similarly, for exponential growth functions, as x approaches negative infinity sx S 2`d , y approaches zero sy S 0d . Examine, for instance, the exponential growth function y 5 1000 ? 2x, also graphed in Figure 5.12. When x 5 0, 21, 22, 23, . . . , the corresponding y values are 1000, 1000 ? 221 5 500, 1000 ? 222 5 250, 1000 ? 223 5 125, . . . . So as x S 2` , the y values come closer and closer to but never reach zero. So exponential growth functions are also asymptotic to the x-axis.
Graphs of Exponential Functions For functions in the form y 5 Ca x, The value of C tells us where the graph crosses the y-axis. The value of a affects the steepness of the graph. Exponential growth sa . 1, C . 0d . The larger the value of a, the more rapid the growth and the more rapidly the graph rises. Exponential decay (0 , a , 1, C . 0). The smaller the value of a, the more rapid the decay and the more rapidly the graph falls. The graphs are asymptotic to the x-axis. Exponential growth: As x S 2`, y S 0 Exponential decay: As x S 1`, y S 0
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Algebra Aerobics 5.4 1. a. Draw a rough sketch of each of the following functions all on the same graph: y 5 1000s1.5d x y 5 1000s1.1d x y 5 1000s1.8d x b. Do the three curves intersect? If so, where? c. In the first quadrant (where x . 0 and y . 0), which curve should be on the top? Which in the middle? Which on the bottom? d. In the second quadrant (where x , 0 and y . 0), which curve should be on the top? Which in the middle? Which on the bottom? 2. a. Draw a rough sketch of each of the following functions all on the same graph: Q 5 250s0.6d t
Q 5 250s0.3d t
and
70 60 50 40
Q 5 250s0.2d t
b. Do the three curves intersect? If so, where? c. In the first quadrant, which curve should be on the top? Which in the middle? Which on the bottom? d. In the second quadrant, which curve should be on the top? Which in the middle? Which on the bottom? 3. a. Draw a rough sketch of each of the following functions on the same graph: P 5 50 ? 3t
5. Which of these functions has the most rapid growth? Which the most rapid decay? a. f(t) 5 100 ? 1.06t b. g(t) 5 25 ? 0.89t c. h(t) 5 4000 ? 1.23t d. r(t) 5 45.9 ? 0.956t e. P(t) 5 32,000 ? 1.092t 6. Examine the graphs of the exponential growth functions ƒ(x) and g(x) in Figure 5.13.
P 5 150 ? 3t
b. Do the curves intersect anywhere? c. Describe when and where one curve lies above the other. 4. Identify any horizontal asymptotes for the following functions: a. Q 5 100 ? 2t c. y 5 100 2 15x 7 r b. gsrd 5 s6 ? 10 d ? s0.95d
30 20
g(x) f(x)
10
x –5
5
Figure 5.13 Graphs of ƒ(x) and
g(x).
a. Which graph has the larger initial value? b. Which graph has the larger growth factor? c. As x S 2` , which graph approaches zero more rapidly? d. Approximate the point of intersection. After the point of intersection, as x S 1` , which function has larger values?
Exercises for Section 5.4 A graphing program is required for Exercises 9 and 10, and recommended for Exercises 11, 15, and 16. 1. Each of the following three exponential functions is in the standard form y 5 C ? a x. y52
x
y55
x
y 5 10
x
a. In each case identify C and a. b. Specify whether each function represents growth or decay. In particular, for each unit increase in x, what happens to y? c. Do all three curves intersect? If so, where? d. In the first quadrant, which curve should be on top? Which in the middle? Which on the bottom? e. Describe any horizontal asymptotes.
f. For each function, generate a small table of values. g. Graph the three functions on the same grid and verify that your predictions in part (d) are correct. 2. Repeat Exercise 1 for the functions y 5 s0.5d2x, y 5 2 ? 2 x, and y 5 5 ? 2x. 3. Repeat Exercise 1 for the functions y 5 3x, y 5 A13B x, and y 5 3 ? A13B x. 4. Match each equation with its graph (at the top of the next page). ƒsxd 5 30 ? 2x
hsxd 5 100 ? 2x
gsxd 5 30 ? s0.5d x
jsxd 5 50 ? s0.5d x
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y A B
100
289
c. As x S 2` , will the function with the largest growth factor approach zero more slowly or more quickly than the other functions? d. For x . 0, the graph of which function remains on top?
D
7. Examine the accompanying graphs of three exponential functions. y
C
h(x) 10
g(x) x –5
0
5
f(x)
5. Below are graphs of four exponential functions. Match each function with its graph. P 5 5 ? s0.7d x
R 5 10 ? s1.8d x
Q 5 5 ? s0.4d x
S 5 5 ? s3d x
5
y
y 30
30
x –5
x
x –2
0 Graph A
2
–2
0 Graph C
y
2
y
30
30
0 Graph B
x
2
–2
0 Graph D
2
6. Examine the accompanying graphs of three exponential growth functions. y 10
f (x)
5
x –1
10
a. Order the functions from smallest decay factor to largest. b. What point do all of the functions have in common? Will they share any other points? c. As x S 1`, will the function with the largest decay factor approach zero more slowly or more quickly than the other functions? d. When x , 0, the graph of which function remains on top?
a. b. c. d.
As x As x As x As x
S S S S
ƒsxd 5 4s3.5d x
gsxd 5 4s0.6d x
hsxd 5 4 1 3x
ksxd 5 4 2 6x
1` , which function(s) approach 1` ? 1` , which function(s) approach 0? 2` , which function(s) approach 2` ? 2` , which function(s) approach 0?
9. (Graphing program required.) Graph the functions ƒ(x) 5 30 1 5x and gsxd 5 3s1.6d x on the same grid. Supply the symbol , or . in the blank that would make the statement true. a. ƒ(0) ______ g(0) e. ƒ(26) ______ g(26) b. ƒ(6) ______ g(6) f. As x S 1` , ƒ(x) ______ g(x) c. ƒ(7) ______ g(7) g. As x S 2` , ƒ(x) ______ g(x) d. ƒ(25) ______ g(25)
g(x)
h(x)
–5
5
8. Generate quick sketches of each of the following functions, without the aid of technology.
x –2
–1
5
a. Order the functions from smallest growth factor to largest. b. What point do all of the functions have in common? Will they share any other points?
10. (Graphing program required.) Graph the functions ƒsxd 5 6s0.7d x and gsxd 5 6s1.3d x on the same grid. Supply the symbol ,, ., or 5 in the blank that would make the statement true. a. ƒ(0) ______ g(0) b. ƒ(5) ______ g(5) c. ƒ(25) ______ g(25) d. As x S 1`, ƒsxd ______ g(x) e. As x S 2`, ƒsxd ______ g(x)
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11. (Graphing program recommended.) a. As x S 1` , which function will dominate, ƒ(x) 5 100 1 500x or gsxd 5 2s1.005d x? b. Determine over which x interval(s) g(x) . ƒ(x). 12. Two cities each have a population of 1.2 million people. City A is growing by a factor of 1.15 every 10 years, while city B is decaying by a factor of 0.85 every 10 years. a. Write an exponential function for each city’s population PA std and PB std after t years. b. For each city’s population function generate a table of values for x 5 0 to x 5 50, using 10-year intervals, then sketch a graph of each town’s population on the same grid. 13. Which function has the steepest graph? F(x) 5 100(1.2) x G(x) 5 100(0.8) x H(x) 5 100(1.2) 2x 14. Examine the graphs of the following functions. y
y
10
10
f(x)
x
x –10
10
–10
10
g(x)
–10
–10
y
y
10
10
a. Which graphs appear to be mirror images of each other across the y-axis? b. Which graphs appear to be mirror images of each other across the x-axis? c. Which graphs appear to be mirror images of each about the origin (i.e., you could translate one into the other by reflecting first about the y-axis, then about the x-axis)? 15. (Graphing program recommended.) On the same graph, sketch ƒsxd 5 3s1.5d x, gsxd 5 23s1.5d x, and hsxd 5 3s1.5d 2x. a. Which graphs are mirror images of each other across the y-axis? b. Which graphs are mirror images of each other across the x-axis? c. Which graphs are mirror images of each other about the origin (i.e., you could translate one into the other by reflecting first about the y-axis, then about the x-axis)? d. What can you conclude about the graphs of the two functions ƒsxd 5 Ca x and gsxd 5 2Ca x? e. What can you conclude about the graphs of the two functions ƒsxd 5 Ca x and gsxd 5 Ca2x? 16. (Graphing program recommended.) Make a table of values and plot each pair of functions on the same coordinate system. a. y 5 2x and y 5 2x for 23 # x # 3 b. y 5 s0.5d x and y 5 0.5x for 23 # x # 3 c. Which of the four functions that you drew in parts (a) and (b) represent growth? d. How many times did the graphs that you drew for part (a) intersect? Find the coordinates of any points of intersection. e. How many times did the graphs that you drew for part (b) intersect? Find the coordinates of any points of intersection.
h (x) x
x
–10
10
–10
10
k(x)
–10
–10
5.5 Exponential Functions: A Constant Percent Change We defined exponential functions in terms of multiplication by a constant (growth or decay) factor. In this section we show how these factors can be translated into a constant percent increase or decrease. While growth and decay factors are needed to construct an equation, journalists are more likely to use percentages.
Exponential Growth: Increasing by a Constant Percent In Section 5.1 we modeled the growth of E. coli bacteria with the exponential function N 5 s1.37 ? 107 d ? 1.5t. The growth factor, 1.5, tells us that for each unit increase in
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time, the initial number of E. coli is multiplied by 1.5. How can we translate this into a statement involving percentages? If Q is the number of bacteria at any point in time, then one time period later, number of E. coli 5 Rewrite 1.5 as a sum 5 Use the distributive property 5 Rewrite 0.5 as a percentage 5
Q ? (1.5) Q(1 1 0.5) 1 ? Q 1 (0.5) ? Q Q 1 50% of Q
So for each unit increase in time, the number of bacteria increases by 50%. The term 50% is called the growth rate in percentage form and 0.5, the equivalent in decimal form, is called the growth rate in decimal form. A quantity that increases by a constant percent over each time interval represents exponential growth. We can represent this growth in terms of growth factors or growth rates. Exponential growth requires a growth factor . 1. If the exponential growth factor 5 1, the object would never grow, but would stay constant through time. So, growth factor 5 1 1 growth rate growth factor 5 1 1 r where r is the growth rate in decimal form. EXAMPLE
1
Converting the growth rate to a growth factor The U.S. Bureau of the Census has made a number of projections for the first half of this century. For each projection, convert the growth rate into a growth factor. a. Disposable income is projected to grow at an annual rate of 2.9%. b. Nonagriculture employment is projected to grow at 1.0% per year. c. Employment in manufacturing is projected to grow at 0.2% per year.
SOLUTION
EXAMPLE
a. A growth rate of 2.9% is 0.029 in decimal form. The growth factor is 1 1 0.029 5 1.029. b. A growth rate of 1.0% is 0.01 in decimal form. The growth factor is 1 1 0.01 5 1.01. c. A growth rate of 0.2% is 0.002 in decimal form. The growth factor is 1 1 0.002 5 1.002.
2
According to UN statistics, in the year 2005 there were an estimated 3.2 billion people living in urban areas. The number of people living in urban areas is projected to grow at a rate of 1.7% per year. If this projection is accurate, how many people will be living in urban areas in 2030?
SOLUTION
If the urban population is increasing by a constant percent each year, the growth is exponential. To construct an exponential function modeling this growth, we need to find the growth factor. We know that the growth rate 5 1.7% or 0.017. So, growth factor 5 1 1 0.017 5 1.017 Given an initial population of 3.2 billion in 2005, our model is U(n) 5 3.2 ? (1.017) n where n 5 number of years since 2005 and U(n) 5 urban population (in billions). Using our model, the urban population in 2030, U(25) < 4.9 billion people.
Exponential Decay: Decreasing by a Constant Percent In Section 5.3, we used the exponential function C 5 C0 ? s0.871d t (where C0 5 the peak amount of caffeine and t 5 number of hours since peak caffeine level) to model the
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amount of caffeine in the body after drinking a cup of coffee. The decay factor, 0.871, tells us that for each unit increase in time, the amount of caffeine is multiplied by 0.871. As with exponential growth, we can translate this statement into one that involves percentages. If Q is the amount of caffeine at any point in time, then one time period later, amount of caffeine Rewrite 0.871 as a difference Use the distributive property Rewrite 0.129 as a percentage
5 5 5 5
Q ? (0.871) Q ? (1 2 0.129) 1 ? Q 2 (0.129) ? Q Q 2 12.9% of Q
So for each unit increase in time, the amount of caffeine decreases by 12.9%. The term 12.9% is called the decay rate in percentage form and the equivalent decimal form, 0.129, is called the decay rate in decimal form. A quantity that decreases by a constant percent over each time interval represents exponential decay. We can represent this decay in terms of decay factors or decay rates. Exponential decay requires a decay factor between 0 and 1. decay factor 5 1 2 decay rate decay factor 5 1 2 r where r is the decay rate in decimal form. Factors, Rates, and Percentages For an exponential function in the form y 5 Cax, where C . 0, The growth factor a . 1 can be represented as a511r where r is the growth rate, the decimal representation of the percent rate of change. The decay factor a, where 0 , a , 1, can be represented as a512r where r is the decay rate, the decimal representation of the percent rate of change. Examples: If a quantity is growing by 3% per month, the growth rate r in decimal form is 0.03, so the growth factor a 5 1 1 0.03 5 1.03. If a quantity is decaying by 3% per month, the decay rate r in decimal form is 0.03, so the decay factor a 5 1 2 0.03 5 0.97.
EXAMPLE
3
Constructing an exponential decay function Sea ice that survives the summer and remains year round, called perennial sea ice, is melting at the alarming rate of 9% per decade, according to a 2006 report by the National Oceanic and Atmospheric Administration. Assuming the percent decrease per decade in sea ice remains constant, construct a function that represents this decline.
SOLUTION
If the percent decrease remains constant, then the melting of sea ice represents an exponential function with a decay rate of 0.09 in decimal form. So, decay factor 5 1 2 0.09 5 0.91 If d 5 number of decades since 2006 and A 5 amount of sea ice (in millions of acres) in 2006, then we can model the amount of sea ice, S(d), by the exponential function: Ssdd 5 As0.91d d
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EXAMPLE
4
SOLUTION
293
Growth or decay? In the following functions, identify the growth or decay factors and the corresponding growth or decay rates in both percentage and decimal form. a. ƒsxd 5 100s0.01d x c. p 5 5.34s0.015d n x b. gsxd 5 230s3.02d d. y 5 8.75s2.35d x a. The decay factor is 0.01 and if r is the decay rate in decimal form, then decay factor 5 1 2 decay rate 0.01 5 1 2 r r 5 0.99 The decay rate is 0.99, or 99% as a percentage. b. The growth factor is 3.02 and if r is the growth rate in decimal form, then growth factor 5 1 1 growth rate 3.02 5 1 1 r 3.02 2 1 5 r r 5 2.02 The growth rate in decimal form is 2.02, or 202% as a percentage. c. The decay factor is 0.015 and the decay rate is 0.985 in decimal form, or 98.5% as a percentage. d. The growth factor is 2.35 and the growth rate is 1.35 in decimal form, or 135% as a percentage.
Revisiting Linear vs. Exponential Functions In Section 5.2 we compared linear vs. exponential functions as additive versus multiplicative. We now have another way to compare them.
Linear vs. Exponential Functions Linear functions represent quantities that increase or decrease by a constant amount. Exponential functions represent quantities that increase or decrease by a constant percent.
EXAMPLE
5
Identifying linear vs. exponential growth According to industry sources, U.S.wireless data service revenues are growing by an amazing 75% annually. India is the biggest growth market, increasing by about 6 million cell phones every month. Which of these two statements represents linear and which represents exponential growth? Justify your answers.
SOLUTION
Quantities that increase by a constant amount represent linear growth. So, an increase of 6 million cell phones every month describes linear growth. Quantities that increase by a constant percent represent exponential growth. So, an increase of 75% annually describes exponential growth.
EXAMPLE
6
Construct an equation for each description of the population of four different cities. a. In 1950, city A had a population of 123,000 people. Each year since 1950, the population decreased by 0.8%.
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b. In 1950, city B had a population of 4,500,000 people. Since 1950, the population has grown by approximately 2000 people per year. c. In 1950, city C had a population of 625,000. Since 1950, the population has declined by about 5000 people each decade. d. In 1950, city D had a population of 2.1 million people. Each decade since 1950, the population has increased by 15%. SOLUTION
EXAMPLE
For each of the following equations, let P 5 population (in thousands) and t 5 years since 1950. a. PA 5 123s0.992d t b. PB 5 4500 1 2t where the unit for rate of change is people (in thousands) per year. c. PC 5 625 2 0.5t where the unit for rate of change is people (in thousands) per year. The rate of change is 20.5 people (in thousands) per year, since the population declined by 5000 people per decade, or 500 people per year. d. growth rate per decade 5 0.15 growth factor per decade 5 1 1 0.15 5 1.15 growth factor per year 5 s1.15d 1/10 < 1.014 PD 5 2100s1.014d t
7
You are offered a job with a salary of $30,000 a year and annual raises of 5% for the first 10 years of employment. Show why the average rate of change of your salary in dollars per year is not constant.
SOLUTION
Let n 5 number of years since you signed the contract and S(n) 5 your salary in dollars after n years. When n 5 0, your salary is represented by S(0) 5 $30,000. At the end of your first year of employment, your salary increases by 5%, so S(1) 5 30,000 1 5% of 30,000 5 30,000 1 0.05 (30,000) 5 30,000 1 1500 5 $31,500 At the end of your second year, your salary increases again by 5%, so S(2) 5 31,500 1 5% of 31,500 5 31,500 1 0.05(31,500) 5 31,500 1 1575 5 $33,075 The average rate of change between years 0 and 1 is S(1) 2 S(0) 5 $31,500 2 $30,000 5 $1500 1 The average rate of change between years 1 and 2 is S(2) 2 S(1) 5 $33,075 2 $31,500 5 $1575 1 So, the average rates of change differ. This is to be expected, since in the first year the 5% raise applies only to the initial salary of $30,000, whereas in the second year the 5% raise applies to both the initial $30,000 salary and to the $1500 raise from the first year.
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Algebra Aerobics 5.5 3. Determine the growth or decay factor and the growth or decay rate in percentage form. a. ƒ(t) 5 100 ? 1.06t b. g(t) 5 25 ? 0.89t c. h(t) 5 4000 ? 1.23t d. r(t) 5 45.9 ? 0.956t e. P(t) 5 32,000 ? 1.092t 4. Find the percent increase (or decrease) if: a. Hourly wages grew from $1.65 per hour to $6.00 per hour in 30 years. b. A player’s batting average dropped from 0.299 to 0.167 in one season. c. The number of new AIDS cases grew from 17 last month to 23 this month.
A calculator that can evaluate powers is recommended for Problem 4. 1. Complete the following statements. a. A growth factor of 1.22 corresponds to a growth rate of ____. b. A growth rate of 6.7% corresponds to a growth factor of ____. c. A decay factor of 0.972 corresponds to a decay rate of ____. d. A decay rate of 12.3% corresponds to a decay factor of ____. 2. Fill in the table below.
Exponential Function
Initial Value
Growth or Decay?
Growth or Decay Factor
Growth or Decay Rate (as %)
A 5 4s1.03d t A 5 10s0.98d t 1000 30 $50,000 200 grams
1.005 0.96 Growth Decay
7.05% 49%
Exercises for Section 5.5 Access to the internet and a graphing program is required for Exercise 26 part (e). A graphing program is recommended for Exercises 16, 24, and 25. 1. Assume that you start with 1000 units of some quantity Q. Construct an exponential function that will describe the value of Q over time T if, for each unit increase in T, Q increases by: a. 300% b. 30% c. 3% d. 0.3% 2. Given each of the following exponential growth functions, identify the growth rate in percentage form: a. Q 5 10,000 ? 1.5T d. Q 5 10,000 ? 2T b. Q 5 10,000 ? 1.05T e. Q 5 10,000 ? 2.5T T c. Q 5 10,000 ? 1.005 f. Q 5 10,000 ? 3T 3. Given the following exponential decay functions, identify the decay rate in percentage form. a. Q 5 400s0.95d t d. y 5 200s0.655d x t b. A 5 600s0.82d e. A 5 10s0.996d T t c. P 5 70,000s0.45d f. N 5 82s0.725d T 4. What is the growth or decay factor for each given time period? a. Weight increases by 0.2% every 5 days. b. Mass decreases by 6.3% every year. c. Population increases 23% per decade. d. Profit increases 300% per year. e. Blood alcohol level decreases 35% per hour.
5. Fill in the following chart and then construct exponential functions for each part (a) to (g).
Initial Value a. b. c. d. e. f. g.
600 1200 6000 1.5 million 1.5 million 7 60
Growth or Decay?
Growth or Decay Factor
Growth or Decay Rate (% form)
2.06 200% 75% 25% 25%
Decay Decay Growth 4.35 0.35
6. Match the statements (a) through (d) with the correct exponential function in (e) through (h). Assume time t is measured in the unit indicated. a. Radon-222 decays by 50% every t days. b. Money in a savings account increases by 2.5% per year. c. The population increases by 25% per decade. d. The pollution in a stream decreases by 25% every year. e. A 5 1000s1.025d t f. A 5 1000s0.75d t g. A 5 1000 s 12 d t h. A 5 1000s1.25d t
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7. Each of two towns had a population of 12,000 in 1990. By 2000 the population of town A had increased by 12% while the population of town B had decreased by 12%. Assume these growth and decay rates continued. a. Write two exponential population models A(T) and B(T) for towns A and B, respectively, where T is the number of decades since 1990. b. Write two new exponential models a(t) and b(t) for towns A and B, where t is the number of years since 1990. c. Now find A(2), B(2), a(20), and b(20) and explain what you have found. 8. On November 25, 2003, National Public Radio did a report on Under Armour, a sports clothing company, stating that their “profits have increased by 1200% in the last 5 years.” a. Let P(t) represent the profit of the company during every 5-year period, with A0 the initial amount. Write the exponential model for the company’s profit. b. Assuming an initial profit of $100,000, what would be the profit in year 5? Year 10? c. Determine the annual growth rate for Under Armour. Each of the tables in Exercises 9–12 represents an exponential function. Construct that function and then identify the corresponding growth or decay rate in percentage form. 9.
10.
11.
12.
x y
0 500.00
1 425.00
2 361.25
3 307.06
x y
0 500.00
1 575.00
2 661.25
3 760.44
x y
x y
0 225.00
0 225.00
1 228.38
2 231.80
3 235.28
1 221.63
2 218.30
3 215.03
13. Generate equations that represent the pollution levels, P(t), as a function of time, t (in years), such that P(0) 5 150 and: a. P(t) triples each year. b. P(t) decreases by twelve units each year. c. P(t) decreases by 7% each year. d. The annual average rate of change of P(t) with respect to t is constant at 1. 14. Given an initial value of 50 units for parts (a)–(d) below, in each case construct a function that represents Q as a function of time t. Assume that when t increases by 1: a. Q(t) doubles b. Q(t) increases by 5% c. Q(t) increases by ten units d. Q(t) is multiplied by 2.5
15. Between 1970 and 2000, the United States grew from about 200 million to 280 million, an increase of approximately 40% in this 30-year period. If the population continues to expand by 40% every 30 years, what will the U.S. population be in the year 2030? In 2060? Use technology to estimate when the population of the United States will reach 1 billion. 16. (Graphing program recommended.) According to Mexico’s National Institute of Geography, Information and Statistics, in 2005 in Mexico’s Quintana Roo state, where the tourist industry in Cancun has created a boom economy, the population was about 1.14 million and growing at a rate of 5.3% per year. In 2005 in Mexico’s Baja California state, where many labor-intensive industries are located next to the California border, the population was about 2.8 million and growing at a rate of 3.3% per year. The nation’s capital, Mexico City, had 19.2 million inhabitants as of 2005 and the population was declining at a rate of 0.012% per year. a. Construct three exponential functions to model the growth or decay of Quintana Roo, Baja California, and Mexico City. Identify your variables and their units. b. Use your functions to predict the populations of Quintana Roo, Baja California, and Mexico City in 2010. 17. A pollutant was dumped into a lake, and each year its amount in the lake is reduced by 25%. a. Construct a general formula to describe the amount of pollutant after n years if the original amount is A0. b. How long will it take before the pollution is reduced to below 1% of its original level? Justify your answer. 18. A swimming pool is initially shocked with chlorine to bring the chlorine concentration to 3 ppm (parts per million). Chlorine dissipates in reaction to bacteria and sun at a rate of about 15% per day. Above a chlorine concentration of 2 ppm, swimmers experience burning eyes, and below a concentration of 1 ppm, bacteria and algae start to proliferate in the pool environment. a. Construct an exponential decay function that describes the chlorine concentration (in parts per million) over time. b. Construct a table of values that corresponds to monitoring chlorine concentration for at least a 2-week period. c. How many days will it take for the chlorine to reach a level tolerable for swimmers? How many days before bacteria and algae will start to grow and you will need to add more chlorine? Justify your answers. 19. a. If the inflation rate is 0.7% a month, what is it per year? b. If the inflation rate is 5% a year, what is it per month? 20. Rewrite each expression so that no fraction appears in the exponent and each expression is in the form a x. a. 3x/4
b. 2x/3
x c. Q12R /4
Hint: Remember that ax/n (a1/n)x
x d. Q14R /2
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21. a. Show that the following two functions are roughly equivalent. ƒsxd 5 15,000s1.2d x/10 and gsxd 5 15,000s1.0184d x b. Create a table of values for each function for 0 , x # 25. c. Would it be correct to say that a 20% increase over 10 years represents a 1.84% annual increase? Explain your answer. 22. The exponential function QsTd 5 600s1.35d T represents the growth of a species of fish in a lake, where T is measured in 5-year intervals. a. Determine Q(1), Q(2), and Q(3). b. Find another function q(t), where t is measured in years. c. Determine q(5), q(10), and q(15). d. Compare your answers in parts (a) and (c) and describe your results. 23. Blood alcohol content (BAC) is the amount of alcohol present in your blood as you drink. It is calculated by determining how many grams of alcohol are present in 100 milliliters (1 deciliter) of blood. So if a person has 0.08 grams of alcohol per 100 milliliters of blood, the BAC is 0.08 g/dl. After a person has stopped drinking, the BAC declines over time as his or her liver metabolizes the alcohol. Metabolism proceeds at a steady rate and is impossible to speed up. For instance, an average (150-lb) male metabolizes about 8 to 12 grams of alcohol an hour (the amount in one bottle of beer3). The behavioral effects of alcohol are closely related to the blood alcohol content. For example, if an average 150-lb male drank two bottles of beer within an hour, he would have a BAC level of 0.05 and could suffer from euphoria, inhibition, loss of motor coordination, and overfriendliness. The same male after drinking four bottles of beer in an hour would be legally drunk with a BAC of 0.10. He would likely suffer from impaired motor function and decision making, drowsiness, and slurred speech. After drinking twelve beers in one hour, he will have attained the dosage for stupor (0.30) and possibly death (0.40). The following table gives the BAC of an initially legally drunk person over time (assuming he doesn’t drink any additional alcohol). Time (hours) BAC (g/dl)
0 0.100
1 0.067
2 0.045
3 0.030
4 0.020
a. Graph the data from the table (be sure to carefully label the axes). b. Justify the use of an exponential function to model the data. Then construct the function where B(t) is the BAC for time t in hours. c. By what percentage does the BAC decrease every hour? d. What would be a reasonable domain for your function? What would be a reasonable range? e. Assuming the person drinks no more alcohol, when does the BAC reach 0.005 g/dl? One drink is equal to 114 oz of 80-proof liquor, 12 oz of beer, or 4 oz of table wine. 3
297
24. (Graphing program recommended.) If you have a heart attack and your heart stops beating, the amount of time it takes paramedics to restart your heart with a defibrillator is critical. According to a medical report on the evening news, each minute that passes decreases your chance of survival by 10%. From this wording it is not clear whether the decrease is linear or exponential. Assume that the survival rate is 100% if the defibrillator is used immediately. a. Construct and graph a linear function that describes your chances of survival. After how many minutes would your chance of survival be 50% or less? b. Construct and graph an exponential function that describes your chances of survival. Now after how many minutes would your chance of survival be 50% or less? 25. (Graphing program recommended.) The infant mortality rate (the number of deaths per 1000 live births) fell in the United States from 7.2 in 1996 to 6.4 in 2006. a. Assume that the infant mortality rate is declining linearly over time. Construct an equation modeling the relationship between infant mortality rate and time, where time is measured in years since 1996. Make sure you have clearly identified your variables. b. Assuming that the infant mortality rate is declining exponentially over time, construct an equation modeling the relationship, where time is measured in years since 1996. c. Graph both of your models on the same grid. d. What would each of your models predict for the infant mortality rate in 2010? 26. [Part (e) requires use of the Internet and technology to find a best-fit function.] A “rule of thumb” used by car dealers is that the trade-in value of a car decreases by 30% each year. a. Is this decline linear or exponential? b. Construct a function that would express the value of the car as a function of years owned. c. Suppose you purchase a car for $15,000. What would its value be after 2 years? d. Explain how many years it would take for the car in part (c) to be worth less than $1000. Explain how you arrived at your answer. e. Internet search: Go to the Internet site for the Kelley Blue Book (www.kbb.com). i. Enter the information about your current car or a car you would like to own. Specify the actual age and mileage of the car. What is the Blue Book value? ii. Keeping everything else the same, assume the car is 1 year older and increase the mileage by 10,000. What is the new value? iii. Find a best-fit exponential function to model the value of your car as a function of years owned. What is the annual decay rate? iv. According to this function, what will the value of your car be 5 years from now?
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5.6 Examples of Exponential Growth and Decay Exponential and linear behaviors are ones you will frequently (though perhaps unknowingly) encounter throughout life. Here we’ll take a look at some examples showing the wide range of applications of exponential functions. EXAMPLE
1
Fitting a curve to data: Medicare costs The costs for almost every aspect of health care in America have risen dramatically over the last 30 years. One of the central issues is the amount of dollars spent on Medicare, a federal program that provides health care for nearly all people age 65 and over. Estimate the average annual percentage increase in Medicare expenses.
SOLUTION
Table 5.10 shows Medicare expenses from 1970 to 2005 as reported by the federal government. In column 2, the years are reinitialized to become the years since 1970. Medicare Expenses
See Excel or graph link file MEDICARE.
Years since 1970
1970 1975 1980 1985 1990 1995 2000 2005
0 5 10 15 20 25 30 35
Medicare Expenses (billions of dollars) 7.7 16.3 37.1 71.5 109.5 184.4 224.3 342.0
Expenditures (in billion of $)
DATA
Year
Medicare Expenses (in billions of $) 500 450 400 350 300 250 200 150 100 50 0
5
10
15 20 25 30 Years since 1970
35
40
Table 5.10
Figure 5.14 Medicare expenses (in
Source: www.census.gov.
billions) with best-fit exponential model.
A curve-fitting program gives a best-fit exponential function as C(n) 5 10.6 ? (1.11) n where n 5 number of years since 1970 and C(n) is the associated cost in billions of dollars. The data and best-fit exponential curve are plotted in Figure 5.14. In our model the initial value (in year 0 or in 1970) is $10.6 billion. This differs from the actual value in 1970 of $7.7 billion, since the model is the best fit to all the data and does not necessarily include any of the original data points. More important, the growth factor is 1.11, so the growth rate is 0.11 in decimal form. So our model estimates that between 1970 and 2005 Medicare expenses were increasing by 11% each year, an amount that far exceeds the general inflation rate. EXAMPLE
2
Changing the time unit: Credit card debt In 2007 the average American household owed a balance of almost $9200 on credit cards.4 For late payments, credit card companies typically charge a high annual percentage rate (APR) that could be anywhere from 10% to 20%. a. Assuming you have a $500 balance on a credit card with a 20% APR, construct a model for your debt over time. b. What is your monthly interest rate? c. If you made no payments and no charges for 6 months, how much would you owe?
SOLUTION
a. Since the debt is increasing by a constant percent, the growth is exponential. Let T 5 number of years and D 5 credit card debt in dollars. An APR of 20% means that 4
www.cardweb.com
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the annual growth rate in decimal form is 0.20, and the annual growth factor is 1.20. So given an initial value of $500, an exponential model would be D 5 500 ? s1.20d T b. We know that the yearly growth factor is 1.20. If a stands for the monthly growth factor, then a12 5 1.20 To find the value for a we can take the twelfth root of both sides. or
(a12 ) 1/12 5 (1.20) 1/12 a < 1.015
Since the monthly growth factor is 1.015, the monthly growth rate is 0.015 in decimal form. So a yearly interest rate of 20% translates into a monthly interest rate of about 1.5%. c. To find your debt after 6 months, we need the monthly exponential growth function. If we let t 5 number of months and 1.015 is the monthly growth factor, then D < 500 ? s1.015d t Note that we have transformed debt (D) as a function of number of years (T) into a function of number of months (t). Assuming you made no payments, after 6 months you would owe D < 500 ? s1.015d 6 5 $547 So you would owe $47 in interest (almost 10% of the initial amount) for the “privilege” of borrowing $500 for 6 months.
Half-Life and Doubling Time Every exponential growth function has a fixed doubling time, and every exponential decay function has a fixed half-life. Doubling Time and Half-Life The doubling time of an exponentially growing quantity is the time required for the quantity to double in size. The half-life of an exponentially decaying quantity is the time required for one-half of the quantity to decay.
EXAMPLE
3
SOLUTION
Constructing an exponential function given the doubling time or half-life In each situation construct an exponential function. a. The doubling time is 3 years; the initial quantity is 1000. b. The half-life is 5 days; the initial quantity is 50. a. A doubling time of 3 years describes an exponentially growing quantity that doubles every 3 years. We can construct an exponential function, P(t) 5 1000at
Since the doubling time is 3 years, then a3 5 2 Taking the cube root of each side a 5 21/3
where a is the yearly growth factor and t 5 number of years.
(1)
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P(t) 5 1000 ? (21/3)t P(t) 5 1000 ? 2t/3
Substituting in (1) Rule 3 for exponents
b. A half-life of 5 days means an exponentially decaying quantity that decreases by half every 5 days. We can construct an exponential decay function, Q(t) 5 50at
where a is the daily decay factor and t 5 number of days.
(2)
Since the half-life is 5 days, then a5 5 12 Taking the fifth root of each side a 5 A12B 1/5 Substituting in (2) and Using Rule 3 for exponents Q(t) 5 50 ? A12B t/5 Writing an exponential function using the doubling time (or half-life) If an exponential growth function G(t) 5 Cat has a doubling time of n, then
so,
an 5 2 a 5 21/n G(t) 5 C ? 2t/n
If an exponential decay function D(t) 5 Cat has a half-life of n, then an 5
1 2
a 5 A12B 1/n so,
D(t) 5 C ? A12B t/n
For example, if G(t) 5 500at has a doubling time of 6 hours, then G(t) 5 500 ? 2t/6, where t 5 number of hours. If D(t) 5 25at has a half-life of 3 minutes, then D(t) 5 25 ? A12B t/3, where t 5 number of minutes.
EXAMPLE
4
Half-life: Radioactive decay One of the toxic radioactive by-products of nuclear fission is strontium-90. A nuclear accident, like the one in Chernobyl, can release clouds of gas containing strontium-90. The clouds deposit the strontium-90 onto vegetation eaten by cows, and humans ingest strontium-90 from the cows’ milk. The strontium-90 then replaces calcium in bones, causing cancer and birth defects. Strontium-90 is particularly insidious because it has a half-life of approximately 28 years. That means that every 28 years about half (or 50%) of the existing strontium-90 has decayed into nontoxic, stable zirconium-90, but the other half still remains. a. Construct a model for the decay of 100 mg of strontium-90 as a function of 28-year time periods. b. Construct a new model that describes the decay as a function of the number of years. Generate a corresponding table and graph.
SOLUTION
a. Strontium-90 decays by 50% every 28 years. If we define our basic time period T as 28 years, then the decay rate is 0.5 and the decay factor is 1 2 0.5 5 0.5 (or 1/2). Assuming an initial value of 100 mg, the function S 5 100 ? s0.5d T gives S, the remaining milligrams of strontium-90 after T time periods (of 28 years).
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b. The number 0.5 is the decay factor over a 28-year period. If we let a 5 annual decay factor, then a28 5 0.5 To find the value for a we can take the 28th root of both sides. (a28 ) 1/28 5 (0.5) 1/28 a < 0.976
or
So if t 5 number of years, the yearly decay factor is 0.976, making our function S < 100 ? s0.976d t We now have an equation that describes the amount of strontium-90 as a function of years, t. The decay factor is 0.976, so the decay rate is 1 2 0.976 5 0.024, or 2.4%. So each year there is 2.4% less strontium-90 than the year before. 120
Time T (28-year time periods)
Time t (years)
Strontium-90 (mg)
0 1 2 3 4 5
0 28 56 84 112 140
100 50 25 12.5 6.25 3.125
Table 5.11
Milligrams of strontium -90
Decay of Strontium-90
100 80 60 40 20
t 0
50
100
150
Years
Figure 5.15 Radioactive decay of
strontium-90.
Table 5.11 and Figure 5.15 show the decay of 100 mg of strontium-90 over time. After 28 years, half (50 mg) of the original 100 mg still remains. It takes an additional 28 years for the remaining amount to halve again, still leaving 25 grams out of the original 100 after 56 years. EXAMPLE
5
Identifying the doubling time or half-life Assuming t is in years, identify the doubling time or half-life for each of the following exponential functions. a. Qstd 5 80s2d t/12 b. P(t) 5 5 ? 106 A12B 4t c. Rstd 5 130s0.5d 3t
SOLUTION
a. When t 5 12 years, the quantity Qs12d 5 80 ? 212/12 5 80 ? 21 5 160 . So the quantity has doubled from the initial value of 80 and the doubling time is 12 years. b. When t 5 1/4, the quantity Ps1/4d 5 s5 ? 106 ds1/2d 4s1/4d 5 s5 ? 106 ds1/2d 1. So the quantity is half the initial value of 5 ? 106 and the half-life is 1/4 years or 3 months. c. When t 5 1/3, the quantity Rstd 5 130s0.5d 3s1/3d 5 130s0.5d 1 5 65. So the quantity is half the initial value of 130 and the half-life is 1/3 year or 4 months.
The “rule of 70”: A rule of thumb for calculating doubling or halving times A simple way to understand the significance of constant percent growth rates is to compute the doubling time. A rule of thumb is that a quantity growing at R% per year
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has a doubling time of approximately 70/R years. If the quantity is growing at R% per month, then 70/R gives its doubling time in months. The same reasoning holds for any unit of time. The rule of 70 is easy to apply. For example, if a quantity is growing at a rate of 2% a year, then the doubling time is about 70/2 or approximately 35 years. The rule of 70 also applies when R represents the percentage at which some quantity is decaying. In these cases, 70/R equals the half-life. For now we’ll take the rule of 70 on faith. It provides good approximations, especially for smaller values of R (those under 10%). When we return to logarithms in Chapter 6, we’ll find out why this rule works.
SOMETHING TO THINK ABOUT
Using a graphing calculator or a spreadsheet, test the rule of 70. Can you find a value for R for which this rule does not work so well?
EXAMPLE
6
Suppose that at age 23 you invest $1000 in a retirement account that grows at 5% per year. a. Roughly how long will it take your investment to double? b. If you retire at age 65, approximately how much will you have in your account?
SOLUTION
a. Using the rule of 70, we get 70/5 5 14, so your investment will double approximately every 14 years. b. If you retire at 65, then 42 years or three doubling periods will have elapsed s3 ? 14 5 42d . So your $1000 investment (disregarding inflation) will have increased by a factor of 23 and be worth about $8000.
EXAMPLE
7
According to Mexico’s National Institute of Geography, Information and Statistics (at www.inegi.gob.mx/difusion/ingles), in 1997 Mexico’s population reached 93.7 million. The annual growth rate is listed as approximately 1.4%. The website states that “if this rate persists, the Mexican population will double in 49.9 years.” Does that time period seem about right?
SOLUTION
Using the rule of 70, the approximate doubling time for a 1.4% annual growth rate would be 70/1.4 5 50 years. So 49.9 is a reasonable value for the doubling time.
The Rule of 70 The rule of 70 states that if a quantity is growing (or decaying) at R% per time period, then the doubling time (or half-life) is approximately 70 R provided R is not much bigger than 10. Examples: If a quantity is growing at 7% per year, the doubling time is approximately 70/7 5 10 years. If a quantity is decaying at 5% per minute, the half-life is approximately 70/5 5 14 minutes.
EXAMPLE
8
Plutonium, the fuel for atomic weapons, has an extraordinarily long half-life, about 24,400 years. Once the radioactive element plutonium is created from uranium, 24,400 years later half the original amount will still remain. You can see why there is concern over stored caches of atomic weapons. Use the rule of 70 to estimate plutonium’s annual decay rate.
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SOLUTION
?
The rule of 70 says Multiply by R Divide and simplify
SOMETHING TO THINK ABOUT
An atomic weapon is usually designed with a 1% mass margin. That is, it will remain functional until the original fuel has decayed by more than 1%, leaving less than 99% of the original amount. Estimate how many years a plutonium bomb would remain functional.
303
70/R 5 24,400 70 5 24,400R R 5 70/24,400 < 0.003
Remember that R is already in percentage, not decimal, form. So the annual decay rate is a tiny 0.003%, or three-thousandths of 1%.
Algebra Aerobics 5.6a A calculator that can evaluate powers is recommended for Problem 2. 1. For each of the following exponential functions (with t in days) determine whether the function represents exponential growth or decay, then estimate the doubling time or half-life. For each function find the initial amount and the amount after one day, one month, and one year. (Note: To make calculations easier, use 360 days in one year.) a. ƒstd 5 300 ? 2t/30 c. P 5 32,000 ? 0.5t b. gstd 5 32 ? 0.5t/2 d. hstd 5 40,000 ? 2t/360 2. Estimate the doubling time using the rule of 70. a. gsxd 5 100s1.02d x, where x is in years.
DATA
CELCOUNT.
9
White Blood Cell Counts On September 27 a patient was admitted to Brigham and Women’s Hospital in Boston for a bone marrow transplant. The transplant was needed to cure myelodysplastic syndrome, in which the patient’s own marrow fails to produce enough white blood cells to fight infection. The patient’s bone marrow was intentionally destroyed using chemotherapy and radiation, and on October 3 the donated marrow was injected. Each day, the hospital carefully monitored the patient’s white blood cell count to detect when the new marrow became active. The patient’s counts are plotted in Figure 5.16. Normal counts for a healthy individual are between 4000 and 10,000 cells per milliliter.
12,000 10,000 Cells per milliliter
EXAMPLE
b. M 5 10,000s1.005d t, where t is in months. c. The annual growth rate is 8.1%. d. The annual growth factor is 1.065. 3. Use the rule of 70 to approximate the growth rate when the doubling time is: a. 10 years b. 5 minutes c. 25 seconds 4. Estimate the time it will take an initial quantity to drop to half its value when: a. hsxd 5 10s0.95d x, where x is in months. b. K 5 1000s0.75d t, where t is in seconds. c. The annual decay rate is 35%.
8000 6000 4000 2000 0 27 Sept
7 Oct
17 Oct Date
Figure 5.16 White blood cell count.
27 Oct
6 Nov
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What was the daily percent increase and estimated doubling time during the period of exponential growth? SOLUTION
Clearly, the whole data set does not represent exponential growth, but we can reasonably model the data between, say, October 15 and 31 with an exponential function. Figure 5.17 shows a plot of this subset of the original data, along with a computer-generated best-fit exponential function, where t number of days after October 15. Wstd 5 105s1.32d t The fit looks quite good. 9000
White blood cells per milliliter
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6000
3000
0
4 8 12 Days since October 15
16
Figure 5.17 White blood cell counts with
a best-fit exponential function between October 15 and October 31.
The initial quantity of 105 is the white blood cell count (per milliliter) for the model, not the actual white blood cell count of 70 measured by the hospital. This discrepancy is not unusual; remember that a best-fit function may not necessarily pass through any of the specific data points. The growth factor is 1.32, so the growth rate is 0.32. So between October 15 and 31, the number of white blood cells was increasing at a rate of 32% a day. According to the rule of 70, the number of white blood cells was doubling roughly every 70/32 5 2.2 days!
Compound Interest Rates Exponential functions occur frequently in finance. Probably the most common application is compound interest. For example, suppose you have $100 in a passbook savings account that returns 1% compounded annually. That means that at the end of the first year, you earn 1% in interest on the initial $100, called the principal. From then on you earn 1% not only on your principal, but also on the interest that you have already earned. This is called compounding and represents exponential growth. If the growth rate is 1%, or 0.01 in decimal form, then the growth factor is 1 1 0.01 5 1.01. So each subsequent year, the current value of your savings account will be multiplied by 1.01. The following function models the growth: P1
5 5
Value of account 5
100 100
? s1.01d t ? s1 1 0.01d t
original a investment b ? (1 1 interest rate)no. of years
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Calculating Compound Interest If P0 5 original investment r 5 interest rate (in decimal form) t 5 time periods at which the interest rate is compounded the resulting value, Pr, of the investment after t time periods is given by the formula Pr 5 P0 ? s1 1 rd t EXAMPLE
?
10
Comparing investments Suppose you have $10,000 that you could put in either a checking account that earns no interest, a mutual fund that earns 5% a year, or a risky stock investment that you hope will return 15% a year. a. Model the potential growth of each investment. b. Compare the investments after 10, 20, 30, and 40 years.
SOLUTION
a. Assuming you make no withdrawals, the money in the checking account will remain constant at $10,000. For both the mutual fund and the stock investment you are expecting a constant annual percent rate, so you can think of them as compounding annually. The mutual fund’s predicted annual growth rate is 5%, so its growth factor would be 1.05. The stock’s annual growth rate (you hope) will be 15%, so its growth factor would be 1.15. The initial value is $10,000, so after t years
SOMETHING TO THINK ABOUT
On March 13, 1997, a resident of Melrose, Massachusetts, “scratched” a $5 state lottery ticket and won $1,000,000. The Massachusetts State Lottery Commission notified her that after deducting taxes, they would send her a check for $33,500 every March 13 for 20 years. Is this fair?
5 10,000 ? s1.05d t value of mutual fund value of stock investment 5 10,000 ? s1.15d t b. Table 5.12 shows the value of each investment over time. Number of Years
Checking Account (0% per year)
Mutual Fund (5% per year)
Stock (15% per year)
0 10 20 30 40
$10,000 $10,000 $10,000 $10,000 $10,000
$10,000 $16,289 $26,533 $43,219 $70,400
$10,000 $40,456 $163,665 $662,118 $2,678,635
Table 5.12
Forty years from now, your checking account will still have $10,000, but your mutual fund will be worth $70,400. Should you have been lucky enough to have invested in the next Microsoft, 40 years from now, with a 15% interest rate compounded annually, your $10,000 would have become well over $2.5 million.
?
SOMETHING TO THINK ABOUT
What happens if the inflation rate is higher than the interest rate on your investments?
EXAMPLE
11
Inflation and the diminishing dollar Compound calculations are the same whether you are dealing with inflation or investments. For example, a 5% annual inflation rate would mean that what cost $1 today would cost $1.05 one year from today. If we think of the percentages in Table 5.14 as representing inflation rates, then something that costs $10,000 today in 10 years will cost $10,000, $16,289, or $40,456 if the annual inflation rate is, respectively, 0%, 5%, or 15%. How much will your money be worth in 20 years? Suppose you have a $100,000 nest egg hidden under your bed, awaiting your retirement. In 20 years, how much will your nest egg be worth in today’s dollars if the annual inflation rate is: a. 3%? b. 7%?
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SOLUTION
a. Given a 3% inflation rate, what costs $1 today will cost $1.03 next year. Using a ratio to compare costs, we get cost now $1.00 5 < 0.971 cost in 1 year $1.03 cost now < 0.971 ? (cost in 1 year) Rewriting Our equation shows that next year’s dollars need to be multiplied by 0.971 to reduce them to the equivalent value in today’s dollars. An annual inflation rate of 3% means that in one year, $1 will be worth only 0.971 ? ($1) 5 $0.971, or about 97 cents in today’s dollars. Each additional year reduces the value of your nest egg by a factor of 0.971; in other words, 0.971 is the decay factor. So the value of your nest egg in real dollars is given by V3% snd 5 $100,000 ? 0.971n where V3% snd is the real value (measured in today’s dollars) of your nest egg after n years with 3% annual inflation. When n 5 20, we have V3% s20d 5 $100,000 ? 0.97120 < $100,000 ? 0.555 5 $55,500 b. Similarly, since $1.00/$1.07 < 0.935, then V7% snd 5 $100,000 ? 0.935n where V7% snd is the real value (measured in today’s dollars) of your nest egg after n years with 7% annual inflation. When n 5 20, we have V7% s20d 5 $100,000 ? 0.93520 < $100,000 ? 0.261 5 $26,100 So assuming 3% annual inflation, after 20 years your nest egg of $100,000 in real dollars would be almost cut in half. With a 7% inflation your nest egg in real dollars would only be about one-quarter of the original amount.
Real or Constant Dollars To make meaningful comparisons among dollar values in different years, economists use real or constant dollars, dollars that are adjusted for inflation. For example, the U.S. Census Bureau reported that in 2004 the median income of households was $44,389 and in 1998 only $38,885 in 1998 dollars. This suggests an increase of about $5500. But measured in 2004 dollars, the median income in 1998 was $45,003. So in real (inflation adjusted) dollars, there was a decrease in median household income.
Algebra Aerobics 5.6b A calculator that can evaluate powers is recommended for Problem 4. 1. Approximately how long would it take for your money to double if the interest rate, compounded annually, were: a. 3%? b. 5%? c. 7%? 2. Suppose you are planning to invest a sum of money. Estimate the rate that you need so that your investment doubles in: a. 5 years b. 10 years c. 7 years 5
3. Construct a function that represents the resulting value if you invested $1000 for n years at an annually compounded interest rate of: a. 4% b. 11% c. 110% 4. In the early 1980s Brazil’s inflation was running rampant at about 10% per month. 5 Assuming this inflation rate continued unchecked, construct a function to describe the purchasing power of 100 cruzeiros after n months. (A cruzeiro is a Brazilian monetary unit.) What would 100 cruzeiros be worth after 3 months? After 6 months? After a year?
In such cases sooner or later the government usually intervenes. In 1985 the Brazilian government imposed an anti-inflationary wage and price freeze. When the controls were dropped, inflation soared again, reaching a high in March 1990 of 80% per month! At this level, it begins to matter whether you buy groceries in the morning or wait until that night. On August 2, 1993, the government devalued the currency by defining a new monetary unit, the cruzeiro real, equal to 1000 of the old cruzeiros. Inflation still continued, and on July 1, 1994, yet another unit, the real, was defined equal to 2740 cruzeiros reales. By 1997, inflation had slowed considerably to about 0.1% per month.
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5. For each of the following equations, determine the initial investment, the growth factor, and the growth rate, and estimate the time it will take to double the investment. Function
Initial Investment
Growth Factor
$50,000 $100,000
307
a. A 5 $10,000 ? 1.065t c. A 5 $300,000 ? 1.11t b. A 5 $25 ? 1.08t d. A 5 $200 ? 1.092t 6. Fill in Table 5.13. Growth Rate
Amount 1 Year Later
Doubling Time
7.2% 5.8%
$3,000
$106,700 $52,500
1.13
Table 5.13
EXAMPLE
12
Musical pitch There is an exponential relationship between musical octaves and vibration frequency. The vibration frequency of the note A above middle C is 440 cycles per second (or 440 hertz). The vibration frequency doubles at each octave. a. Construct a function that gives the vibration frequency as a function of octaves. Construct a corresponding table and graph. b. Rewrite the function as a function of individual note frequencies. (Note: There are twelve notes to an octave.)
SOLUTION
a. Let F be the vibration frequency in hertz (Hz) and N the number of octaves above or below the chosen note A. Since the frequency doubles over each octave, the growth factor is 2. If we set the initial frequency at 440 Hz, then we have F 5 440 ? 2N Table 5.14 and the graph in Figure 5.18 show a few of the values for the vibration frequency. Musical Octaves Vibration Frequency (Hz), F
23 22 21 0 1 2 3
55 110 220 440 880 1760 3520
Frequency (Hz)
“E6: Musical Keyboard Frequencies” in Exponential & Log Functions offers a multimedia demonstration of this function.
4000
Number of Octaves above or below A (at 440 Hz), N
3000
2000
1000
0 –3 –2 –1 1 2 3 Octaves above or below A (at 440 Hz)
Figure 5.18 Octaves versus
Table 5.14
frequency.
b. There are twelve notes in each octave. If we let n 5 number of notes above or below A, then we have n 5 12N. Solving for N, we have N 5 n/12. By substituting this expression for N, we can define F as a function of n. F 5 440 5 440 5 440 < 440
? ? ? ?
2n/12 2(1/12)?n (21/12 ) n (1.059) n
So each note on the “even-tempered” scale has a frequency 1.059 times the frequency of the preceding note.
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The Malthusian Dilemma The most famous attempt to predict growth mathematically was made by a British economist and clergyman, Thomas Robert Malthus, in an essay published in 1798. He argued that the growth of the human population would overtake the growth of food supplies, because the population size was multiplied by a fixed amount each year, whereas food production only increased by adding a fixed amount each year. In other words, he assumed populations grew exponentially and food supplies grew linearly as functions of time. He concluded that humans are condemned always to breed to the point of misery and starvation, unless the population is reduced by other means, such as war or disease. EXAMPLE
13
Constructing the Malthusian equations Malthus believed that the population of Great Britain, then about 7,000,000, was growing by 2.8% per year. He counted food supply in units that he defined to be enough food for one person for a year. At that time the food supply was adequate, so he assumed that Britons were producing 7,000,000 food units. He predicted that food production would increase by about 280,000 units a year. a. Construct functions modeling Britain’s population size and the amount of food units over time. b. Plot the functions from part (a) on the same grid. Estimate when the population would exceed the food supply.
SOLUTION
a. Since the population is assumed to increase by a constant percent each year, an exponential model is appropriate. Given an initial size of 7,000,000 and an annual growth rate of 2.8%, the exponential function Pstd 5 7,000,000 ? s1.028d t describes the population growth over time, where t is the number of years and P(t) is the corresponding population size. Since the food units are assumed to grow by a constant amount each year, a linear model is appropriate. Given an initial value of 7,000,000 food units and an annual constant rate of change of 280,000, the linear function Fstd 5 7,000,000 1 280,000t describes the food unit increase over time, where t 5 number of years and F(t) 5 number of food units in year t . Figure 5.19 reveals that if the formulas were good models, then after about 25 years the population would start to exceed the food supply, and some people would starve. 30.00
25.00 Population or food units (in millions)
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Population 20.00 Food units 15.00
10.00
5.00
0
10
20
30
40
Years
Figure 5.19 Malthus’s predictions for
population and food.
50
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The two centuries since Malthus published his famous essay have not been kind to his theory. The population of Great Britain in 2002 was about 58 million, whereas Malthus’s model predicted over 100 million people before the year 1900. Improved food production techniques and the opening of new lands to agriculture have kept food production in general growing faster than the population. The distribution of food is a problem and famines still occur with unfortunate regularity in parts of the world, but the mass starvation Malthus predicted has not come to pass.
Forming a Fractal Tree Tree structures offer another useful way of visualizing exponential growth. A computer program can generate a tree by drawing two branches at the end of a trunk, then two smaller branches at the ends of each of those branches, and two smaller branches at the ends of the previous branches, and so on until a branch reaches twig size. This kind of structure produced from self-similar repeating scaled graphic operations is called a fractal (Figure 5.20). There are many examples of fractal structures in nature, such as ferns, coastlines, and human lungs.
Figure 5.20 A fractal tree.
The fractal tree shown is drawn in successive levels (Figure 5.21): • At level 0 the program draws the trunk, one line. • At level 1 it draws two branches on the previous one, for a total of two new lines. • At level 2 it draws two branches on each of the previous two, for a total of four new lines. • At level 3 it draws two branches on each of the previous four, for a total of eight new lines.
Level 0
Level 1
Figure 5.21 Forming a fractal tree.
Level 2
Level 3
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Table 5.15 shows the relationship between the level, L, and the number of new lines, N, at each level. The formula N 5 2L describes the relationship between level, L, and the number of new lines, N. For example, at the fifth level, there would be 25 5 32 new lines.
SOMETHING TO THINK ABOUT
Level, L
If you look back at your family tree forty generations ago (roughly 800 years), you had 240 ancestors. This number is larger than all the people that ever lived on the surface of the earth. How can that be?
0 1 2 3 4
?
New Lines, N 2 2 2 2
? ? ? ?
1 2 4 8
5 5 5 5
1 2 4 8 16
N as Power of 2 1 5 20 2 5 21 4 5 2 ? 2 5 22 8 5 2 ? 2 ? 2 5 23 16 5 2 ? 2 ? 2 ? 2 5 24
Table 5.15
EXAMPLE
14
You can think of Figure 5.20 as depicting your family tree. Each level represents a generation. The trunk is you (at level 0). The first two branches are your parents (at level 1). The next four branches are your grandparents (at level 2), and so on. How many ancestors do you have ten generations back?
SOLUTION
The answer is 210 5 1024; that is, you have 1024 great-great-great-great-great-greatgreat-great-grandparents.
EXAMPLE
15
An information system that is similar to this process is an emergency phone tree, in which one person calls two others, each of whom calls two others, until everyone in the organization has been called. How many levels of phone calls would be needed to reach an organization with 8000 people?
SOLUTION
If we think of Figure 5.20 and Table 5.15 as representing this phone tree, then each new line (or branch) represents a person. We need to count not just the number of new people N at each level L, but also all the previous people called. At level 0, there is the one person who originates the phone calls. At level 1, there is the original person plus the two he or she called, for a total of 1 1 2 5 3 people. At level 2, there are 1 1 2 1 4 5 7 people, etc. At level 11, there are 1 1 2 1 4 1 8 1 16 1 32 1 64 1 128 1 256 1 512 1 1024 1 2048 5 4095 people who have been called. At level 12, there would be 212 5 4096 new people called, for a total of 4095 1 4096 5 8191 people called. So it would take twelve levels of the phone tree to reach 8000 people.
Algebra Aerobics 5.6c 1. Identify the value of x that would make each of the following equations a true statement. a. 2x 5 32 d. 2x 5 2 g. 2x 5 18 b. 2x 5 256
e. 2x 5 1
h. 2x 5 "2
c. 2x 5 1024 f. 2x 5 12 2. Assume the tree-drawing process was changed to draw three branches at each level. a. Draw a trunk and at least two levels of the tree.
b. What would the general formula be for N, the number of new lines, as a function of L, the level? 3. In an emergency phone tree in which one person calls three others, each of whom calls three others, and so on, until everyone in the organization has been called, how many levels of phone calls are required for this phone tree to reach an organization of 8000 people?
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Exercises for Section 5.6 Technology for finding a best-fit function is required for Exercises 7, 9, 17, and 18. Internet access is required for Exercises 9 part (c), 29 part (c), and 32 part (e). 1. Which of the following functions have a fixed doubling time? A fixed half-life? a. y 5 6(2) x
c. Q 5 300 A12B
b. y 5 5 1 2x
d. A 5 10s2d t/5
T
2. Identify the doubling time or half-life of each of the following exponential functions. Assume t is in years. [Hint: What value of t would give you a growth (or decay) factor of 2 (or 1/2)?]
b. Q 5 1000s2d t/50
e. N 5 550s2d t/10
c. Q 5 300 A2B
f. N 5 50 A12B
t/250
t/20
3. Fill in the following chart. (The first column is done for you.)
a.
b.
c.
d.
50
1000
4
5000
30 days
7 years
25 minutes
18 months
Initial Value Doubling time
Exponential ƒ(t) 5 50(2) t/30 function sƒstd 5 Ca t/n d Growth factor per unit of time
Population of the United States, 1790–2000
2.34% per day
4. Make a table of values for the dotted points on each graph. Using your table, determine if each graph has a fixed doubling time or a fixed half-life. If so, create a function formula for that graph. y
2500 2000 1500 1000 500
320 240 160 80 0
100 80 60 40 20
0
5
10 15 Graph A
20
x
25
y
0
30 45 Graph B
60
x
75
y
Year
Millions
Year
Millions
1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890
3.9 5.3 7.2 9.6 12.9 17.1 23.2 31.4 39.8 50.2 63.0
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
76.2 92.2 106.0 123.2 132.2 151.3 179.3 203.3 226.5 248.7 281.4
Source: U.S. Bureau of the Census, www.census.gov. 300 250
0
2000 1600 1200 800 400 15
Radioactive uranium-238 decays sequentially into thirteen other lighter elements until it stabilizes at lead-206. The half-lives of the fifteen different elements in this decay chain vary from 0.000 164 seconds (from polonium-214 to lead-210) all the way up to 4.47 billion years (from uranium-238 to thorium-234). a. Find the decay rate per billion years for uranium-238 to decay into thorium-234. b. Find the decay rate per second for polonium-214 to decay into lead-210. 7. (Requires technology to find a best-fit function.) We have seen the accompanying table and graph of the U.S. population at the beginning of Chapter 2.
21/30 5 1.0234 per day
Growth rate per unit of time (in percentage form)
6. Lead-206 is not radioactive, so it does not spontaneously decay into lighter elements. Radioactive elements heavier than lead undergo a series of decays, each time changing from a heavier element into a lighter or more stable one. Eventually, the element decays into lead-206 and the process stops. So, over billions of years, the amount of lead in the universe has increased because of the decay of numerous radioactive elements produced by supernova explosions.
10
20 30 Graph C
40
x
50
y
0
Population in millions
1 t
b. 50 A12B x/20 _____ 50(0.9659)x d. 750 A12B x/165 _____ 750(0.911)x
f. N 5 50 A12B t/20
d. Q 5 100 A12B
a. 3(2)x/5 _____ 3s1.225d x
c. 200(2)x/8 _____ 200s1.0905d x
e. P 5 500 2 12 T
a. Q 5 70s2d t
5. Insert the symbol . , , , or < (approximately equal) to make the statement true. Assume x . 0.
200 150 100 50
50
x
100 150 200 250 Graph D
0 1780
1820
1860
1900 Year
1940
1980
DATA
USPOP
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a. Find a best-fit exponential function. (You may want to set 1790 as year 0.) Be sure to clearly identify the variables and their units for time and population. What is the annual growth factor? The growth rate? The estimated initial population? b. Graph your function and the actual U.S. population data on the same grid. Describe how the estimated population size differs from the actual population size. In what ways is this exponential function a good model for the data? In what ways is it flawed? c. What would your model predict the population to be in the year 2010? In 2025? 8. (Requires results from Exercise 7.) According to a letter published in the Ann Landers column in the Boston Globe on Friday, December 10, 1999, “When Elvis Presley died in 1977, there were 48 professional Elvis impersonators. In 1996, there were 7,328. If this rate of growth continues, by the year 2012, one person in every four will be an Elvis impersonator.” a. What was the growth factor in the number of Elvis impersonators for the 19 years between 1977 and 1996? b. What would be the annual growth factor in number of Elvis impersonators between 1977 and 1996? c. Construct an exponential function that describes the growth in number of Elvis impersonators since 1977. d. Use your function to estimate the number of Elvis impersonators in 2012. e. Use your model for the U.S. population from Exercise 7 to determine if in 2012 one person out of every four will be an Elvis impersonator. Explain your reasoning. 9. (Requires technology to find a best-fit function.) Reliable data on Internet use are hard to find, but World Telecommunications Indicators cites estimates of 3 million U.S. users in 1991, 30 million in 1996, 166 million in 2002, 199 million in 2004 and 232 million in 2007. a. Use technology to plot the data, and generate a best-fit linear and a best-fit exponential function for the data. Which do you think is the better model? b. What would the linear model predict for Internet usage in 2010? What would the exponential model predict? c. Internet use: Go online and see if you can find the number of current internet users in the U.S. Which of your models turned out to be more accurate? 10. China is the most populous country in the world. In 2000 it had about 1.262 billion people. By 2005 the population had grown to 1.306 billion. Use this information to construct models predicting the size of China’s population in the future. a. Identify your variables and units. b. Construct a linear model. c. Construct an exponential model. d. What will China’s population be in 2050 according to each of your models? 11. Tritium, the heaviest form of hydrogen, is a critical element in a hydrogen bomb. It decays exponentially with a half-life of about 12.3 years. Any nation wishing to maintain a viable hydrogen bomb has to replenish its tritium supply roughly every 3 years, so world tritium supplies are closely watched.
Construct an exponential function that shows the remaining amount of tritium as a function of time as 100 grams of tritium decays (about the amount needed for an average size bomb). Be sure to identify the units for your variables. 12. (Graphing program recommended.) Cosmic ray bombardment of the atmosphere produces neutrons, which in turn react with nitrogen to produce radioactive carbon-14. Radioactive carbon-14 enters all living tissue through carbon dioxide (via plants). As long as a plant or animal is alive, carbon-14 is maintained in the organism at a constant level. Once the organism dies, however, carbon-14 decays exponentially into carbon-12. By comparing the amount of carbon-14 to the amount of carbon-12, one can determine approximately how long ago the organism died. Willard Libby won a Nobel Prize for developing this technique for use in dating archaeological specimens. The half-life of carbon-14 is about 5730 years. In answering the following questions, assume that the initial quantity of carbon-14 is 500 milligrams. a. Construct an exponential function that describes the relationship between A, the amount of carbon-14 in milligrams, and t, the number of 5730-year time periods. b. Generate a table of values and plot the function. Choose a reasonable set of values for the domain. Remember that the objects we are dating may be up to 50,000 years old. c. From your graph or table, estimate how many milligrams are left after 15,000 years and after 45,000 years. d. Now construct an exponential function that describes the relationship between A and T, where T is measured in years. What is the annual decay factor? The annual decay rate? e. Use your function in part (d) to calculate the number of milligrams that would be left after 15,000 years and after 45,000 years. 13. The body eliminates drugs by metabolism and excretion. To predict how frequently a patient should receive a drug dosage, the physician must determine how long the drug will remain in the body. This is usually done by measuring the half-life of the drug, the time required for the total amount of drug in the body to diminish by one-half. a. Most drugs are considered eliminated from the body after five half-lives, because the amount remaining is probably too low to cause any beneficial or harmful effects. After five half-lives, what percentage of the original dose is left in the body? b. The accompanying graph shows a drug’s concentration in the body over time, starting with 100 milligrams. Remaining Drug Dose in Milligrams 120 Drug (in milligrams)
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100 80 60 40 20 0
2
4 6 8 Time (hours)
10
12
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Use the given graph to answer the following questions. i. Estimate the half-life of the drug. ii. Construct an equation that approximates the curve. Specify the units of your variables. iii. How long would it take for five half-lives to occur? Approximately how many milligrams of the original dose would be left then? iv. Write a paragraph describing your results to a prospective buyer of the drug. 14. Estimate the doubling time using the rule of 70 when: a. P 5 2.1s1.0475d t, where t is in years b. Q 5 2.1s1.00475d T , where T is in years 15. Use the rule of 70 to approximate the growth rate when the doubling time is: a. 5730 years c. 5 seconds b. 11,460 years d. 10 seconds 16. Estimate the time it will take an initial quantity to drop to half its value when: a. P 5 3.02s0.998d t, with t in years b. Q 5 12s0.75d T , with T in decades 17. (Requires technology to find a best-fit function.) Estimates for world population vary, but the data in the accompanying table are reasonable estimates.
DATA
WORLDPOP
World Population Year
Total Population (millions)
1800 1850 1900 1950 1970 1980 1990 2000 2005
980 1260 1650 2520 3700 4440 5270 6080 6480
Source: United Nations Population Division, www.undp.org/popin.
a. Enter the data table into a graphing program (you may wish to enter 1800 as 0, 1850 as 50, etc.) or use the data file WORLDPOP in Excel or in graph link form. b. Generate a best-fit exponential function. c. Interpret each term in the function, and specify the domain and range of the function. d. What does your model give for the growth rate? e. Using the graph of your function, estimate the following: i. The world population in 1750, 1920, 2025, and 2050 ii. The approximate number of years in which world population attained or will attain 1 billion (i.e., 1000 million), 4 billion, and 8 billion f. Estimate the length of time your model predicts it takes for the population to double from 4 billion to 8 billion people.
313
18. (Requires technology to find a best-fit function.) In DATA 1911, reindeer were introduced to St. Paul Island, one of the Pribilof Islands, off the coast of Alaska in the Bering Sea. There was plenty of food and no REINDEER hunting or reindeer predators. The size of the reindeer herd grew rapidly for a number of years, as given in the accompanying table. Population of Reindeer Herd Year
Population Size
Year
Population Size
1911 1912 1913 1914 1915 1916 1917 1918 1919 1920 1921 1922 1923 1924
17 20 42 76 93 110 136 153 170 203 280 229 161 212
1925 1926 1927 1928 1929 1930 1931 1932 1933 1934 1935 1936 1937 1938
246 254 254 314 339 415 466 525 670 831 1186 1415 1737 2034
Source: V.B. Scheffer, “The rise and fall of a reindeer herd,” Scientific Monthly, 73:356–362, 1951.
a. Use the reindeer data file (in Excel or graph link form) to plot the data. b. Find a best-fit exponential function. c. How does the predicted population from part (b) differ from the observed ones? d. Does your answer in part (c) give you any insights into why the model does not fit the observed data perfectly? e. Estimate the doubling time of this population. 19. In medicine and biological research, radioactive substances are often used for treatment and tests. In the laboratories of a large East Coast university and medical center, any waste containing radioactive material with a half-life under 65 days must be stored for 10 half-lives before it can be disposed of with the non-radioactive trash. a. By how much does this policy reduce the radioactivity of the waste? b. Fill out the accompanying chart and develop a general formula for the amount of radioactive pollution at any period, given an initial amount, A0. Number of Half-Life Periods 0 1 2 3 4 Period n
Pollution Amount A0, original amount A1 5 0.5A0 A2
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20. Belgrade, Yugoslavia (from USA Today, September 21, 1993): A 10 billion dinar note hit the streets today. . . . With inflation at 20% per day, the note will soon be as worthless as the 1 billion dinar note issued last month. A year ago the biggest note was 5,000 dinars. . . . In addition to soaring inflation, which doubles prices every 5 days, unemployment is at 50%. In the excerpt, inflation is described in two very different ways. Identify these two descriptions in the text and determine whether they are equivalent. Justify your answer. 21. It takes 3 months for a malignant lung tumor to double in size. At the time a lung tumor was detected in a patient, its mass was 10 grams. a. If untreated, determine the size in grams of the tumor at each of the listed times in the table. Find a formula to express the tumor mass M (in grams) at any time t (in months). t, Time (months)
M, Mass (g)
0 3 6 9 12
10
b. Lung cancer is fatal when a tumor reaches a mass of 2000 grams. If a patient diagnosed with lung cancer went untreated, estimate how long he or she would survive after the diagnosis. c. By what percentage of its original size has the 10-gram tumor grown when it reaches 2000 grams? 22. It is now recognized that prolonged exposure to very loud noise can damage hearing. The accompanying table gives the permissible daily exposure hours to very loud noises as recommended by OSHA, the Occupational Safety and Health Administration. Sound Level, D (decibels)
Maximum Duration, H (hours)
120 115 110 105 100 95 90
0 0.25 0.5 1 2 4 8
a. Examine the data for patterns. How is D progressing? How is H progressing? Do the data represent a growth or a decay phenomenon? Explain your answer. b. Find a formula for H as a function of D. Fit the data as closely as possible. Graph your formula and the data on the same grid.
23. a. Construct a function that would represent the resulting value if you invested $5000 for n years at an annually compounded interest rate of: i. 3.5% ii. 6.75% iii. 12.5% b. If you make three different $5000 investments today at the three different interest rates listed in part (a), how much will each investment be worth in 40 years? 24. A bank compounds interest annually at 4%. a. Write an equation for the value V of $100 in t years. b. Write an equation for the value V of $1000 in t years. c. After 20 years will the total interest earned on $1000 be ten times the total interest earned on $100? Why or why not? 25. (Graphing program recommended.) You have a chance to invest money in a risky investment at 6% interest compounded annually. Or you can invest your money in a safe investment at 3% interest compounded annually. a. Write an equation that describes the value of your investment after n years if you invest $100 at 6% compounded annually. Plot the function. Estimate how long it would take to double your money. b. Write an equation that describes the value of your investment after n years if you invest $200 at 3% compounded annually. Plot the function on the same grid as in part (a). Estimate the time needed to double your investment. c. Looking at your graph, indicate whether the amount in the first investment in part (a) will ever exceed the amount in the second account in part (b). If so, approximately when? 26. According to the Arkansas Democrat Gazette (February 27, 1994): Jonathan Holdeen thought up a way to end taxes forever. It was disarmingly simple. He would merely set aside some money in trust for the government and leave it there for 500 or 1000 years. Just a penny, Holdeen calculated, could grow to trillions of dollars in that time. But the stash he had in mind would grow much bigger—to quadrillions or quintillions—so big that the government, one day, could pay for all its operations simply from the income. Then taxes could be abolished. And everyone would be better off. a. Holdeen died in 1967, leaving a trust of $2.8 million that is being managed by his daughter, Janet Adams. In 1994, the trust was worth $21.6 million. The trust was debated in Philadelphia Orphans’ Court. Some lawyers who were trying to break the trust said that it is dangerous to let it go on, because “it would sponge up all the money in the world.” Is this possible? b. After 500 years, how much would the trust be worth? Would this be enough to pay off the current national debt (over $7 trillion in 2004)? What about after 1000 years? Describe the model you used to make your predictions. 27. Describe how a 6% inflation rate will erode the value of a dollar over time. Approximately when would a dollar be worth only 50 cents? This is the half-life of the dollar’s buying power under 6% inflation.
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a. Describe a formula that will tell you how many female ticks there will be in n years, if you start with one impregnated tick. b. How many male ticks will there be in n years? How many total ticks in n years? c. If the surface of Earth is approximately 5.089 ? 1014 square meters and an adult tick takes up 0.5 square centimeters of land, approximately how long would it take before the total number of ticks stemming from one generation of ticks would cover the surface of Earth?
28. The future value V of a savings plan, where regular payments P are made n times to an account in which the interest rate, i, is compounded each payment period, can be calculated using the formula V5P?
s1 1 id n 2 1 i
The total number of payments, n, equals the number of payments per year, m, times the number of years, t, so n5m?t The interest rate per compounding period, i, equals the annual interest rate, r, divided by the number of compounding periods a year, m, so i 5 r/m a. Substitute n 5 m ? t and i 5 r/m in the formula for V, getting an expression for V in terms of m, t, and r. b. If a parent plans to build a college fund by putting $50 a month into an account compounded monthly with a 4% annual interest rate, what will be the value of the account in 17 years? c. Solve the original formula for P as a function of V, i, and n. d. Now you are able to find how much must be paid in every month to meet a particular final goal. If you estimate the child will need $100,000 for college, what monthly payment must the parent make if the interest rate is the same as in part (b)? 29. [Part (c) requires use of the Internet, and technology to find a best-fit function is recommended.] The following data show the total government debt for the United States from 1950 to 2006. DATA
FEDDEBT
Year
Debt ($ billions)
1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 2006
257 274 291 322 381 542 909 1818 3207 4921 5674 7933 8507
a. By hand or with technology, plot the data in the accompanying table and sketch a curve that approximates the data. b. Construct an exponential function that models the data. What would your model predict for the current debt? c. Use the “debt clock” at www.brillig.com/debt_clock to find the current debt. How accurate was your prediction in part (b)? 30. The average female adult Ixodes scapularis (a deer tick that can carry Lyme disease) lives only a year but can lay up to 10,000 eggs right before she dies. Assume that the ticks all live to adulthood, and half are females that reproduce at the same rate and half the eggs are male.
315
31. MCI, a phone company that provides long-distance service, introduced a marketing strategy called “Friends and Family.” Each person who signed up received a discounted calling rate to ten specified individuals. The catch was that the ten people also had to join the “Friends and Family” program. a. Assume that one individual agrees to join the “Friends and Family” program and that this individual recruits ten new members, who in turn each recruit ten new members, and so on. Write a function to describe the number of new people who have signed up for “Friends and Family” at the nth round of recruiting. b. Now write a function that would describe the total number of people (including the originator) signed up after n rounds of recruiting. c. How many “Friends and Family” members, stemming from this one person, will there be after five rounds of recruiting? After ten rounds? d. Write a 60-second summary of the pros and cons of this recruiting strategy. Why will this strategy eventually collapse? 32. In a chain letter one person writes a letter to a number of other people, N, who are each requested to send the letter to N other people, and so on. In a simple case with N 5 2, let’s assume person A1 starts the process. A1
B1
B2
C1
D1
C2
D2
D3
C3
D4
D5
C4
D6
D7
D8
A1 sends to B1 and B2; B1 sends to C1 and C2; B2 sends to C3 and C4; and so on. A typical letter has listed in order the chain of senders who sent the letters. So D7 receives a letter that has A1, B2, and C4 listed. If these letters request money, they are illegal. A typical request looks like this: • When you receive this letter, send $10 to the person on the top of the list. • Copy this letter, but add your name to the bottom of the list and leave off the name at the top of the list. • Send a copy to two friends within 3 days. For this problem, assume that all of the above conditions hold.
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e. Internet search: Chain letters are an example of a “pyramid growth” scheme. A similar business strategy is multilevel marketing. This marketing method uses the customers to sell the product by giving them a financial incentive to promote the product to potential customers or potential salespeople for the product. (See Exercise 31.) Sometimes the distinction between multilevel marketing and chain letters gets blurred. Search the U.S. Postal Service website (www.usps.gov) for “pyramid schemes” to find information about what is legal and what is not. Report what you find.
a. Construct a mathematical model for the number of new people receiving letters at each level L, assuming N 5 2 as shown in the above tree. b. If the chain is not broken, how much money should an individual receive? c. Suppose A1 sent out letters with two additional phony names on the list (say A1a and Alb) with P.O. box addresses she owns. So both B1 and B2 would receive a letter with the list A1, A1a, Alb. If the chain isn’t broken, how much money would A1 receive? d. If the chain continued as described in part (a), how many new people would receive letters at level 25?
5.7 Semi-Log Plots of Exponential Functions With exponential growth functions, we often face the same problem that we did in Chapter 4 when we tried to compare the size of an atom with the size of the solar system. The numbers go from very small to very large. In our E. coli model, for example, the number of bacteria started at 100 and grew to over 1 billion in twenty-four time periods (see Table 5.1). It is virtually impossible to display the entire data set on a standard graph. Whenever we need to graph numbers of widely varying sizes, we turn to a logarithmic scale. Previously, using standard linear scales on both axes, we could graph only a subset of the E. coli data in order to create a useful graph (see Figure 5.1). However, if we convert the vertical axis to a logarithmic scale, we can plot the entire data set (Figure 5.22). When one axis uses a logarithmic scale and the other a linear scale, the graph is called a semi-log (or log-linear) plot.
Semi-Log Plot
N 10,000,000,000 1,000,000,000 100,000,000
No. of bacteria
10,000,000 1,000,000 100,000 10,000 1,000 100 10 1
0
5
15 10 Time periods
20
25
Figure 5.22 Semi-log plot of E. coli models data over twenty-
four time periods.
t
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Why Does the Graph Appear as a Straight Line? On a semi-log plot, moving n units horizontally to the right is equivalent to adding n units, but moving up vertically n units is equivalent to multiplying by a factor of n. To stay on a line with slope m, we need to move vertically m units for each unit we move horizontally. So the line tells us that each time we increase the time period by 1, the number of E. coli is multiplied by a constant (namely 2). That’s precisely the definition of an exponential function. In general, the graph of any exponential function on a semi-log plot will be a straight line. We’ll take a closer look at why this is true in Chapter 6. Since most graphing software easily converts standard linear plots to log-linear or semi-log plots, this is one of the simplest and most reliable ways to recognize exponential growth in a data set.
When an exponential function is plotted using a standard linear scale on the horizontal axis and a logarithmic scale on the vertical axis, its graph is a straight line. This type of graph is called a log-linear or semi-log plot.
EXAMPLE
1
To learn more about Gordon Moore’s predictions, read his original paper and a more current interview in Wired.
Moore’s law In 1965 Gordon Moore, cofounder of Intel, made his famous prediction that the number of transistors per integrated circuit would increase exponentially over time (doubling every 18 months). This became known as Moore’s Law. Figure 5.23 shows the actual increase in the number of transistors over time. Do the data justify his claim of exponential growth?
Moore’s Law 10,000,000,000 Dual-Core Intel® Itanium® 2 processor 1,000,000,000
Intel® Itanium® 2 processor Intel® Itanium® processor
100,000,000
®
Intel Pentium® 4 processor Intel® Pentium® III processor
10,000,000
Pentium® processor
Intel 486TM processor TM
Intel 386
1,000,000
Transistors
Intel® Pentium® II processor Intel®
processor 286
100,000
8086 10,000
8080 8008 4004 1970
1975
1980
1985
1990
1995
2000
2005
1,000 2010
Figure 5.23 The semi-log plot of number of transistors (per circuit board) over time. Source: Intel’s website at www.intel.com.
SOLUTION
Figure 5.23 shows a semi-log plot of the number of transistors (per integrated circuit) over time. The plot looks basically linear, indicating that the data are exponential in nature.
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Algebra Aerobics 5.7 1. a. Using the graph in Figure 5.22, estimate the time it takes for the E. coli to increase by a factor of 10. b. From the original expression for the population, N 5 100 ? 2t , when does the population increase by a factor of 8? c. By a factor of 16? d. Are these three answers consistent with each other? 2. What would you expect the graph of y 5 25 ? 10 x to look like on a semi-log plot? Construct a table of values for x equal to 0, 1, 2, 3, 4 and 5. Plot the graph of this function on the accompanying semi-log grid. Does your graph match your prediction? 3. Use your graph for Problem 2 to estimate the values of y when x 5 3.5 and when x 5 7.
100,000,000
y
10,000,000 1,000,000 100,000 10,000 1,000 100 10 1
x 0
2
4
6
8
Exercises for Section 5.7 Technology for finding a best-fit function is required for Exercises 5 and 6.
3. The three accompanying graphs are all of the same function, y 5 1000s1.5d x.
1. (Graphing program recommended.) Below is a table of values for y 5 500s3d x and for log y. x
y
log y
0 5 10 15 20 25 30
500 121,500 29,524,500 7.17 ? 109 1.74 ? 1012 4.24 ? 1014 1.03 ? 1017
2.699 5.085 7.470 9.856 12.241 14.627 17.013
1000
1
2
4
6
8
10
12
x
14
0
2
4
6
Graph A 100,000,000
8
10
12
x
14
Graph C
y 100,000
y
10,000
1,000,000
1,000
10,000
100 100 10
10 0
2
4
6
8
x
10 12 14
Graph B
0 1 2 3 4 5 6 7 8
x
Graph B
y 5 4 3 2 1
y
0.1 0.01
10
x
Graph C
1
0
100
x
10
10
1,000
0 1 2 3 4 5 6 7 8
100 100
10,000
Graph A
2. Match each function with its semi-log plot. a. y 5 200s1.5d x c. y 5 200s0.9d x x b. y 5 200s2.5d d. y 5 200s0.5d x y
100,000
0 1 2 3 4 5 6 7 8
a. Plot y vs. x on a linear scale. Remember to identify the largest number you will need to plot before setting up axis scales. b. Plot log y vs. x on a semi-log plot with a log scale on the vertical axis and a linear scale on the horizontal axis. c. Rewrite the y-values as powers of 10. How do these values relate to log y?
1000
y
y 18,000 16,000 14,000 12,000 10,000 8000 6000 4000 2000
1
0
2
4
6
8
x
10 12 14
Graph D
a. Which graph uses a linear scale for y on the vertical axis? A power-of-10 scale on the vertical axis? Logarithms on the vertical axis? b. Why do graphs B and C look the same? c. For graphs A and B, estimate the number of units needed on the horizontal scale for the value of y 5 1000 to increase by a factor of 10. d. On which graph is it easier to determine when the function has increased by a factor of 10? e. On which graph is it easier to determine when the function has doubled?
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U.S. International Trade (Billions of Dollars)
Annual probability of dying (%)
4. According to Rubin and Farber’s Pathology, “Smoking tobacco is the single largest preventable cause of death in the United States, with direct health costs to the economy of tens of billions of dollars a year. Over 400,000 deaths a year— about one sixth of the total mortality in the United States— occur prematurely because of smoking.” The accompanying graph compares the risk of dying for smokers, ex-smokers, and nonsmokers. It shows that individuals who have smoked for 2 years are twice as likely to die as a nonsmoker. Someone who has smoked for 14 years is three times more likely to die than a nonsmoker. Log scale 7.0 6.0
4.0 Ex-smokers (stopped 5–9 years)
2.0 Nonsmokers 1.0 2
4
6
8 Years
10
12
Year
Total Exports
Total Imports
1960 1965 1970 1975 1980 1985 1990 1995 2000 2005
25.9 35.3 56.6 132.6 271.8 288.8 537.2 793.5 1070.6 1275.2
22.4 30.6 54.4 120.2 291.2 410.9 618.4 891.0 1448.2 1992.0
Source: U.S. Department of Commerce, Bureau of Economic Analysis, U.S. Bureau of the Census, Statistical Abstract of the United States: 2006.
Smokers (1–2 packs/day)
5.0
3.0
319
14
16
Source: E. Rubin and J. L. Farber, Pathology, 3rd ed. (Philadelphia: Lippincott-Raven, 1998), p. 310. Copyright © 1998 by LippincottRaven. Reprinted by permission.
a. The graphs for the smokers (one to two packs per day), ex-smokers, and nonsmokers all appear roughly as straight lines on this semi-log plot. What, then, would be appropriate functions to use to model the increased probability of dying over time for all three groups? b. The plots for smokers and ex-smokers appear roughly as two straight lines that start at the same point, but the graph for smokers has a steeper slope. How would their two function models be the same and how would they be different? c. The plots of ex-smokers and nonsmokers appear as two straight lines that are roughly equidistant. How would their two function models be the same and how would they be different? 5. (Requires technology to find a best-fit function.) DATA a. Load the file CELCOUNT, which contains all of the white blood cell counts for the bone marrow transplant patient. Now graph the data on a semi-log plot. Which section(s) of the curve represent exponential growth or decay? Explain how you can tell. b. Load the file ECOLI, which contains the E. coli counts for twenty-four time periods. Graph the E. coli data on a semi-log plot. Which section of this curve represents exponential growth or decay? Explain your answer. 6. (Requires technology to find a best-fit function.) The accompanying table shows the U.S. international trade in goods and services.
DATA
USTRADE
a. U.S. imports and exports both expanded rapidly between 1960 and 2005. Use technology to plot the total U.S. exports and total U.S. imports over time on the same graph. b. Now change the vertical axis to a logarithmic scale and generate a semi-log plot of the same data as in part (a). What is the shape of the data now, and what does this suggest would be an appropriate function type to model U.S. exports and imports? c. Construct appropriate function models for total U.S. imports and for total exports. d. The difference between the values of exports and imports is called the trade balance. If the balance is negative, it is called a trade deficit. The balance of trade has been an object of much concern lately. Calculate the trade balance for each year and plot it over time. Describe the overall pattern. e. We have a trade deficit that has been increasing rapidly in recent years. But for quantities that are growing exponentially, the “relative difference” is much more meaningful than the simple difference. In this case the relative difference is exports 2 imports exports This gives the trade balance as a fraction (or if you multiply by 100, as a percentage) of exports. Calculate the relative difference for each year in the above table and graph it as a function of time. Does this present a more or less worrisome picture? That is, in particular over the last decade, has the relative difference remained stable or is it also rapidly increasing in magnitude? 7. A Fidelity Investments report included the graph, on the following page, illustrating how $10,000 invested in a Fidelity Fund © created on December 1992 would have grown over 10 years. The graph also includes the Standard & Poor’s 500 Index6 (S&P 500) for comparison. a. What sort of plot is this? b. The growth from 1993 to 2000 in both the Fidelity Fund and the S&P 500 Index appears roughly linear. What does that tell you? 6
The Standard & Poor’s 500 Index is an index of 500 stocks that is used to measure the performance of the entire U.S. domestic stock market.
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c. Between 2000 and 2002, the values of the Fund and 500 Index have a roughly linear decline. What does that tell you?
d. Give at least two reasons why Fidelity would publish this graph.
$10,000 Over 10 Years Let's say hypothetically that $10,000 was invested in Fidelity Fund on December 31, 1992. The chart shows how the value of your investment would have grown, and also shows how the Standard & Poor's 500 Index did over the same period. Fidelity Fund
S&P 500
$45,000 $40,000 $35,000 $30,000 $25,000
$20,000
$15,000
$10,000
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
C H A P T E R S U M M A RY Exponential Functions
Graphs of Exponential Functions
The general form of the equation for an exponential function is
For functions in the form y 5 Ca x: The value of C tells us where the graph crosses the y-axis. The value of a affects the steepness of the graph.
y 5 Ca x
sa . 0 and a 2 1d,
where
C is the initial value or y-intercept a is the base and is called the growth (or decay) factor If C . 0 and a . 1, the function represents growth 0 , a , 1, the function represents decay
Linear vs. Exponential Growth Exponential growth is multiplicative, whereas linear growth is additive. Exponential growth involves multiplication by a constant factor for each unit increase in input. Linear growth involves adding a fixed amount for each unit increase in input. In the long run, any exponential growth function will eventually dominate any linear growth function.
Exponential growth (a . 1, C . 0). The larger the value of a, the more rapid the growth and the more rapidly the graph rises. Exponential decay (0 , a , 1, C . 0). The smaller the value of a, the more rapid the decay and the more rapidly the graph falls. The graphs of both exponential growth and decay functions are asymptotic to the x-axis. Exponential growth: As x S 2`, y S 0. Exponential decay: As x S 1`, y S 0. y
y
y
Exponential
Linear
x x
Exponential growth y = Cax (a > 1)
x Exponential decay y = Cax (0 < a < 1)
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321
Factors, Rates, and Percents
Properties of Exponential Functions
An exponential function can be represented as a constant percent change. Growth and decay factors can be translated into constant percent increases or decreases, called growth or decay rates. For an exponential function in the form y 5 Ca x,
The doubling time of an exponentially growing quantity is the time required for the quantity to double in size. The half-life of an exponentially decaying quantity is the time required for one-half of the quantity to decay. The rule of 70 offers a simple way to estimate the doubling time or the half-life. If a quantity is growing at R% per year, then its doubling time is approximately 70/R years. If a quantity is decaying at R% per month, then 70/R gives its half-life in months.
growth factor 5 1 1 growth rate a511r where r is the growth rate in decimal form. decay factor 5 1 2 decay rate a512r
Semi-Log Plots of Exponential Functions When an exponential function is plotted using a standard scale on the horizontal axis and a logarithmic scale on the vertical axis, its graph is a straight line. This is called a loglinear or semi-log plot.
where r is the decay rate in decimal form.
C H E C K Y O U R U N D E R S TA N D I N G I. Is each of the statements in Problems 1–24 true or false? Give an explanation for your answer.
years since 1990, then in 1990 expenditures were about $134 billion and were increasing by 4.1% per year.
1. If y 5 ƒ(x) is an exponential function and if increasing x by 1 increases y by a factor of 3, then increasing x by 2 increases y by a factor of 6.
11. Increasing $1000 by $100 per year for 10 years gives you more than increasing $1000 by 10% per year for 10 years.
2. If y 5 ƒ(x) is an exponential function and if increasing x by 1 increases y by 20%, then increasing x by 3 increases y by about 73%.
12. The value of the dollar with inflation at 2% per month is the same as the value of the dollar with inflation at 24% per year.
3. y 5 x3 is an exponential function.
13. If M dollars is invested at 6.25% compounded annually, the amount of money, A, in 14 years is A 5 Ms0.0625d 14.
4. The graph of y 5 10 ? 2x is decreasing. 5. The average rate of change between any two points on the graph of y 5 32.5s1.06d x is constant.
Problems 14 and 15 refer to the following graph.
6. Of the three exponential functions y1 5 5.4s0.8d , y2 5 5.4s0.7d x, and y3 5 5.4s0.3d x, y3 decays the most rapidly. x
D
B 70
yC A
7. The graph of the exponential function y 5 1.02x lies below the graph of the line y 5 x for x . 5. 8. If y 5 100s0.976d x, then as x increases by 1, y decreases by 97.6%. 9. The function in the accompanying figure represents exponential decay with an initial population of 150 and decay factor of 0.8. x D
–4
0
4
150
14. Of the functions plotted in the figure, graph C best describes the exponential function ƒsxd 5 20s3d x.
(2, 96)
100
15. Of the functions graphed in the figure, graph B best describes the exponential function gsxd 5 20s3d 2x.
50
t 0
5
10. If the exponential function that models federal budget expenditures (E in billions of dollars) for a particular department is E 5 134s1.041d t , where t 5 number of
16. If B 5 100s0.4d x, then as x increases by 1, B decreases by 60%. 17. If y 5 ƒ(x) is an exponential function and if increasing x by 1 increases y by 2, then increasing x by 5 increases y by 10.
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For Problems 18 and 19, refer to the following table. Year Population (000s)
1990 100
1995 127.6
2000 162.9
2005 207.9
18. An exponential function would be a reasonable model for the population data. 19. Assuming this population grows exponentially, the annual growth rate is 6.29% because 162.9 population in year 2000 5 5 1.629 population in year 1990 100 a growth of 62.9% in 10 years, or 62.9%/10 years 5 6.29% per year. 20. After 30 years, the amount of interest earned on $5000 invested at 5% interest compounded annually will be half as much as the amount of interest earned on $10,000 invested at the same rate. 21. If a population behaves exponentially over time and if population in year 3 5 0.98, then after 3 years the population in year 0 population will have decreased by 2%.
29. Two exponential decay functions with the same initial population but with the first decaying at twice the rate of the second. 30. Two functions, one linear and one exponential, that pass through the data points (0, 5) and (3, 0.625). 31. An exponential function that describes the value of a dollar over time with annual inflation of 3%. 32. An exponential function that has a doubling time of 14 years. 33. A function that describes the amount remaining of an initial amount of 200 mg of a substance with a half-life of 20 years. III. Is each of the statements in Problems 34–43 true or false? If a statement is false, give a counter-example. 34. Exponential functions are multiplicative and linear functions are additive. 35. All exponential functions are increasing. 36. If P 5 Cat describes a population P as a function of t years since 2000, then C is the initial population in the year 2000.
22. If the half-life of a substance is 10 years, then three halflives of the substance would be 30 years.
37. Exponential functions y 5 Cat have a fixed or constant percent change per year.
23. If the doubling time of $100 invested at an interest rate r compounded annually is 9 years, then in 27 years the amount of the investment will be $600.
38. Quantities that increase by a constant amount represent linear growth, whereas quantities that increase by a constant percent represent exponential growth.
24. The doubling time for the function in the accompanying graph is approximately 10 years.
39. The graph of an exponential growth function plotted on a semi-log plot is concave upward. 40. Eventually, the graph of every exponential function meets the horizontal axis.
Q 25
41. For exponential functions, the growth rate is the same as the growth factor.
t 0
15
II. In Problems 25–33, give an example of a function with the specified properties. Express your answer using formulas, and specify the independent and dependent variables.
42. Exponential functions of the form y 5 C ? a x (where a . 0, a 2 1) are always asymptotic to the horizontal axis. 43. Of the two functions in the accompanying figure, only A is decreasing at a constant percent.
25. An exponential function that has an initial population of 2.2 million people and increases 0.5% per year.
25
26. An exponential function that has an initial population of 2.2 million people and increases 0.5% per quarter.
A
v
B
27. An exponential function that has an initial value of $1.4 billion and decreases 2.3% per decade. 28. An exponential function that passes through (2, 125) and (4, 5).
–5
0
t 20
CHAPTER 5 REVIEW: PUTTING IT ALL TOGETHER Internet access is required for Exercise 9 part (c). A calculator that can evaluate powers is recommended for Exercises 18, 25 part (c), and 27. 1. In each case, generate equations that represent the population, P, as a function of time, t (in years), such that when t 5 0, P 5 150 and:
a. b. c. d.
P doubles each year. P decreases by twelve units each year. P increases by 5% each year. The annual average rate of change of P with respect to t is constant at 12.
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2. Describe in words the following functions by identifying the type of function, the initial value, and what happens with each unit increase in x. a. f(x) 5 100(4)x b. g(x) 5 100 2 4x
6. A polluter dumped 25,000 grams of a pollutant into a lake. Each year the amount of pollutant in the lake is reduced by 4%. Construct an equation to describe the amount of pollutant after n years.
3. Match each of the following functions with the appropriate graph. b. y 5 3x c. y 5 4x a. y 5 2x
7. You have a sore throat, so your doctor takes a culture of the bacteria in your throat with a swab and lets the bacteria grow in a Petri dish. If there were originally 500 bacteria in the dish and they grow by 50% per day, describe the relationship between G(t), the number of bacteria in the dish and t, the time since the culture was taken (in days).
16
y A
8. Which of the following functions are exponential and which are linear? Justify your answers.
B
C
x –2
2
4. Sketch graphs of the following functions on the same grid: b. y 5 0.2x c. y 5 0.3x a. y 5 0.1x 5. Studies have shown that lung cancer is directly correlated with smoking. The accompanying graph shows for each year after a smoker has stopped smoking his or her relative risk of lung cancer. The relative risk is the number of times more likely a former smoker is to get lung cancer than someone who has never smoked. For example, if the relative risk is 4.5, then that patient is 4.5 times more likely to get lung cancer than a life-long non-smoker. Time is the measured in number of years since a smoker stopped smoking. a. According to the graph, how many times more likely (the relative risk) is a male who just stopped smoking to get lung cancer than a lifelong nonsmoker? How many times more likely is a female who stopped smoking 12 years ago to get lung cancer than someone who never smoked? b. Why is it reasonable that the relative risk is declining? c. What does it mean if the relative risk is 1? Would you expect the relative risk to go below 1? Explain.
x
f(x)
g(x)
h(x)
j(x)
0 1 2 3 4 5
2500 2200 100 400 700 1000
1 2.5 6.25 15.625 39.0625 97.65625
10.25 8.25 6.25 4.25 2.25 0.25
1 0.5 0.25 0.125 0.0625 0.03125
9. The United Nations Department of Economic and Social Affairs reported in 1999 that “the world’s population stands at 6 billion and is growing at 1.3% per year, for an annual net addition of 78 million people.” a. This statement actually contains two contradictory descriptions for predicting world population growth. What are they? b. In order to decide which of the statements is more accurate, use the information provided to construct a linear model and an exponential model for world population growth. c. The U.S. Census Bureau has a website that gives estimates for the current world population at http://www.census.gov/ ipc/www/popclockworld.html. Look up the current population. Which of your models is a more accurate predictor of the current world population? 10. Professional photographers consider exposure value (a combination of shutter speed and aperture) in relation to luminance (the amount of light that falls on a certain region) in setting camera controls to produce a desired effect. The data shown in the table give the relationship of the exposure value, Ev , to the luminance, L, for a particular type of film. Find a formula giving L as a function of Ev .
Lung cancer mortality relative risk
25.00 Male Female 20.00
15.00
10.00
5.00
0.00 0
5
10
15
20
25
30
Time since occasion (years)
35
40
45
Ev, Exposure Value
L, Luminance (candelas/sq. meter)
0 1 2 3 4 5 6 7 8 9 10 11 12
0.125 0.25 0.5 1 2 4 8 16 32 64 128 256 512
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11. According to Optimum Population Trust, the number of motor vehicles in the United Kingdom has nearly doubled every 25 years since 1925. Use the “rule of 70” to estimate the annual percent increase.
DHS used an annual percent growth rate from the projected estimate for the initial value in 2006. 15. The Energy Information Administration has made the following projections for energy consumption for the world:
12. Mirex, a pesticide used to control fire ants, is rather longlived in the environment. It takes 12 years for half of the original amount to break down into harmless products. Use the “rule of 70” to estimate the annual percent decrease. 13. The U.S. Department of Human Services reported, “Health care spending in the United States is projected to grow 7.4 percent and surpass $2 trillion in 2005, down from the 7.9 percent growth experienced in 2004.” The Department also published the following data about national health care expenditures and projections. Find the annual growth factor and the annual percent growth rate for health care since 2003. Show your calculations, and check to see if they agree with percent growth reported by the U.S. Department of Human Services.
2003
Health care expenditures (billions)
$1,740.6
2004
2005
Year Quadrillion Btu
2015 563
2020 613
2006
$1,877.6 $2,016.0 $2,169.5
Source: http://www.cms.hhs.gov/NationalHealthExpendData/.
14. The U.S. Department of Human Services projected that in 2010 national health care expenditures would be approximately $2,887.3 billion. Use your results from Problem 13 to estimate expenditures in 2010, assuming the initial value is $2,169.5 billion (the projected estimate for 2006) and the annual percent growth rate is equal to: a. The annual percent growth rate from 2005 to 2006. b. The annual percent growth rate from 2004 to 2005. c. How close is each projection for 2010 from parts (a) and (b) to the DHS projection? From your calculations, estimate the percent growth rate used by DHS. Assume
1972
1976
1980
2030 722
Source: www.eia.doe.gov/iea/.
Would a linear or an exponential function be an appropriate model for these data? Construct a function that models the data. 16. a. Generate a series of numbers N using N 5 20.5n, for integer values of n from 0 to 10. Use the value of 20.5 ≈ 1.414 and the rules of exponents to calculate values for N. b. Rewrite the N formula using a square root sign.
18. (A calculator that can evaluate powers is recommended.) The following graph shows median house prices in the United States from 1968 to 2004. a. Estimate from the graph the median price paid for a home in 1968 and in 1993. Use these data points to construct a linear and an exponential model to represent the growth in median price from 1968 to 1993. b. Which model is a better predictor of the national median price of a home in 2004? Would you use this model to make predictions for next year? Why or why not?
USA Median Prices $190,000 $180,000 $170,000 $160,000 $150,000 $140,000 $130,000 $120,000 $110,000 $100,000 $90,000 $80,000 $70,000 $60,000 $50,000 $40,000 $30,000 $20,000 $10,000 $0 1968
2025 665
17. NUA.com estimated that there were 605 million Internet users worldwide in 2002. Useit.com estimated that the number of Internet users will reach 2 billion in 2015 and 3 billion in 2040. a. Given NUA’s 2002 estimate of 605 million users, what annual growth rate would give 2 billion users in 2015? b. Given Useit’s estimate of 2 billion users in 2015, what annual growth rate would give 3 billion users in 2040?
Projected Year
World Energy Consumption
1984
1988
Source: National Association of Realtors www.realtor.org.
1992
1996
2000
2004
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19. Radioactive iodine (I-131), used to test for thyroid problems, has a half-life of 8 days. If you start with 20 grams of radioactive iodine, describe the relationship between A(t), the amount of I-131 (in grams), and t, the time (in days) since you first measured the sample. 20. Xenon gas (Xe-133) is used in medical imaging to study blood flow in the heart and brain. When inhaled, it is quickly absorbed into the bloodstream, and then gradually eliminated from the body (through exhalation). About 5 minutes after inhalation, there is half as much xenon left in the body (i.e., Xe-133 has a biological half-life of about 5 minutes). If you originally breathe in and absorb 3 ml. of xenon gas, describe the relationship between X, the amount of xenon gas in your body (in ml), and t, the time (in minutes) since you inhaled it. 21. In the accompanying chart, which graphs approximate exponential growth?
1,000,000,000 Pages
100,000,000 Internet traffic 10,000,000 1,000,000 100,000 10,000 1,000
Internet hosts
Gopher traffic
WWW traffic
Web sites
100 10 1 Jan Jul- Jan Jul- Jan Jul- Jan Jul- Jan Jul- Jan Jul- Jan -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97
Source: http://www.useit.com/alertbox/9509.html. 22. In 2005 the Government Accountability Office issued a study of textbook prices. The report noted that in the preceding two decades textbook prices had been rising at the rate of 6% per year, roughly double the annual inflation rate of about 3% a year. a. If a textbook cost $30 in 1985, what would it cost in 2005? b. Assuming textbook prices continue to rise at 6% per year, if a textbook costs $80 in 2005, what will it cost in 2015? c. A textbook that was first published in 1990 costs $120 in 2007. Since 1990 its price has increased at the rate of 6% per year. What was its price in 1990? 23. You may have noticed that when you take a branch from certain trees, the branch looks like a miniature version of the tree. When you break off a piece of the branch, that looks like the tree too. Mathematicians call this property self-similarity. The village of Bourton-on-the-Water in southwest England has a wonderful example of selfsimilarity: it has a 1/10 scale model of itself. Because the
325
1/10 scale model is a complete model of the town, it must contain a model of itself (that is, a 1/100 scale model of Bourton), and because the 1/100 scale model is also a complete model of Bourton, it also contains a scale model (that is, a 1/1000 scale model of Bourton.) a. If Ao is the area of the actual village, how does the area of the first 1/10 scale model (where each linear dimension is one-tenth of the actual size) compare with Ao? b. If the scale models are made of the same materials as the actual village, how does the weight of the building materials in the 1/100 scale model church compare with the weight Wo of the original church? c. If the actual village is called a level 0 model, the 1/10 model is level 1, the 1/100 model is level 2, and so on, find a formula for the area An at any level n of the model as a function of Ao and for the weight Wn at any level n of the model as a function of Wo. 24. Computer worms are annoying and potentially very destructive. They are self-reproducing programs that run independently and travel across network connections. For some worms, such as SoBig, if you are sent an e-mail and you open up an attachment that contains SoBig, two things happen. First, SoBig installs a program that can be remotely activated in the future to send spam messages or shut down your computer. Then SoBig e-mails itself to everyone in your address book. a. Construct a function that could model the spread of SoBig. Assume that everyone has in his or her address book ten people, none of whom have received SoBig. b. Assume everyone in America has a computer. How many levels in a fractal tree would it take for SoBig to affect 300,000,000 Americans? (Hint: To find the total number of people who receive SoBig you need to add those at level 0, level 1, level, 2, etc.) c. Why is a worm like SoBig so dangerous? 25. [A calculator that can evaluate powers is recommended for part (c).] Lung cancer is one of the leading causes of death, especially for smokers. There are various forms of cancer that attack the lungs, including cancers that start in some other organ and metastasize to the lungs. Doubling times for lung cancers vary considerably but are likely to fall within the range of 2 to 8 months. Although individual cancers are unpredictable and may speed up or slow down, the following examples give an idea of the range of time possibilities. a. Two people are found to have lung cancer tumors whose volumes are each estimated at 0.5 cubic centimeters. A former asbestos worker has a tumor with an expected doubling time of 8 months. A heavy smoker has a tumor with an expected doubling time of 2 months. Write two tumor growth exponential functions, A(t) and S(t), for the asbestos worker and smoker, respectively, where t = number of 2-month time periods. b. Graph S(t) and A(t) over 12 time periods on the same grid and compare the graphs. c. Assuming both tumors grow in a spherical shape, what would be the diameter of each tumor after one year? (Note: The volume of a sphere V 5 4/3πr 3.)
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26. Which of the following graphs suggests a linear model for the original data? Which suggests an exponential model for the original data? Justify your answers. Human Life Expectancy 80 75
Life expectancy (Years)
70
27. (A calculator that can evaluate powers is recommended.) In February 2007 the U.S. Intergovernmental Panel on Climate Change reported that human activities have increased greenhouse gases in the atmosphere, resulting in global warming and other climate changes. Methane (a greenhouse gas) in the atmosphere has more than doubled since preindustrial times from about 750 parts per billion in 1850 to 1,750 in 2005. The following graph shows the increase in methane gas since 1850.
65
Methane in the Atmosphere (parts per billion)
60
2,000 55
1,750 50
1,500 45
1,250 40
1,000 35 1840 1860 1880 1900 1920 1940 1960 1980 2000 2020
750 1850
1900
1950
2000
Source: The New York Times International, February 3, 2007.
Total U.S. GDP In Constant Dollars 10,100
a. Construct an exponential equation, M(t) that could be used to model the increase in methane gas in the atmosphere, where t = number of years since 1850. b. From the graph, estimate the approximate doubling time. Now calculate the approximate doubling time using the “rule of 70.” How close are your answers? c. Describe the exponential growth in methane gas since 1850 in two different ways.
9,100
Year 2000 dollars (billions)
8,100 7,100 6,100 5,100 4,100 3,100 2,100 1,100 100 1915 1925 1935 1945 1955 1965 1975 1985 1995 2005
Internet Backbone Bandwidth 10,000,000,000
100,000,000
SONET (Fiber Optic)
T3
10,000,000 T1 1,000,000 100,000 MODEM
Source: http://www.kurzweilai.net.
2000
1998
1996
1994
1992
1990
1988
1986
1984
1,000
1982
10,000 1980
Bits per second
1,000,000,000
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E X P L O R AT I O N 5 . 1 Properties of Exponential Functions
Objectives • explore the effects of a and C on the graph of the exponential function in the form y 5 Ca x
where
a . 0 and a 2 0
Material/Equipment • computer and software “E3: y 5 Ca x Sliders” in Exponential & Log Functions, or graphing calculator • graph paper
Procedure We start by choosing values for a and C and graphing the resulting equations by hand. From these graphs we make predictions about the effects of a and C on the graphs of other equations. Take notes on your predictions and observations so you can share them with the class. Work in pairs and discuss your predictions with your partner. Making Predictions 1. Start with the simplest case, where C 5 1. The equation will now have the form y 5 ax Make a data table and by hand sketch on the same grid the graphs for y 5 2 x (here a 5 2) and y 5 3 x (here a 5 3). Use both positive and negative values for x. Predict where the graphs of y 5 2.7 x and y 5 5 x would be located on your graph. Check your work and predictions with your partner. x
y 5 2x
y 5 3x
22 21 0 1 2 3
How would you describe your graphs? Do they have a maximum or a minimum value? What happens to y as x increases? What happens to y as x decreases? Which graph shows y changing faster compared with x? 2. Now create two functions in the form y 5 a x where 0 , a , 1. Create a data table and graph your functions on the same grid. Make predictions for other functions where C 5 1 and 0 , a , 1. 3. Now consider the case where C has a value other than 1 for the general exponential function y 5 Ca x Create a table of values and sketch the graphs of y 5 0.5(2x) (in this case C 5 0.5 and a 5 2) and y 5 3(2x) (in this case C 5 3 and a 5 2). What do all these graphs have in common? What do you think will happen when a 5 2 and C 5 10? What do you think will happen to the graph if a 5 2 and C 523? Check your predictions with your partner. 327
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x
y 5 0.5(2x)
y 5 3(2x)
y 5 23(2x)
22 21 0 1 2 3
How would you describe your graphs? Do they have a maximum or a minimum value? What happens to y as x increases? What happens to y as x decreases? What is the y-intercept for each graph? Testing Your Predictions Now test your predictions by using a program called “E3: y 5 Ca x Sliders” in the Exponential & Log Functions software package or by creating graphs using technology. 1. What effect does a have? Make predictions when a . 1 and when 0 , a , 1, based on the graphs you constructed by hand. Explore what happens when C 5 1 and you choose different values for a. Check to see whether your observations about the effect of a hold true when C 2 1. How does changing a change the graph? When does y 5 ax describe growth? When does it describe decay? When is it flat? Write a rule that describes what happens when you change the value for a. You only have to deal with cases where a . 0. 2. What effect does C have? Make a prediction based on the graphs you constructed by hand. Now choose a value for a and create a set of functions with different C values. Graph these functions on the same grid. How does changing C change the graph? What does the value of C tell you about the graphs of functions in the form y 5 Cax? Describe your graphs when C . 0 and when C , 0. Use technology to test your generalizations. Exploration-Linked Homework Write a 60-second summary of your results, and present it to the class.
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LOGARITHMIC LINKS: LOGARITHMIC AND EXPONENTIAL FUNCTIONS OVERVIEW If we know a specific output for an exponential function, how can we find the associated input? To answer this question, we can use logarithmic functions, the close relatives of exponential functions. We return to the E. coli model to calculate the doubling time. After reading this chapter you should be able to • use logarithms to solve exponential equations • apply the rules for common and natural (base e) logarithms • create an exponential model for continuous compounding • understand the properties of logarithmic functions • describe the relationship between logarithmic and exponential functions • find the equation of an exponential function on a semi-log plot
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6.1 Using Logarithms to Solve Exponential Equations Estimating Solutions to Exponential Equations So far, we have found a function’s output from particular values of the input. For example, in Chapter 5, we modeled the growth of E. coli bacteria with the function N 5 100 ? 2 t where N 5 number of E. coli bacteria and t 5 time (in 20-minute periods). If the input t is 5, then the corresponding output N is 100 ? 25 5 100 ? 32 5 3200 bacteria. Now we do the reverse, that is, find a function’s input when we know the output. Starting with a value for N, the output, we find a corresponding value for t, the input. For example, at what time t will the value for N, the number of E. coli, be 1000? This turns out to be a harder question to answer than it might first appear. First we will estimate a solution from a table and a graph, and then we will learn how to find an exact solution using logarithms. From a data table and graph In Table 6.1, when t 5 3, N 5 800. When t 5 4, N 5 1600. Since N is steadily increasing, if we know N 5 1000, then the value of t is somewhere between 3 and 4.
Values for N 5 100 ? 2 t t
N 2000
N
0 1 2 3 4 5 6 7 8 9 10
100 200 400 800 1,600 3,200 6,400 12,800 25,600 51,200 102,400
Table 6.1
(3.3, 1000)
1000
t 0
1
2
3
4
5
Figure 6.1 Estimating a value
for t when N 5 1000 on a graph of N 5 100 ? 2t.
We can also estimate the value for t when N 5 1000 by looking at a graph of the function N 5 100 ? 2t (Figure 6.1). By locating the position on the vertical axis where N 5 1000, we can move over horizontally to find the corresponding point on the function graph. By moving from this point vertically down to the t-axis, we can estimate the t value for this point. The value for t appears to be approximately 3.3, so after about 3.3 time periods (or 66 minutes), the number of bacteria is 1000. From an equation An alternative strategy is to set N 5 1000 and solve the equation for the corresponding value for t. Start with the equation Set N 5 1000 Divide both sides by 100
N 5 100 ? 2 t 1000 5 100 ? 2 t 10 5 2 t
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Then we are left with the problem of finding a solution to the equation 10 5 2 t We can estimate the value for t that satisfies the equation by bracketing 2t between consecutive powers of 2. Since 23 5 8 and 24 5 16, then 23 , 10 , 24. So 23 , 2 t , 24, and therefore t is between 3 and 4, which agrees with our previous estimates. Using this strategy, we can find an approximate solution to the equation 1000 5 100 ? 2 t. Strategies for finding exact solutions to such equations require the use of logarithms.
Algebra Aerobics 6.1a 1. Given the equation M 5 250 ? 3 t , find values for M when: a. t 5 0 b. t 5 1 c. t 5 2 d. t 5 3 2. Use Table 6.1 to determine between which two consecutive integers the value of t lies when N is: a. 2000 b. 50,000 3. Use the graph of y 5 3x in Figure 6.2 to estimate the value of x in each of the following equations. a. 3x 5 7 b. 3x 5 0.5
4. Use Table 6.2 to determine between which two consecutive integers the value of x in each of the following equations lies. a. 5x 5 73 b. 5x 5 0.36
y 10
x
5x
22 21 0 1 2 3 4
0.04 0.2 1 5 25 125 625
Table 6.2
5. For each of the following, find two consecutive integers for the exponents a and b that would make the statement true. a. 2a , 13 , 2b c. 5a , 0.24 , 5b b. 3a , 99 , 3b d. 10a , 1500 , 10b
x –2
2 –2
Figure 6.2 Graph of y 5 3x.
Rules for Logarithms In Chapter 4 we defined logarithms. Recall that:
The logarithm base 10 of x is the exponent of 10 needed to produce x. log 10 x 5 c
means that 10 c 5 x
where x . 0
Logarithms base 10 are called common logarithms and log 10 x is written as log x.
So, 105 5 100,000 23
10
5 0.001
is equivalent to saying that
log 100,000 5 5
is equivalent to saying that
log 0.001 5 23
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Using a calculator we can solve equations such as Rewrite it in equivalent form use a calculator to compute the log
10 x 5 80 log 80 5 x 1.903 < x
But to solve equations such as 2 t 5 10 that involve exponential expressions where the base is not 10, we need to know more about logarithms. As the following expressions suggest, the rules of logarithms follow directly from the definition of logarithms and from the rules of exponents.
Rules for Exponents If a is any positive real number and p and q are any real numbers, then 1. ap ? aq 5 a p1q 3. sap d q 5 ap?q 2. ap /aq 5 ap2q 4. a 0 5 1 Corresponding Rules for Logarithms If A and B are positive real numbers and p is any real number, then 1. log sA ? Bd 5 log A 1 log B 3. log (Ap ) 5 p log A 2. log sA/Bd 5 log A 2 log B 4. log 1 5 0 ssince 100 5 1d
?
SOMETHING TO THINK ABOUT
Try expressing in words all the other rules for logarithms in terms of exponents.
Finding the common logarithm of a number means finding its exponent when the number is written as a power of 10. So when you see “logarithm,” think “exponent,” and the rules of logarithms make sense. As we learned in Chapter 4, the log of 0 or a negative number is not defined. But when we take the log of a number, we can get 0 or a negative value. For example, log 1 5 0 and log 0.1 5 21. We will list a rationale for each of the rules of logarithms and prove Rule 1. We leave the other proofs as exercises. Rule 1
log sA ? Bd 5 log A 1 log B
Rationale Rule 1 of exponents says that when we multiply two terms with the same base, we keep the base and add the exponents; that is, ap ? aq 5 ap1q. Rule 1 of logs says that if we rewrite A and B each as a power of 10, then the exponent of A ? B is the sum of the exponents of A and B. Proof If we let log A 5 x, then and if log B 5 y, then We have two equal products by laws of exponents Taking the log of each side by definition of log Substituting log A for x and log B for y we arrive at our desired result.
10 x 5 10 y 5 A?B5 A?B5 log(A ? B) 5 log(A ? B) 5 log (A ? B) 5
A B 10 x ? 10 y 10 x1y log (10 x1y ) x1y log A 1 log B
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Rule 2
333
log sA/Bd 5 log A 2 log B
Rationale Rule 2 of exponents says that when we divide terms with the same base, we keep the base and subtract the exponents, that is, ap /aq 5 ap2q. Rule 2 of logs says that if we write A and B each as a power of 10, then the exponent of A/B equals the exponent of A minus the exponent of B.
EXAMPLE
1
Using the rules for logs Indicate whether or not the statement is true using the rules of exponents and the definition of logarithms. a. log s102 ? 103 d 0 log 102 1 log103 b. log s105 ? 1027 d 0 log 105 1 log1027 c. log s102 /103 d 0 log 102 2 log103 d. log s103 /1023 d 0 log 103 /log 1023
SOLUTION
To see if an expression is a true statement, we must verify that the two sides of the expression are equal. a. log s102 ? 103 d 0 log 102 1 log103 Rule 1 of exponents definition of log
log 105 0 log102 1 log 103 55213 555
Since the values on the two sides of the equation are equal, our original equation is a true statement. b. log s105 ? 1027 d 0 log 105 1 log 1027 Rule 1 of exponents definition of log
log 1022 0 log105 1 log 1027 22 5 5 1 (27) 22 5 22
Since the values on the two sides of the equation are equal, our original equation is a true statement. c. log s102 /103 d 0 log 102 2 log 103 Rule 2 of exponents definition of log
log s1021 d 0 log 102 2 log 103 21 5 2 2 3 21 5 21
Since the values on the two sides of the equation are equal, our original equation is a true statement. d. log (103 /1023 ) 0 log 103 / log 1023 Rule 2 of exponents combine terms in exponent definition of log
log 1032(23) 0 log 103 / log 1023 log 106 0 log 103 / log 1023 6 0 3/(23) 6 2 21
Since the values on the two sides of the equation are not equal, our original equation is not a true statement.
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log Ap 5 p log A
Rule 3
SOMETHING TO THINK ABOUT
Why does log "AB lie halfway between log A and log B?
Rationale Since log A2 5 log (A ? A) 5 log A 1 log A 5 2 log A log A3 5 log (A ? A ? A) 5 log A 1 log A 1 log A 5 3 log A log A4 5 log (A ? A ? A ? A) 5 4 log A it seems reasonable to expect that in general log Ap 5 p log A log 1 5 0
Rule 4 Rationale
Since 100 5 1 by definition, the equivalent statement using logarithms is log 1 5 0. EXAMPLE
2
Simplifying expressions with logs Simplify each expression, and if possible evaluate with a calculator. log 103 5 3 log 10 5 3 ? 1 5 3 log 23 5 3 log 2 < 3 ? 0.301 < 0.903 log "3 log x21 log 0.01a (log 5) ? (log 1)
EXAMPLE
3
5 5 5 5
log 31/2 5 12 log 3 < 12 ? 0.477 < 0.239 (21) ? log x 5 2log x a log 0.01 5 a ? (22) 5 22a (log 5) ? 0 5 0
Expanding expressions with logs Use the rules of logarithms to write the following expression as the sum or difference of several logs. log a
SOLUTION
By Rule 2 by Rule 1 and exponent notation by Rule 3
log a
xsy 2 1d 2 b "z
x(y 2 1) 2 b 5 log x(y 2 1) 2 2 log "z "z 5 log x 1 log (y 2 1) 2 2 log z1/2 5 log x 1 2 log (y 2 1) 2 12 log z
We call this process expanding the expression. EXAMPLE
4
Contracting expressions with logs Use the rules of logarithms to write the following expression as a single logarithm. 2 log x 2 log(x 2 1)
SOLUTION
By Rule 3 by Rule 2
2 log x 2 log (x 2 1) 5 log x2 2 log(x 2 1) 5 log a
We call this process contracting the expression.
x2 b x21
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Common Error Probably the most common error in using logarithms stems from confusion over the division property. log A 2 log B 5 log AABB
but
log A 2 log B 2
log A log B
For example, log 100 2 log 10 5 log a but
EXAMPLE
5
SOLUTION
log 100 2 log 10 2
100 b 5 log 10 5 1 10
log 100 52 log 10
Solving equations that contain logs Solve for x in the equation log x 1 log x 2 5 15 log x 1 log x 2 5 15 log x 3 5 15 3 log x 5 15 log x 5 5
Given Rule 1 for logs Rule 3 for logs divide by 3 rewrite using definition of logs
x 5 105 or 100,000
Algebra Aerobics 6.1b 1. Using only the rules of exponents and the definition of logarithm, verify that: a. log s105 /107 d 5 log 105 2 log107 b. log [105 ? s107 d 3] 5 log 105 1 3 log 107 2. Determine the rule(s) of logarithms that were used in each statement. a. log 3 5 log 15 2 log 5 b. log 1024 5 10 log 2 c. log "31 5 12 log 31 d. log 30 5 log 2 1 log 3 1 log 5 e. log 81 2 log 27 5 log 3 3. Determine if each of the following is true or false. For the true statements tell which rule of logarithms was used. a. log (x 1 y) 0 log x 1 log y b. log (x 2 y) 0 log x 2 log y c. 7 log x 0 log x 7 d. log 101.6 0 1.6 log 7 0 e. log 7 2 log 3 log 3 f.
log 7 0 log (7 2 3) log 3
4. Expand, using the properties of logarithms: x "x 1 1 2x 2 1 a. log c. log Å x 1 1 sx 2 1d 2 xy x 2 sy 2 1d b. log d. log z y 3z 5. Contract, expressing your answer as a single logarithm: 13 [log x 2 log (x 1 1)]. 6. Use rules of logarithms to combine into a single logarithm (if necessary), then solve for x. a. log x 5 3 b. log x 1 log 5 5 2 c. log x 1 log 5 5 log 2 d. log x 2 log 2 5 1 e. log x 2 log (x 2 1) 5 log 2 f. log (2x 1 1) 2 log (x 1 5) 5 0 log 103 7. Show that log 103 2 log 102 2 log 102
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Solving Exponential Equations Answering our original question: Solving 1000 = 100 • 2 t Remember the question that started this chapter? We wanted to find out how many time periods it would take 100 E. coli bacteria to become 1000. To find an exact solution, we need to solve the equation 1000 5 100 ? 2t or, dividing by 100, the equivalent equation, 10 5 2t. We now have the necessary tools. 10 5 2t log 10 5 log 2t log 10 5 t log 2 log 10 5t log 2 1
Given take the logarithm of each side use Rule 3 of logs divide both sides by log 2 use a calculator divide
which is consistent with our previous estimates of a value for t between 3 and 4, approximately equal to 3.3. In our original model the time period represents 20 minutes, so 3.32 time periods represents 3.32 ? s20 minutesd 5 66.4 minutes. So the bacteria would increase from the initial number of 100 to 1000 in a little over 66 minutes. EXAMPLE
6
Doubling your money As we saw in Chapter 5, the equation P 5 250(1.05)n gives the value of $250 invested at 5% interest (compounded annually) for n years. How many years does it take for the initial $250 investment to double to $500?
SOLUTION
a. Estimating the answer: If R 5 5% per year, then the rule of 70 (discussed in Section 5.6) estimates the doubling time as 70/R 5 70/5 5 14 years b. Calculating a more precise answer: We can set P 5 500 and solve the equation. 500 5 250(1.05)n 2 5 (1.05)n log 2 5 log (1.05)n log 2 5 n log 1.05 log 2 5n log 1.05 0.3010
Divide both sides by 250 take the log of both sides use Rule 3 of logs divide by log 1.05 evaluate with a calculator divide and switch sides
So the estimate of 14 years using the rule of 70 was pretty close. EXAMPLE
7
Solve the following equation for x in two ways. 8x 5 2x11
SOLUTION
Method 1: Make the bases the same. Make the base the same set exponents equal combine terms solve for x
8x 5 2x11 23x 5 2x11 3x 5 x 1 1 2x 5 1 1 x5 2
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Method 2: Use the rules for logarithms. Take logs of both sides Rule 3 of logs distributive law subtract x log 2 from both sides factor out x Rule 2 of logs divide by log 4 We can double check by both evaluating 8x at x 5 12 or evaluating 2x11 at x 5 EXAMPLE
8
SOLUTION
1 2
8x 5 2x11 log 8x 5 log 2x11 x log 8 5 (x 1 1) log 2 x log 8 5 x log 2 1 log 2 x log 8 2 x log 2 5 log 2 x (log 8 2 log 2) 5 log 2 x log 4 5 log 2 log 2 log 2 log 2 1 x5 5 5 5 log 4 log 22 2 log 2 2 81/2 < 2.8284
2(1/2)11 5 23/2 5 (23 ) 1/2 5 81/2 < 2.8284
Time to decay to a specified amount In Chapter 5 we used the function ƒ(t) 5 100(0.976) t to measure the remaining amount of radioactive material as 100 milligrams (mg) of strontium-90 decayed over time t (in years). How many years would it take for there to be only 10 mg of strontium-90 left? Set ƒ(t) 5 10 and solve the equation. Divide both sides by 100 take the log of both sides use Rule 3 of logs divide by log 0.9755 evaluate logs divide and switch sides
10 5 100(0.976) t 0.1 5 (0.976) t log 0.1 5 log (0.976) t log 0.1 5 t log 0.976 log 0.1 5t log 0.976 21
So it takes almost a century for 100 mg of strontium-90 to decay to 10 mg.
Algebra Aerobics 6.1c These problems require a calculator that can evaluate logs. 1. Solve the following equations for t. a. 60 5 10 ? 2t c. 80s0.95d t 5 10 b. 500s1.06d t 5 2000 2. Using the model N 5 100 ? 2t for bacteria growth, where t is measured in 20-minute time periods, how long will it take for the bacteria count: a. To reach 7000? b. To reach 12,000? 3. First use the rule of 70 to estimate how long it would take $1000 invested at 6% compounded annually to double to $2000. Then use logs to find a more precise answer. 4. Use the function in Example 8 to determine how long it will take for 100 milligrams of strontium-90 to decay to 1 milligram.
5. Solve each equation for t (in years). Which equation(s) asks you to find the time necessary for the initial amount to double? For the initial amount to drop to half? a. 30 5 60s0.95d t c. 500 5 200s1.045d t t b. 16 5 8s1.85d 6. Find the half-life of a substance that decays according to the following models. a. A 5 120s0.983d t (t in days) t b. A 5 0.5s0.92d (t in hours) c. A 5 A0 s0.89d t (t in years)
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Exercises for Section 6.1 Many of these problems (and those in later sections) require a calculator that can evaluate powers and logs. Some require a graphing program. 1. Use the accompanying table to estimate the number of years it would take $100 to become $300 at the following interest rates compounded annually. a. 3% b. 7% Compound Interest over 40 Years Number of Years
Value of $100 at 3% ($)
0 10 20 30 40
Value of $100 at 7% ($)
100 134 181 243 326
100 197 387 761 1497
2. a. Generate a table of values to estimate the half-life of a substance that decays according to the function y 5 100(0.8) x, where x is the number of time periods, each time period is 12 hours, and y is in grams. b. How long will it be before there is less than 1 gram of the substance remaining? 3. The accompanying graph shows the concentration of a drug in the human body as the initial amount of 100 mg dissipates over time. Estimate when the concentration becomes: a. 60 mg b. 40 mg c. 20 mg Remaining Drug Dosage 120 Concentration (milligrams)
100 80
6. Given that log 5 < 0.699, without using a calculator determine the value of: 1 a. log 25 b. log c. log 1025 d. log 0.0025 25 Check your answers with a scientific calculator. 7. Expand each logarithm using only the numbers 2, 3, log 2, and log 3. a. log 9 b. log 18 c. log 54 8. Identify the rules of logarithms that were used to expand each expression. a. log 14 5 log 2 1 log 7 b. log 14 5 log 28 2 log 2 c. log 36 5 2 log 6 d. log 9z3 5 2 log 3 1 3 log z e. log 3x4 5 log 3 1 4 log x 16 f. log a b 5 4 log 2 2 (log 3 1 log x) 3x 9. Use the rules of logarithms to find the value of x. Verify your answer with a calculator. a. log x 5 log 2 1 log 6 b. log x 5 log 24 2 log 2 c. log x2 5 2 log 12 d. log x 5 4 log 2 2 3 log 2 10. Use the rules of logarithms to show that the following are equivalent. a. log 144 5 2 log 3 1 4 log 2 b. 7 log 3 1 5 log 3 5 12 log 3 16 c. 2(log 4 2 log 3) 5 log a b 9 d. 24 log 3 1 log 3 5 log a
60
1 b 27
11. Prove Rule 2 of logarithms:
40
log(A/B) 5 log A 2 log B (A, B . 0)
20 0
2
4 6 Time (hours)
8
10
12. Expand, using the rules of logarithms. a. log Qx 2y 3 "z 2 1R
4. (Requires a graphing program.) Assume throughout that x represents time in seconds. a. Plot the graph of y 5 6(1.3)x for 0 # x # 4. Estimate the doubling time from the graph. b. Now plot y 5 100(1.3)x and estimate the doubling time from the graph. c. Compare your answers to parts (a) and (b). What does this tell you? 5. Without a calculator, determine x if we know that log x equals: a. 23 c. 0 e. 21 b. 6 d. 1
b. log
4 c. log Qt 2 ? "tp 3R
A "BC 3
13. Contract, using the rules of logarithms, and express your answer as a single logarithm. a. 3 log K 2 2 log(K 1 3) b. 2log m 1 5 log(3 1 n) c. 4 log T 1 21 log T d. 13 (log x 1 2 log y) 2 3(log x 1 2 log y) 14. For each of the following equations either prove that it is correct (by using the rules of logarithms and exponents) or else show that it is not correct (by finding numerical values
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for the variables that make the values of the two sides of the equation different). log x x a. log a b 5 y log y x b. log x 2 log y 5 log a b y c. log(2x) 5 2 log x d. 2 log x 5 log(x2) x11 b 5 log (x 1 1) 2 log (x 1 3) e. log a x13 f. log Qx "x 2 1 1R 5 log x 1 12 (x2 1 1) g. log(x2 1 1) 5 2 log x 1 log 1 15. Prove Rule 3 of common logarithms: log Ap 5 p log A (where A . 0). 16. Solve for t. a. 10 t 5 4 b. 3(2 t ) 5 21 c. 1 1 5 t 5 3
d. 102t 5 5 e. 5 t 5 7 t11 f. 6 ? 2 t 5 3 t21
17. Solve for x. a. log x 5 3 b. log(x 1 1) 5 3 c. 3 log x 5 5 18. Solve for x. a. 2x 5 7 x11
d. log(x 1 1) 2 log x 5 1 e. log x 2 log(x 1 1) 5 1
d. log(x 1 3) 1 log 5 5 2
b. Q "3R 5 9 c. 12(1.5) x11 5 13
2x21
e. log(x 2 1) 5 2
19. In Exercise 1, we estimated the number of years it would take $100 to become $300 at each of the interest rates listed below, compounded annually. Now calculate the number of years by constructing and solving the appropriate equations. a. 3% b. 7% 20. Solve for the indicated variables. a. 825 5 275 ? 3T d. 100 5 25(1.18)t b. 45,000 5 15,000 ? 1.04t e. 32,000 5 8000(2.718)t 10 8 x c. 12 ? 10 5 (6 ? 10 ) ? 5 f. 8 ? 105 5 4 ? 105(2.5)x 21. (Requires a graphing program.) Let ƒ(x) 5 500(1.03) x and g(x) 5 4500. a. Using technology, graph the two functions on the same screen. b. Estimate the point of intersection. c. Solve the equation 4500 5 500(1.03)x using logarithms. d. Compare your answers. 22. (Requires a graphing program.) Let ƒ(x) 5 100(0.8)x and g(x) 5 10. a. Using technology, graph the two functions on the same screen. b. Estimate the point of intersection.
339
c. Solve the equation 10 5 100(0.8)x using logarithms. d. Compare your answers. 23. Find the doubling time or half-life for each of the following functions (where x is in years). a. ƒ(x) 5 100 ? 4 x x b. g(x) 5 100 ? A14B c. h(x) 5 A(4) x (Hint: Set h(x) 5 2A)
d. j(x) 5 A ? A14B
x
(Hint: Set j(x) 5 12 A)
24. The yearly per capita consumption of whole milk in the United States reached a peak of 40 gallons in 1945, at the end of World War II. It has been steadily decreasing at a rate of about 2.8% per year. a. Construct an exponential model M(t) for per capita whole milk consumption (in gallons) where t 5 years since 1945. b. Use your model to estimate the year in which per capita whole milk consumption dropped to 7 gallons per person. How does this compare with the actual consumption of 7 gallons per person in 2005? c. What might have caused this decline? 25. Wikipedia is a popular, free online encyclopedia (at en.wikipedia.org) that anyone can edit. (So articles should be taken with “a grain of salt.”) One Wikipedia article claims that the number of articles posted on Wikipedia has been growing exponentially since October 23, 2002. At that date there were approximately 90 thousand articles posted, and the growth rate was about 0.2% per day. a. Create an exponential model for the growth in Wikipedia articles. b. What is the doubling time? Interpret your answer. 26. If the amount of drug remaining in the body after t hours is given by ƒ(t) 5 100 A12B t/2(graphed in Exercise 3), then calculate: a. The number of hours it would take for the initial 100 mg to become: i. 60 mg ii. 40 mg iii. 20 mg b. The half-life of the drug. 27. In Chapter 5 we saw that the function N 5 N 0 ? 1.5 t described the actual number N of E. coli bacteria in an experiment after t time periods (of 20 minutes each) starting with an initial bacteria count of N0. a. What is the doubling time? b. How long would it take for there to be ten times the original number of bacteria? 28. (Requires a graphing program.) A woman starts a training program for a marathon. She starts in the first week by doing 10-mile runs. Each week she increases her run length by 20% of the distance for the previous week. a. Write a formula for her run distance, D, as a function of week, W. b. Use technology to graph your function, and then use the graph to estimate the week in which she will reach a marathon length of approximately 26 miles.
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c. Now use your formula to calculate the week in which she will start running 26 miles.
H
31. A department store has a discount basement where the policy is to reduce the selling price, S, of an item by 10% of its current price each week. If the item has not sold after the tenth reduction, the store gives the item to charity. a. For a $300 suite, construct a function for the selling price, S, as a function of week, W. b. After how many weeks might the suite first be sold for less than $150? What is the selling price at which the suite might be given to charity? 32. If you drop a rubber ball on a hard, level surface, it will usually bounce repeatedly. (See the accompanying graph at the top of the next column.) Each time it bounces, it rebounds to a height that is a percentage of the previous height. This percentage is called the rebound height. a. Assume you drop the ball from a height of 5 feet and that the rebound height is 60%. Construct a table of values that shows the rebound height for the first four bounces. b. Construct a function to model the ball’s rebound height, H, on the nth bounce. c. How many bounces would it take for the ball’s rebound height to be 1 foot or less? d. Construct a general function that would model a ball’s rebound height H on the nth bounce, where H 0 is the initial height of the ball and r is the ball’s rebound height.
n 1
2 3 4 5 Number of bounces
6
33. The accompanying graph shows fish and shellfish production (in million of pounds) by U.S. companies. Fish and Shellfish Production by U.S. Companies 4000 3500 3000 Millions of pounds
30. The atmospheric pressure at sea level is approximately 14.7 lb/in.2, and the pressure is reduced by half for each 3.6 miles above sea level. a. Construct a model that describes the atmospheric pressure as a function of miles above sea level. b. At how many miles above sea level will the atmospheric pressure have dropped to half, i.e., to 7.35 lb/in.2?
Height
29. The half-life of bismuth-214 is about 20 minutes. a. Construct a function to model the decay of bismuth-214 over time. Be sure to specify your variables and their units. b. For any given sample of bismuth-214, how much is left after 1 hour? c. How long will it take to reduce the sample to 25% of its original size? d. How long will it take to reduce the sample to 10% of its original size?
2500 2000 1500 1000 500 0 1940
1950
1960
1970
1980
1990
2000
2010
Year
Source: U.S. Department of Agriculture, Economic Research Service, Data set on food availability, www.ers.usda.gov.
The growth appears to be exponential between 1970 and 1995. The exponential function F(t) models that growth, where F(t) 5 1355(1.036)t where F(t) represents millions of pounds of fish and shellfish produced by U.S. companies since 1970 (where 0 # t # 25). a. Suppose we used the model to predict when fish production would reach 4000 million pounds. What year would that be? Using the graph, estimate the actual production in that year. b. What might have caused the decline in U.S. fish production after 1995? Do you think that America’s appetite for fish has waned?
6.2 Base e and Continuous Compounding What is e? Any positive number can be used as the base for an exponential or logarithmic expression. However, there is a “natural” base called e that is used in scientific applications. This number was named after Euler, a Swiss mathematician, and its value
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is approximately 2.71828. (You can use 2.72 as an estimate for e in most calculations. A scientific calculator has an ex key for more accurate computations.) The number e is irrational; it cannot be written as the quotient of two integers or as a repeating decimal. Like p, the number e is a fundamental mathematical constant. The irrational number e is a fundamental mathematical constant whose value is approximately 2.71828.
We first learn why e is important, and then how to write any exponential function using e.
Continuous Compounding The number e arises naturally in cases of continuous growth at a specified rate. For example, suppose we invest $100 (called the principal) in a bank account that pays interest of 6% per year. To compute the amount of money we have at the end of 1 year, we must also know how often the interest is credited (added) to our account, that is, how often it is compounded. What is compound interest? Compounding Annually. If we invest $100 in a bank account that pays 6% interest per year, then at the end of one year we would have $100(1.06) 5 $106. When the interest is applied to your account once a year, we say that the interest is compounded annually. Compounding Twice a Year. Now suppose that the interest is compounded twice a year. This means that instead of applying the annual rate of 6% once, it is divided by 2 and applied twice, at the end of each 6-month period. At the end of the first 6 months we earn 6%/2 or 3% interest, so our balance is $100(1.03) 5 $103. At the end of the second 6 months we earn 3% interest on our new balance of $103. So, after 1 year we have $100 at 6% interest compounded twice a year 5 $100 ? (1.03) ? (1.03) 5 $100 ? (1.03) 2 5 $106.09 We earn 9 cents more when interest is credited twice per year than when it is credited once per year. The difference is a result of the interest earned during the second halfyear on the $3 in interest credited at the end of the first half-year. In other words, we’re starting to earn interest on interest. To earn the same amount with only annual compounding, we would need an interest rate of 6.09%. When 6% interest is compounded twice a year, then 6%
is the nominal interest rate (in name only) or the annual percentage rate (APR)
and 6.09%
is the effective interest rate or the annual percentage yield (APY)
The effective interest rate is how much you actually earn (or pay) on an account. Banks and credit card companies are required by law to list both the nominal (the APR) and effective (the APY) interest rates. Compounding Four Times a Year. Next, suppose that interest is compounded quarterly, or four times per year. In each quarter, we receive one-quarter of 6%, or 1.5% interest. Each quarter, our investment is multiplied by 1 1 0.015 5 1.015 and,
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after the first quarter, we earn interest on the interest we have already received. At the end of 1 year we have received interest four times, so our initial $100 investment has become $100 ? (1.015)4 < $106.14 In this case, the effective interest rate (or annual percentage yield) is about 6.14%. Compounding n Times a Year. We may imagine dividing the year into smaller and smaller time intervals and computing the interest earned at the end of 1 year. The effective interest rate will be slightly more each time (Table 6.3). Investing $100 for One Year at a Nominal Interest Rate of 6% Number of Times Interest Computed During the Year 1 2 4 6 12 24 ? ? ? n
Value of $100 at End of One Year ($) 100(1 1 0.06) 5 100(1.06) 5 106.00 100(1 1 0.06/2)2 5 100(1.03)2 5 100(1.0609) 5 106.09 100(1 1 0.06/4)4 5 100(1.015)4 < 100(1.0614) < 106.14 100(1 1 0.06/6)6 5 100(1.010)6 < 100(1.0615) < 106.15 100(1 1 0.06/12)12 5 100(1.005)12 < 100(1.0617) < 106.17 100(1 1 0.06/24)24 5 100(1.0025)24 < 100(1.0618) < 106.18
Effective Annual Interest Rate (%) 6.00 6.09 6.14 6.15 6.17 6.18
100(1 1 0.06/n)n
Table 6.3
In general, if we calculate the interest on $100 n times a year when the nominal interest rate is 6%, we get 0.06 n $100a1 1 b n At the End of t Years. What if we invest $100 for t years at a nominal interest rate of 6% compounded n times a year? The annual growth factor is (1 1 0.06/n)n; that is, every year the $100 is multiplied by (1 1 0.06/n)n. After t years the $100 is multiplied by (1 1 0.06/n)n a total of t times, or equivalently, multiplied by [(1 1 0.06/n)n]t 5 (1 1 0.06/n)nt. So $100 will be worth $100(1 1 0.06/n)nt We can generalize our results: Compounding n Times a Year for t Years The value P of P0 dollars (called the principal) invested at a nominal interest rate r (expressed in decimal form) compounded n times a year for t years is r nt P 5 P0 a1 1 b n Continuous compounding using e Imagine increasing the number of periods, n, without limits, so that interest is computed every week, every day, every hour, every second, and so on. The surprising result is that the term by which $100 gets multiplied, namely, a1 1
0.06 n b n
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does not get arbitrarily large. Examine Table 6.4. Value of (1 + 0.06/n)n as n Increases Compounding Period
n (Number of Compoundings per Year)
Once a day Once an hour Once a minute Once a second
365 365 ? 24 5 8,760 365 ? 24 ? 60 5 525,600 365 ? 24 ? 60 ? 60 5 31,536,000
Approximate Value of (1 1 0.06/n)n 1.0618313 1.0618363 1.0618365 1.0618365
Table 6.4
Where does the irrational constant e fit in? As n, the number of compounding periods per year, increases, the value of (1 1 0.06/n)n approaches 1.0618365 < e0.06. You can confirm this on your calculator. As n gets arbitrarily large, we can think of the compounding occurring at each instant. We call this continuous compounding. If we invest $100 at 6% continuously compounded, at the end of 1 year we will have $100 ? e0.06 < $100 ? 1.0618365 5 $106.18365 When 6% annual interest is compounded continuously, then 6% and
6.18365%
is the nominal interest rate (in name only) or the annual percentage rate (APR) is the effective interest rate or the annual percentage yield (APY), which tells you how much interest you will earn after one year.
At the End of t Years. If the interest is compounded continuously, the annual growth factor is e 0.06 ; that is, every year the $100 is multiplied by e 0.06 . After t years the $100 is multiplied by e0.06 a total of t times, or equivalently multiplied by (e0.06)t 5 e0.06t. So $100 will be worth $100e0.06t Hence, if we invest $100 over t years at a nominal interest rate of 6%, we will have $100(1 1 0.06/n)nt $100e0.06t
if the interest is compounded n times a year if the interest is compounded continuously
Just as (1 1 0.06/n)n approaches e0.06 as n gets very large, (1 1 r/n)n approaches er. So if P0 dollars are invested at an annual interest rate r (in decimal form) compounded continuously, then after t years we have P0 ? (er ) t 5 P0 ? ert
Compounding Continuously for t Years The value P of P0 dollars invested at a nominal interest rate r (expressed in decimal form) compounded continuously for t years is P 5 P0 ? e rt
EXAMPLE
1
If you have $250 to invest, and you are quoted a nominal interest rate of 4%, construct the equations that will tell you how much money you will have if the interest is compounded once a year, quarterly, once a month, or continuously. In each case calculate the value after 10 years.
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SOLUTION
See Table 6.5. Investing $250 at a Nominal Interest Rate of 4% for Different Compounding Intervals Number of Compoundings per Year 1 4 12 Continuous
$ Value after t Years 250 ? (1 1 0.04)t 5 250 ? (1.04)t 250 ? (1 1 0.04/4)4t 5 250 ? (1.01)4t 250 ? (1 1 0.04/12)12t < 250 ? (1.00333)12t 250 ? e0.04t
Approximate $ Value When t 5 10 Years 370.06 372.22 372.71 372.96
Table 6.5
EXAMPLE
2
Continuously compounding debt Suppose you have a debt on which the nominal annual interest rate (APR) is 7% compounded continuously. What is the effective interest rate (APY)?
SOLUTION
The nominal interest rate (APR) of 7% is compounded continuously, so the equation P 5 P0 e0.07t describes the amount P that an initial debt P0 becomes after t years. Using a calculator, we find that e0.07 < 1.073. The equation could be rewritten as P 5 P0 (1.073) t. So the effective interest rate (APY) on your debt is about 7.3% per year.
EXAMPLE
3
You have a choice between two bank accounts. One is a passbook account in which you receive simple interest of 5% per year, compounded once per year. The other is a 1-year certificate of deposit (CD), which pays interest at the rate of 4.9% per year, compounded continuously. Which account is the better deal?
SOLUTION
Since the interest on the passbook account is compounded once a year, the nominal and effective interest rates are both 5%. The equation P 5 P0 s1.05d t can be used to describe the amount P that the initial investment P0 is worth after t years. The 1-year certificate of deposit has a nominal interest rate of 4.9%. Since this rate is compounded continuously, the equation P 5 P0 e 0.049t describes the amount P that the initial investment P0 is worth after t years. Since e0.049 < 1.0502, the equation can also be written as P 5 P0 (1.0502)t, and the effective interest rate is 5.02%. So the CD is a better deal.
Exponential Functions Base e The notation of continuous compounding is useful in scientific as well as financial contexts. We can convert any exponential function in the form ƒ(t) 5 Cat into a continuous growth (or decay) function using a power of e as the base. Since a . 0, we can always find a value for k such that a 5 ek So we can rewrite the function f as ƒ(t) 5 C(e k) t 5 Ce kt In general applications, we call k the instantaneous or continuous growth (or decay) rate. The value of k may be given as either a decimal or a percent. If k . 0, then the function represents exponential growth. Why is this true? If an exponential function represents growth, then the growth factor a . 1. If we rewrite a as e k and 1 as e0, then ek . e0, so k . 0. For example, the equation P(t) 5 100 e0.06 t could describe the growth of 100 cells with a continuous growth rate of 0.06 or 6% per time period t.
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If k , 0, then the function represents exponential decay. For exponential decay, the decay factor a is such that 0 , a ,1. Rewriting a as e k and 1 as e0, we have 0 , e k , e0. We know the value of e k is always . 0 since e is a positive number. But if e k , e0, then k , 0. For example, the function Q(t) 5 50e20.03 t could describe the decay of 50 cells with a continuous decay rate of 0.03 or 3% per time period t. Continuous Growth and Decay If Q(t) 5 Ce kt, then k is called the instantaneous or continuous growth (or decay) rate. For exponential growth, k is positive. For exponential decay, k is negative.
EXAMPLE
4
Continuous growth or decay rates Identify the continuous growth (or decay) rate for each of the following functions and graph each function using technology. ƒ(t) 5 100 ? e0.055t g(t) 5 100 ? e0.02t h(t) 5 100 ? e20.055t j(t) 5 100 ? e20.02t
SOLUTION
The function f has a continuous growth rate of 0.055 or 5.5% and g has a continuous growth rate of 0.02 or 2%. The function h has a continuous decay rate of 0.055 or 5.5% and j has a continuous decay rate of 0.02 or 2%. The graphs of these four functions are shown in Figure 6.3. y 200
100
g
j f
h t
–100
0
100
Figure 6.3 Graphs of four
exponential functions.
Converting ek into a Using the rules of exponents, we can rewrite e kt as (e k)t. When we know the value of k, we can calculate the value of e k. For example, Rule 3 of exponents use a calculator to evaluate e0.06
P 5 100 ? e0.06t 5 100 ? se0.06 d t < 100 ? 1.0618t
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The two functions P 5 100 ? e0.06t
and
P 5 100 ? 1.0618 t
are equivalent. The first function (base e) suggests growth that occurs continuously throughout a time period, so we call 0.06 or 6% the continuous growth rate per time period t. The other function suggests growth that happens all at once at the end of each time period, so 0.0618 or 6.18% is just called the growth rate per time period t. To do the reverse, that is, convert any base a into e k, we need to know about logarithms base e, which we’ll meet in the next section.
EXAMPLE
5
For each of the following functions, identify the continuous growth (or decay) rate per year and the growth (or decay) rate based on the growth (or decay) factor. Assume t is measured in years. a. ƒ(t) 5 240 ? e0.127t b. g(t) 5 5700 ? e20.425t
SOLUTION
a. The function ƒ(t) has a continuous growth rate of 0.127 or 12.7% per year. To find the growth factor, we need to convert ƒ(t) into the form ƒ(t) 5 Cat. To do this we need to evaluate e0.127. Using a calculator, we find that e0.127 < 1.135. So ƒ(t) 5 240 ? e0.127t can be rewritten as ƒ(t) 5 240 ? 1.135 t Since the growth factor a 5 1.135, the growth rate is 0.135 or 13.5% per year. b. The function g(t) has a continuous decay rate of 0.425 or 42.5% per year. To find the decay factor, we need to convert g(t) into the form g(t) 5 Ca t . To do this, we need to evaluate e20.425. Using a calculator, we find that e20.425 < 0.654. So g(t) 5 5700 ? e20.425t can be rewritten as g(t) 5 5700 ? 0.654 t So the decay factor a 5 0.654 and the decay rate is 1 2 0.654 5 0.346 or 34.6% per year.
EXAMPLE
6
The cost of bottled water In 1976 approximately 0.28 billion gallons of bottled water were sold in the United States, according to Beverage Marketing Corp., a New York research and consulting firm. Between 1976 and 2004 the bottled water industry in the United States had a continuous growth rate of about 11.5% a year. a. Construct a model that represents the continuous growth of bottled water sales between 1976 and 2004. b. Beverage Marketing later reported that in 2005 sales of bottled water were nearly 7.5 billion gallons. If we extrapolate, what would the model predict for sales in 2005? How does this compare with the actual sales?
SOLUTION
a. To represent continuous growth, we construct an exponential function using base e and a continuous growth rate of 11.5% a year. If we have an initial value of 0.28 billion gallons and t 5 number of years since 1976, then ƒ(t) 5 0.28 ? e0.115t
(0 # t # 28)
models the continuous growth of bottled water sales in the United States between 1976 and 2004.
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b. If we extrapolate our model to predict sales for the year 2005, then t 5 2005 2 1976 5 29 years and ƒ(29) 5 0.28 ? e(0.115)(29) < 0.28 ? 28.1 < 7.86 billion gallons Our model’s prediction is somewhat over the actual sales of nearly 7.5 billion gallons in 2005. You can check the current sales of bottled water at www.beveragemarketing.com.
Algebra Aerobics 6.2
3. Assume that each of the following describes the value of an investment, A, over t years. Identify the principal, nominal rate, effective rate, and number of interest periods per year. d. A 5 50,000 ? 1.0252t a. A 5 6000 ? 1.05t 4t b. A 5 10,000 ? 1.02 e. A 5 125e0.076t c. A 5 500 ? 1.0112t 4. Fill in the missing values, translating from ek to a (the growth or decay factor). a. 5e0.03t 5 5( ___ )t b. 3500e0.25t 5 3500( ___ )t c. 660e1.75t 5 660( ___ )t d. 55,000e20.07t 5 55,000( ___ )t e. 125,000e20.28t 5 125,000( ___ )t 5. The value for e is often defined as the number that (1 1 1/n)n approaches as n gets arbitrarily large. Use your calculator to complete Table 6.6 at the bottom of the page. Use your exponent key (x y or y x) to evaluate the last column. Is your value consistent with the approximate value for e of 2.71828 given in the text?
6. If a principal of $10,000 is invested at the rate of 12% compounded quarterly, the amount accumulated at the end of t years is given by the formula A 5 10,000 ? a1 1
$50,000
A
$40,000 $30,000 $20,000 $10,000
t 0
5
10
15
Years
Figure 6.4 Graph of A 5 10,000(1.03)4t.
a. 1 year b. 5 years c. 10 years d. Use the graph to estimate the number of years it will take to double the original investment. e. Use the equation to calculate the amount A after the years specified in parts (a)–(c) and the doubling time for A.
n
1/n
1 1 (1/n)
1 100 1,000 1,000,000 1,000,000,000
1 0.01
11152 1 1 0.01 5 1.01
Table 6.6
0.12 4t b 5 10,000s1.03d 4t 4
The graph of this function is given in Figure 6.4. Use the graph to estimate for parts (a)–(c) the amount, A, accumulated after:
Value of investment
Most of these problems require a calculator that can evaluate powers of e and, for Problem 4, evaluate logs. 1. Find the amount accumulated after 1 year on an investment of $1000 at 8.5% compounded: a. Annually b. Quarterly c. Continuously 2. Find the effective interest rate for each given nominal interest rate that is compounded continuously. a. 4% b. 12.5% c. 18%
[1 1 (1/n)]n 21 5 2 (1.01)100 < 2.7048138
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7. At birth, Maria’s parents set aside $8000 in an account designated to help pay for her college education. How much will Maria’s account be worth by her 18th birthday if the interest rate was: a. 8% compounded quarterly? b. 8% compounded continuously? c. 8.4% compounded annually? 8. Suppose Maria (from the previous problem) earns a full scholarship and is able to save all the money in her
college account. How much would be in her account on her 30th birthday for each of the rates above? 9. For each of the following functions, identify the continuous growth rate and then determine the effective growth rate. Assume t is measured in years. a. A(t) 5 Pe0.6t b. N(t) 5 N0 e2.3t 10. For each of the following functions, identify the continuous decay rate and then determine the effective decay rate. Assume t is measured in years. b. P(t) 5 P0 e20.15t a. Q(t) 5 Q0 e20.055t
Exercises for Section 6.2 Many of these exercises require a calculator that can evaluate powers (including powers of base e) and logs. One (Exercise 14) requires a graphing program. 1. Use a calculator to find the value of each expression to four decimal places. 0.035 2 a. a1 1 d. e0.035 b 2 b. a1 1 c. a1 1
0.035 4 b 4 0.035 b 12
e. How are these values related?
12
b. $25,000 is invested at 5.75% compounded continuously. c. What is the amount in each account at the end of 5 years? d. Explain, using the concept of effective rates, why one amount is larger than the other. 6. Fill in the following chart assuming that the principal is $10,000 in each case.
Nominal Interest Rate (APR) 5.25%
2. Assume y 5 Pe rx represents P dollars invested at an annual interest rate r (in decimal form) compounded continuously for x years. Then for each of the following, calculate the nominal and effective rates in percentage form.
Compounding Period
Expression for the Value of Your Account after t Years
Effective Interest Rate (APY)
Monthly 10,000(1 1 0.045/4)4t Semiannually
3.25%
8.16%
Daily 10,000(1.02)t
Nominal Rate (APR)
Effective Rate (APY) (Hint: Evaluate at ek 5 a)
a. y 5 Pe0.025x b. y 5 Pe0.039x c. y 5 Pe0.062x
7. Insert the symbol .,,, or < to make the statement true. d. e20.10 ___ 0.90 a. e0.045 ___ 1.046 0.068 e. 0.8607 ___ e20.15 b. 1.068 ___ e c. 1.269 ___ e0.238
3. Assume $10,000 is invested at a nominal interest rate of 8.5%. Write the equations that give the value of the money after n years and determine the effective interest rate if the interest is compounded: a. Annually c. Quarterly b. Semiannually d. Continuously
8. Assume that $5000 was put in each of two accounts. Account A gives 4% interest compounded semiannually. Account B gives 4% compounded continuously. a. What are the total amounts in each of the accounts after 10 years? b. Show that account B gives 0.04% annually more interest than account A.
4. Assume you invest $2000 at 3.5% compounded continuously. a. Construct an equation that describes the value of your investment at year t. b. How much will $2000 be worth after 1 year? 5 years? 10 years? 5. Construct functions for parts (a) and (b) and compare them in parts (c) and (d). a. $25,000 is invested at 5.75% compounded quarterly.
9. The half-life of uranium-238 is about 5 billion years. Assume you start with 10 grams of U-238 that decays continuously. a. Construct an equation to describe the amount of U-238 remaining after x billion years. b. How long would it take for 10 grams of U-238 to become 5 grams? 10. You want to invest money for your newborn child so that she will have $50,000 for college on her 18th birthday. Determine
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how much you should invest if the best annual rate that you can get on a secure investment is: a. 6.5% compounded annually b. 9% compounded quarterly c. 7.9% compounded continuously 11. Determine the doubling time for money invested at the rate of 12% compounded: a. Annually b. Quarterly 12. a. Phosphorus-32 is used to mark cells in biological experiments. If phosphorus-32 has a continuous daily decay rate of 0.0485 or 4.85%, what is its half-life? (Hint: Rewrite the function as y Cax and set y 0.5C.) b. Phosphorus-32 can be quite dangerous to work with if the experimenter fails to use the proper shields, since its highenergy radiation extends out to 610 cm or about 20 feet. Because disposal of radioactive wastes is increasingly difficult and expensive, laboratories often store the waste until it is within acceptable radioactive levels for disposal with non-radioactive trash. For instance, the rule of thumb for the laboratories of a large East Coast university and medical center is that any waste containing radioactive material with a half-life under 65 days must be stored for 10 half-lives before disposal with the non-radioactive trash. i. For how many days would phosphorus-32 have to be stored? ii. What percentage of the original phosphorus-32 would be left at that time? 13. A city of population 1.5 million is expected to experience a 15% decrease in population every 10 years. a. What is the 10-year decay factor? What is the yearly decay factor? The yearly decay rate? b. Use part (a) to create an exponential population model g(t) that gives the population (in millions) after t years. c. Create an exponential population model h(t) that gives the population (in millions) after t years, assuming a 1.625% continuous yearly decrease.
349
d. Compare the populations predicted by the two functions after 20 years. What can you conclude? 14. (Requires a graphing program.) Using technology, graph the functions ƒ(x) 5 15,000e0.085x and g(x) 5 100,000 on the same grid. a. Estimate the point of intersection. (Hint: Let x go from 0 to 60.) b. If ƒ(x) represents the amount of money accumulated by investing at a continuously compounded rate (where x is in years), explain what the point of intersection represents. 15. Rewrite each continuous growth function in its equivalent form ƒ(t) 5 Cat. In each case identify the continuous growth rate, and the effective growth rate. (Assume that t is in years.) c. Q(t) 5 45e0.06t a. P(t) 5 500e0.02t 1.5t b. N(t) 5 3000e d. G(t) 5 750e0.035t 16. Rewrite each continuous decay function in its equivalent form ƒ(t) 5 Cat. In each case identify the continuous decay rate and the effective decay rate. (Assume that t is in years.) c. Q(t) 5 7145e20.06t a. P(t) 5 600e20.02t 20.5t b. N(t) 5 30,000e d. G(t) 5 750e20.035t 17. Find the nominal interest rate (APR) if a bank advertises that the effective interest rate (APY) on an account compounded continuously is: a. 3.43% on a checking account b. 4.6% on a savings account 18. An investment pays 6% compounded four times a year. a. What is the annual growth factor? b. What is the annual growth rate? c. Develop a formula to represent the total value of the investment after each compounding period. d. If you invest $2000 for a child’s college fund, how much will it total after 15 years? e. For how many years would you have to invest to increase the total to $5000?
6.3 The Natural Logarithm The common logarithm uses 10 as a base. The natural logarithm uses e as a base and is written ln x rather than loge x. Scientific calculators have a key that computes ln x.
The Natural Logarithm The logarithm base e of x is the exponent of e needed to produce x. Logarithms base e are called natural logarithms and are written as ln x. ln x 5 c
means that
ec 5 x (x . 0)
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The properties for natural logarithms (base e) are similar to the properties for common logarithms (base 10). Like the common logarithm, ln A is not defined when A # 0. If A and B are positive real numbers and p is any real number, then the following rules hold.
Rules of Common Logarithms 1. log(A ? B) 5 log A 1 log B 2. log(A/B) 5 log A 2 log B
3. log Ap 5 p log A 4. log 1 5 0 (since 100 5 1)
Rules of Natural Logarithms 1. ln(A ? B) 5 ln A 1 ln B 2. ln(A/B) 5 ln A 2 ln B
3. ln Ap 5 p ln A 4. ln 1 5 0 (since e0 5 1)
In the following examples we show how to use natural logarithms to manipulate expressions and solve exponential equations.
EXAMPLE
1
Effective vs. nominal rates If the effective annual interest rate on an account is 5.21%, estimate the nominal annual interest rate that is compounded continuously.
SOLUTION
A 5.21% effective interest rate is 0.0521 in decimal form. So the equation P 5 P 0 (1.0521) x represents the value P of P 0 dollars after x years. To find the nominal interest rate that is continuously compounded, we must convert (1.0521)x into the form e rx or (e r ) x . So e r must equal 1.0521. We need to solve for r in the equation. Take ln of both sides use a calculator definition of ln
1.05215 er ln 1.0521 5 ln er 0.0508 < ln er 0.0508 < r
So 1.0521 < e0.0508. A nominal interest rate of 0.0508 or 5.08% compounded continuously is equivalent to an effective interest rate of 0.0521 or 5.21%.
EXAMPLE
2
SOLUTION
EXAMPLE
3
Expand, using the laws of logarithms, the expression: ln
x13 . 22
Åx
Rewrite using exponents
x 1 3 1/2 x13 5 lna b Åx 2 2 x22
ln
Rule 3 of ln
5
1 x13 lna b 2 x22
Rule 2 of ln
5
1 [ln(x 1 3) 2 ln(x 2 2)] 2
Contract, expressing the answer as a single logarithm:
1 3
lnsx 2 1d 1
1 3
lnsx 1 1d
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SOLUTION
EXAMPLE
4
SOLUTION
lnsx 1 1d 5
1 3
[lnsx 2 1d 1 lnsx 1 1d]
Rule 1 of ln
5
1 3
ln[sx 2 1dsx 1 1d]
Rule 3 of ln
5 ln[sx 2 1dsx 1 1d]1/3
multiply binomials
3 2 5 lnsx 2 2 1d 1/3 or ln " x 21
Distributive property
1 3
lnsx 2 1d 1
1 3
351
Solve the equation 10 5 e t for t. 10 5 e t ln 10 5 ln e t ln 10 5 t t < 2.303
Given take ln of both sides definition of ln evaluate and switch sides
Algebra Aerobics 6.3 Problems 4, 5, 6, and 8 require a calculator that can evaluate natural logs. 1. Evaluate without a calculator: 1 a. ln e2 d. ln 2 e b. ln 1 e. ln "e 1 c. ln e 2. Expand the following: a. ln "xy
c. ln A sx 1 yd 2 sx 2 yd B
3x 2 "x 1 2 b d. ln y3 xsx 2 1d 3. Contract, expressing your answer as a single logarithm: a. ln x 1 ln(x 2 1) d. 12 ln sx 1 yd b. ln(x 1 1) 2 ln x e. ln x 2 2 ln(2x 2 1) c. 2 ln x 2 3 ln y b. lna
4. Find the nominal rate on an investment compounded continuously if the effective rate is 6.4%. 5. Determine how long it takes for $10,000 to grow to $50,000 at 7.8% compounded continuously. 6. Solve the following equations for x. a. e x11 5 10 b. e x22 5 0.5 7. Determine which of the following are true statements. If the statement is false, rewrite the right-hand side so that the statement becomes true. a. ln 81 0 4 ln 3 d. 2 ln 10 0 ln 20 ln 14 b. ln 7 0 e. ln "e 0 12 ln 2 c. ln 35 0 ln 5 1 ln 7 f. 5 ln 2 0 ln 25 8. Use the rules of logarithms to contract each expression into a single logarithm (if necessary), then solve for x. a. ln 2 1 ln 6 5 x c. ln (x 1 1) 5 0.9 b. ln 2 1 ln x 5 2.48 d. ln 5 2 ln x 5 20.06
Exercises for Section 6.3 Some of these exercises require a calculator that can evaluate powers and logs. 1. Determine the rule(s) of logarithms that were used to expand each expression. a. ln 15 5 ln 3 1 ln 5 b. ln 15 5 ln 30 2 ln 2 c. ln 49 5 2 ln 7 d. ln 25z3 5 2 ln 5 1 3 ln z e. ln 5x4 5 ln 5 1 4 ln x 125 f. ln a b 5 3 ln 5 2 (ln 3 1 ln x) 3x
2. Expand each logarithm using only the numbers 2, 5, ln 2, and ln 5. a. ln 25 b. ln 250 c. ln 625 3. Write an equivalent expression using exponents. a. n 5 log 35 c. ln x 5 34 b. ln 75 5 x
d. ln QNN0R 5 2kt
4. Write an equivalent equation in logarithmic form. c. e3x 5 27 a. N 5 102t /c 2k/x b. I 5 I0 ? e d. 12 5 e2k t
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5. Use rules of logarithms to find the value of x. Verify your answer with a calculator. a. ln x 5 ln 2 1 ln 5 b. ln x 5 ln 24 2 ln 2 c. ln x2 5 2 ln 11 d. ln x 5 3 ln 2 1 2 ln 6 e. ln x 5 6 ln 2 2 2 ln 3 f. ln x 5 4 ln 2 2 3 ln 2 6. Use rules of logarithms to contract to a single logarithm. Use a calculator to verify your answer. a. 2 ln 3 1 4 ln 2 c. 2(ln 4 2 ln 3) b. 3 ln 7 2 5 ln 3 d. 24 ln 3 1 ln 3 7. Use rules of logarithms to expand. 3 " 2x b a. ln Q "4xyR b. ln a 4
4 c. ln Q3 ? "x 3R
8. Use rules of logarithms to contract to a single logarithm. a. 12 ln x 2 5 ln y c. ln 2 1 31 ln x 2 4 ln y 4 2 b. 5 ln x 1 5 ln y 9. Contract, expressing your answer as a single logarithm. a. 41 ln (x 1 1) 1 14 ln (x 2 3) b. 3 ln R 2 12 ln P c. ln N 2 2 ln N0 10. Expand: a. ln
3
x2 2 1 12
Çx
b. ln a
x y "2
b
2
c. ln a
K 2L b M11
11. Solve for the indicated variable, by changing to logarithmic form. Round your answer to three decimal places. a. er 5 1.0253
c. 12 5 e 3x
b. 3 5 e0.5t
12. Solve for x by changing to exponential form. Round your answer to three decimal places. a. ln 3x 5 1 b. 3 ln x 5 5 c. ln 3 1 ln x 5 1.5 13. Solve for x. a. e x 5 10 b. 10 x 5 3
c. 2 1 4x 5 7 d. ln x 5 5
14. Solve for t. a. 5t11 5 6t 2
b. e t 5 4 c. 5 ? 22t 5 4
e. ln(x 1 1) 5 3 f. ln x 2 ln(x 1 1) 5 4
t b 51 t22 e. ln t 2 ln(t 2 2) 5 1
d. ln a
15. Prove that ln(A ? B) 5 ln A 1 ln B, where A and B are positive real numbers. 16. For each of the following equations, either prove that it is correct (by using the rules of logarithms and exponents) or else show that it is not correct (by finding numerical values for x that make the values on the two sides of the equation different). a. e x1ln x 5 x ? ex sx 1 1d 2 b. ln a b 5 2 ln(x 1 1) 2 ln x x c. ln a
x ln x b 5 x11 lnsx 1 1d
d. ln(x 1 x2) 5 ln x 1 ln x2 e. ln(x 1 x2) 5 ln x 1 ln(x 1 1) 17. How long would it take $15,000 to grow to $100,000 if invested at 8.5% compounded continuously? 18. The effective annual interest rate on an account compounded continuously is 4.45%. Estimate the nominal interest rate. 19. The effective annual interest rate on an account compounded continuously is 3.38%. Estimate the nominal interest rate. 20. A town of 10,000 grew to 15,000 in 5 years. Assuming exponential growth: a. What is the annual growth rate? b. What was its annual continuous growth rate? 21. Solve the following for r. b. 12 5 er
c. 1.08 5 er
22. Solve the following for r. b. 2 5 er a. 0.9 5 er
c. 0.75 5 er
a. 1.025 5 er
23. Without using a calculator, identify which functions represent growth and which decay. Then, using a calculator, find the corresponding instantaneous growth or decay rate. a. ƒ(t) 5 50e(ln 1.45) t d. j(p) 5 2000e(ln 0.3) p (ln 3.15) x b. g(x) 5 125e e. k(v) 5 600e(ln 1.75)v c. h(m) 5 1500e(ln 0.83) m f. l(x) 5 e(ln 0.75) x 24. Use a calculator to determine the value of each expression. Then rewrite each expression in the form eln a. b. e20.025 c. e20.35 d. e0.83 a. e0.083
6.4 Logarithmic Functions
Exploration 6.1 and course software “E8: Logarithmic Sliders” in Exponential & Log Functions will help you understand the properties of logarithmic functions.
Up until now we have been dealing with logarithms of specific numbers, such as log 2 or ln 10. But since for any x . 0 there is a unique corresponding value of log x or ln x, we can define two logarithmic functions: y 5 log x
and
y 5 ln x
where x . 0
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The Graphs of Logarithmic Functions What will the graphs look like? We know something about the graphs since Properties of logarithms If x . 1, log x and ln x are both positive If x 5 1, log 1 5 0 and ln 1 5 0 If 0 , x , 1, log x and ln x are both negative If x # 0, neither logarithm is defined Table 6.7 and Figure 6.5 show some data points and the graphs of y 5 log x and y 5 ln x. Evaluating log x and ln x x 0.001 0.01 0.1 1 2 3 4 5 6 7 8 9 10
y 5 log x 23.000 22.000 21.000 0.000 0.301 0.477 0.602 0.699 0.778 0.845 0.903 0.954 1.000
y 5 ln x 26.908 24.605 22.303 0.000 0.693 1.099 1.386 1.609 1.792 1.946 2.079 2.197 2.303
Table 6.7
3
y
y = ln x
y = log x x –1
10
–2
Figure 6.5 Graphs of y 5 log x
and y 5 ln x.
The graphs of common and natural logarithms share a distinctive shape. They are both defined only when x . 0 and they are both concave down. Both graphs increase throughout with no maximum or minimum value, although they grow more slowly when x . 1. Vertical asymptotes The graphs of both y 5 log x and y 5 ln x are vertically asymptotic to the y-axis. This means that as x approaches 0 (through positive values), the graphs come closer and closer to the vertical y-axis but never touch it. For both functions, as x ➝ 0, the values for y get increasingly negative, plunging down near the y-axis toward 2∞.
The Graphs of Logarithmic Functions The graphs of both y 5 log x and y 5 ln x • • • • •
lie to the right of the y-axis since they are defined only for x . 0 share a horizontal intercept of (1, 0) are concave down increase throughout with no maximum or minimum are asymptotic to the y-axis
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EXAMPLE
1
Match each function with the appropriate graph in Figure 6.6. a. y 5 log (3x) b. y 5 3 1 log x c. y 5 log (x3) y
y
y
4
4
y
4
4
x
x –1
–1
7
d. y 5 3 log x
7
x –1
7
x –1
–4
–4
–4
–4
Graph A
Graph B
Graph C
Graph D
7
Figure 6.6 Four graphs involving logarithms.
SOLUTION
EXAMPLE
2
SOLUTION
a. Graph A. When x 5 1/3, then y 5 log (3 ? 1/3) 5 log 1 5 0, so the horizontal intercept is (1/3, 0). b. Graph C. When x 5 1, then y 5 3 1 log 1 5 3 1 0 5 3, so the graph passes through (1, 3). c. and d. Graph B. Since log (x3) 5 3 log x, the graphs for functions (c) and (d) are the same. When x 5 1, then y 5 3 log 1 5 0, so the horizontal intercept is (1, 0). There is no match for Graph D. Graph y 5 ln(x 2 2). Describe its relationship to y 5 ln x. Figure 6.7 shows the two graphs. y 3
y = ln x y = ln (x –2) x –1
10
–3
Figure 6.7 Graphs of y 5 ln x and y 5 ln(x22).
The function y 5 ln(x 2 2) tells us to subtract 2 from x and then apply the function ln. So the graph of y 5 ln(x 2 2) is the graph of y 5 ln x shifted two units to the right. For y 5 ln x, the horizontal intercept is at 1 since ln(1) 5 0. For y 5 ln(x 2 2), the horizontal intercept is at 3, since ln(3 2 2) 5 ln 1 5 0.
The Relationship between Logarithmic and Exponential Functions Logarithmic vs. exponential growth Logarithmic and exponential growth are both unbounded. They both increase forever, never reaching a maximum value. But exponential growth is rapid—not only increasing, but doing so at a rate that is speeding up (accelerating). Logarithmic growth is slow—increasing, but at rate that is slowing down (decelerating). For example, if we let t 5 1, 3, and 5, then 10t is respectively 10, 1000, and 100,000, but log(t) is respectively 0, 0.477, and 0.699.
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Logarithmic and exponential functions are inverses of each other What does it mean for two functions to be inverses of each other? It means that what one function does, the other undoes. Log and exponential functions are inverses of each other. For example, log (10 x) 5 x
and
10 log x 5 x
Why are these true? Rationale for log (10 x ) 5 x Finding the logarithm of a number base 10 involves finding the exponent of 10 needed to produce the number. Since 10x is already written as 10 to the power x, then log 10x 5 x. Rationale for 10 log x 5 x By definition, log x is the number such that when 10 is raised to that power the result is x. Proof
Let rewrite using definition of logarithm substitute log x for y
y 5 log x 10 y 5 x 10 log x 5 x
The two functions y 5 10 x
and
y 5 log x
are inverses of each other. Similarly, for natural logarithms ln(e x) 5 x So the functions
y 5 ex
and and
eln x 5 x y 5 ln x
are also inverses of each other.
More Rules for Common and Natural Logarithms 5. log(10 x ) 5 x and ln(e x ) 5 x log x 6. 10 5x and e ln x 5 x
The graphs of two inverse functions are mirror images across the diagonal line y = x The graphs of two inverse functions such as y 5 10 x and y 5 log x are mirror images across the line y 5 x. If you imagine folding the graph along the dotted line y 5 x, the two curves would lie right on top of each other. See Figure 6.8 on the following page. Why are these graphs mirror images? Choose any point (a, b) on the graph of y 5 log x. To reach that point you would need to move a units horizontally (on the x-axis) and b units vertically (on the y-axis). Now imagine folding at the dotted line y 5 x. What are the coordinates of the point’s mirror image? You would need to move a units vertically (on the y-axis) and b units horizontally (on the x-axis). The mirrorimage coordinates would be (b, a). So the points (a, b) and (b, a) are mirror images across the line y 5 x. Table 6.8 lists some pairs of mirror-image points that lie on the graphs of y 5 10 x and y 5 log x.
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10
y y = 10 x
y=x
Mirror-Image Points on y 5 10 x and y 5 log x
(b, a)
(a, b)
y = log x x 10
–2 –2
Figure 6.8 The graphs of y 5 log x and y 5 10 x are mirror images across the dotted line y 5 x.
y 5 10 x
y 5 log x
(0, 1) (1, 10) (2, 100) (21, 0.1) (22, 0.01)
(1, 0) (10, 1) (100, 2) (0.1, 21) (0.01, 22)
Table 6.8
We can use similar arguments to show that the graphs of y 5 ex and y 5 ln x are mirror images as well. Inverse Functions and Their Graphs The functions in each pair y 5 10 x y 5 ex
and and
y 5 log x y 5 ln x
are inverses of each other. What one does, the other undoes. The graphs in each pair are mirror images across the line y 5 x.
3
Graph y 5 ln x and y 5 e x. Identify three pairs of points on the function graphs that are mirror images across the line y 5 x.
SOLUTION
Figure 6.9 shows the graphs of y 5 ln x and y 5 e x, and Table 6.9 contains some pairs of points on y 5 e x and y 5 ln x.
EXAMPLE
10
y = ex
y=x
Mirror-Image Points on y 5 e x and y 5 ln x y = ln x
10
–2 –2
Figure 6.9 Graphs of y 5 ln x and
y 5 e x are mirror images across the dotted line y 5 x.
y 5 ex
y 5 ln x
(0, 1) (1, e) (22, e22)
(1, 0) (e, 1) (e22, 22)
Table 6.9
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Applications of Logarithmic Functions Just as with logarithmic scales, logarithmic functions are used in dealing with quantities that vary widely in size. We’ll examine two such functions used to measure acidity and noise levels. Measuring acidity: The pH scale Chemists use a logarithmic scale called pH to measure acidity. pH values are defined by the function pH 5 2log[H1] where [H1] designates the concentration of hydrogen ions. Chemists use the symbol H1 for hydrogen ions (hydrogen atoms stripped of their one electron), and the brackets [ ] mean “concentration of.” Ion concentration [H1] is measured in moles per liter, M, where one mole equals 6.022 ? 10 23 or Avogadro’s number of ions. A pH value is the negative of the logarithm of the number of moles per liter of hydrogen ions. Table 6.10 and Figure 6.10 show a set of values and a graph for pH. The graph is the standard logarithmic graph flipped over the horizontal axis because of the negative sign in front of the log.
10 pH > 7, basic 8
Calculating pH Values
6
pH = 7, neutral (pure water)
1
[H ] (moles per liter) 10215 10210 1 5 10 Table 6.10
pH 15.000 10.000 0.000 20.699 21.000
pH
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4
pH < 7, acidic
2 5
10
0 –2
Concentration of H+ ions (moles per liter)
Figure 6.10 Graph of the pH
function.
Typically, pH values are between 0 and 14, indicating the level of acidity. Pure water has a pH of 7.0 and is considered neutral. A substance with a pH , 7 is called acidic. A substance with a pH . 7 is called basic or alkaline.1 The lower the pH, the more acidic the substance. The higher the pH, the less acidic and the more alkaline the substance. Table 6.11 shows approximate pH values for some common items. Most foods have a pH between 3 and 7. Substances with a pH below 3 or above 12 can be dangerous to handle with bare hands. Remember that pH is the negative of a logarithmic function. So the larger the hydrogen ion concentration, the smaller the pH. (See Figure 6.10.) Multiplying the hydrogen ion concentration by 10 decreases the pH by 1. So vinegar (with a pH of 3) has a hydrogen ion concentration 10 times greater than that of wine (pH 4) and 10,000 times greater than that of pure water (pH 7).
1
Things that are alkaline tend to be slimy and sticky, like a bar of soap. If you wash your hands with soap and don’t rinse, there will be an alkaline residue. Acids can be used to neutralize alkalinity. For example, shampoo often leaves a sticky alkaline residue. So we use acidic conditioners to neutralize the alkalinity.
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pH of Various Substances Acidic (more hydrogen ions than pure water)
pH
Neutral
pH
Gastric juice Coca-Cola, vinegar Grapes, wine Coffee, tomatoes Bread Beef, chicken
2 3 4 5 5.5 6
Pure water
7
Basic or Alkaline (fewer hydrogen ions than pure water) Egg whites, sea water Soap, baking soda Detergents, toothpaste, ammonia Household cleaner Caustic oven cleaner
pH 8 9 10 12 13
Table 6.11
EXAMPLE
4
Comparing hydrogen ion concentrations a. Compare the hydrogen ion concentration of Coca-Cola with the hydrogen ion concentrations of coffee and ammonia. b. Calculate the hydrogen ion concentration of Coca-Cola.
SOLUTION
a. Table 6.11 gives the pH of Coca-Cola as 3, coffee as 5 (so both are acidic), and ammonia as 10 (which is alkaline). Increasing the pH by 1 corresponds to decreasing the hydrogen ion concentration by a factor of 10. So Coca-Cola will have more ions than coffee, and even more hydrogen than ammonia. The hydrogen ion concentration of Coca-Cola is 10523 5 102 5 100 times more than that of coffee, and 101023 5 107 5 10,000,000 times more than that of ammonia! b. To calculate the hydrogen ion concentration of Coca-Cola, we need to solve the equation Multiply by 21 rewrite as powers of 10 evaluate and use Rule 6 of logs
3 5 2log[H1] 23 5 log[H1] 1023 5 10log[H1] 0.001 5 [H1]
So the hydrogen ion concentration of Coca-Cola is 1023 or 0.001 moles per liter. That means that each liter of Coca-Cola contains 1023 ? s6.022 ? 1023 d 5 6.022 ? 102323 5 6.022 ? 1020 hydrogen ions.
EXAMPLE
5
SOLUTION
Calculating the pH level Sulfuric acid has a hydrogen ion concentration [H1] of 0.109 moles per liter. Calculate its pH. The pH of sulfuric acid equals 2log 0.109 < 0.96.
The rain in many parts of the world is becoming increasingly acidic. The burning of fossil fuels (such as coal and oil) by power plants and automobile emissions release gaseous impurities into the air. The impurities contain oxides of sulfur and nitrogen that combine with moisture in the air to form droplets of dilute sulfuric and nitric acids. An acid releases hydrogen ions in water. High concentrations of hydrogen ions damage plants and water resources (such as the lakes of New England and Sweden) and erode structures (such as the Parthenon in Athens) by removing oxygen molecules. Some experts feel that the acid rain dilemma may be one of the greatest environmental problems facing the world in the near future.
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EXAMPLE
6
SOLUTION
359
The pH function is not linear It is critical that health care professionals understand the nonlinearity of the pH function. An arterial blood pH of 7.35 to 7.45 for a patient is quite normal, whereas a pH of 7.1 means that the patient is severely acidotic and near death. a. Determine the hydrogen ion concentration, [H1], first for a patient with an arterial blood pH of 7.4, then for one with a blood pH of 7.1. b. How many more hydrogen ions are there in blood with a pH of 7.1 than in blood with a pH of 7.4? To find [H1] if the pH is 7.4, we must solve the equation Multiply both sides by 21 then these powers of 10 are equal use Rule 6 of logs evaluate and switch sides
7.4 5 2log[H1] 27.4 5 log[H1] 1 1027.4 5 10log[H ] 1027.4 5 [H1] [H 1] < 0.000 000 040 < 4.0 ? 1028 M
where M is in moles per liter. Similarly, if the pH is 7.1, we can solve the equation to get
7.1 5 2log[H1] [H 1] < 7.9 ? 1028 M
Comparing the two concentrations gives us [H1] in blood with a pH of 7.1 7.9 ? 1028 M 7.9 < 5 <2 1 [H ] in blood with a pH of 7.4 4.0 ? 1028 M 4.0 So there are approximately twice as many hydrogen ions in blood with a pH of 7.1 than in blood with a pH of 7.4.
?
SOMETHING TO THINK ABOUT
A noise emission statute enforced by the Massachusetts Registry of Motor Vehicles requires that the noise level of a motorcycle not exceed 82–86 decibels. Jay McMahon, chairman of the Modified Motorcycle Association, says, “It’s just another way for the Registry to stick it to motorcycle operators. No one complains when a Maserati or a tractor trailer drives by.” Does the statute seem fair to you? Why?
Measuring noise: The decibel scale The decibel scale was designed to reflect the human perception of sounds.2 When it is very quiet, it is easy to notice a small increase in sound intensity. The same increase in intensity in a noisy environment would not be noticed; it would take a much bigger change to be detected by humans. The same is true for light. If a 50-watt light bulb is replaced with a 100-watt light bulb, it is easy to notice the difference in brightness. But if you replaced 500 watts with 550 watts, it would be very hard to distinguish the 50-watt difference. The decibel scale, like the pH scale for acidity or the Richter scale for earthquakes, is logarithmic; that is, it measures order-of-magnitude changes. Noise levels are measured in units called decibels, abbreviated dB. The name is in honor of the inventor of the telephone, Alexander Graham Bell. If we designate I0 as the intensity of a sound at the threshold of human hearing (10216 watts/cm2) and we let I represent the intensity of an arbitrary sound (measured in watts/cm2), then the noise level N of that sound measured in decibels (dB) is defined to be I N 5 10 log a b I0
2
Two scientists, Weber and Fechner, studied the psychological response to intensity changes in stimuli. Their discovery, that the perceived change is proportional to the logarithm of the intensity change of the stimulus, is called the Weber-Fechner stimulus law.
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The unitless expression I/I 0 gives the relative intensity of a sound compared with the reference value of I0. For example, if I/I0 5 100, then the noise level, N, is equal to N 5 10 log(100) 5 10 log(102) 5 10(2) 5 20 dB Table 6.12 shows relative intensities, the corresponding noise levels (in decibels), and how people perceive these noise levels. Note how much the relative intensity (the ratio I/I0) of a sound source must increase for people to discern differences. Each time we add 10 units on the decibel scale, we multiply the relative intensity by 10, increasing it by one order of magnitude.
How Decibel Levels Are Perceived Relative Intensity I/I0 1 10 100 1,000 10,000 100,000 1,000,000 10,000,000 100,000,000 1,000,000,000 10,000,000,000 100,000,000,000 1,000,000,000,000 10,000,000,000,000
Decibels (dB) 0 10 20 30 40 50 60 70 80 90 100 110 120 130
Average Perception Threshold of hearing Soundproof room, very faint Whisper, rustle of leaves Quiet conversation, faint Quiet home, private office Average conversation, moderate Noisy home, average office Average radio, average factory, loud Noisy office, average street noise Loud truck, police whistle, very loud Loud street noise, noisy factory Elevated train, deafening Thunder of artillery, nearby jackhammer Threshold of pain, ears hurt
Table 6.12
EXAMPLE
7
How loud is a rock band? What is the decibel level of a typical rock band playing with an intensity of 1025 watts/cm2? How much more intense is the sound of the band than an average conversation?
SOLUTION
Given I0 5 10216 watts/cm2 and letting I 5 1025 watts/cm2 and N represent the decibel level, by definition substitute for I and I0 Rule 2 of exponents Rule 5 of logs
I N 5 10 log a b I0 5 10 log(1025/10216) 5 10 log(1011) 5 10 ? 11 5 110 decibels
So the noise level of a typical rock band is about 110 decibels. According to Table 6.12, an average conversation measures about 50 decibels. So the noise level of the rock band is 60 decibels higher. Each increment of 10 decibels corresponds to a one-order-of-magnitude increase in intensity. So the sound of a rock
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band is about six orders of magnitude, or 106 (a million times), more intense than an average conversation. EXAMPLE
8
Perceiving sound What’s wrong with the following statement? “A jet airplane landing at the local airport makes 120 decibels of noise. If we allow three jets to land at the same time, there will be 360 decibels of noise pollution.”
SOLUTION
There will certainly be three times as much sound intensity, but would we perceive it that way? According to Table 6.12, 120 decibels corresponds to a relative intensity of 1012. Three times that relative intensity would equal 3 ? 1012. So the corresponding decibel level would be Rule 1 of logs use a calculator combine multiply and round off
N 5 10 log s3 ? 1012 d 5 10(log 3 1 log(1012)) < 10s0.477 1 12d < 10s12.477d < 125 decibels
So three jets landing will produce a decibel level of 125, not 360. We would perceive only a slight increase in the noise level.
Algebra Aerobics 6.4 A graphing program is recommended for Problem 3 and a scientific calculator for Problem 7. 1. How would the graphs of y 5 log x2 and y 5 2 log x compare? 2 Draw a rough sketch of the graph y 5 2ln x. Compare it with the graph of y 5 ln x. 3. Compare the graphs of ƒ(x) 5 log x and g(x) 5 ln x. a. Where do they intersect? b. Where does each have an output value of 1? Of 2? c. Describe each graph for values of x such that 0 , x , 1. d. Describe each graph for values of x where x . 1. 4. Given the accompanying graph of a function f, sketch the graph of its inverse. y 3
x –3
3
–3
5. Use the rules for logarithms to evaluate the following expressions. a. log 10 3 e. ln e 5 b. log 10 25 f. ln e 0.07 0.09 c. 3 log 10 g. ln e 3.02 1 ln e –0.27 d. 10 log 3.4 h. e ln 0.9 6. A typical pH value for rain or snow in the northeastern United States is about 4. Is this basic or acidic? What is the corresponding hydrogen ion concentration? How does this compare with the hydrogen ion concentration of pure water? 7. What is the decibel level of a sound whose intensity is 1.5 ? 10212 watts/cm2? 8. If the intensity of a sound increases by a factor of 100, what is the increase in the decibel level? What if the intensity is increased by a factor of 10,000,000?
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Exercises for Section 6.4 Some of these exercises require a calculator that can evaluate powers and logs. Exercise 6 requires a graphing program. 1. Use the rules of logarithms to explain how you can tell which graph is y 5 log x and which is y 5 log(5x).
y 2
A
4. The functions ƒ(x) 5 ln x, g(x) 5 ln (x 1 1) , and h(x) 5 ln(x 1 2) are graphed below. a. Match each function with its graph. b. Find the value of x for each function that makes that function equal to zero. c. Identify the coordinates of the horizontal intercept for each function. d. Assuming k . 0, describe how the graph of ƒ(x) moves if you replace x by (x 1 k). e. Determine the y-intercept, if possible, for each function.
B 3
x –1
y
3
–1
x –5
2. Use the rules of logarithms to explain how you can tell which graph is y 5 log x and which is y 5 log(x/5).
1
A B
.
y A x
–3
a. Which function graphs are mirror images of each other across the y-axis? b. Which function graphs are mirror images of each other across the x-axis?
B –1
y
–2
y 3
3
3. The functions ƒ(x) 5 log x, g(x) 5 log(x 2 1), and h(x) 5 log(x 2 2) are graphed below. a. Match each function with its graph. b. Find the value of x for each function that makes that function equal to zero. c. Identify the x-intercept for each function. d. Assuming k . 0, describe how the graph of ƒ(x) moves if you replace x by (x 2 k).
f(x)
h(x)
x –1
5
3
y
3
g(x)
x
B
C
1
–3
y
k(x)
5
A
x –5
–3
y
–1
–2
C
5. Examine the following graphs of four functions. 3
–1
1
5
x –1
–3
5
x –5
1
–3
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6. (Requires a graphing program.) On the same grid graph y1 5 ln(x), y2 5 2ln(x), y3 5 ln(2x) and y4 5 2ln(2x). a. Which pairs of function graphs are mirror images across the y-axis? b. Which pairs of function graphs are mirror images across the x-axis? c. What predictions would you make about the graphs of the functions ƒ(x) 5 a ln x and g(x) 5 2a ln x? Using technology, test your predictions for different values of a. d. What predictions would you make about the graphs of the functions ƒ(x) 5 a ln x and g(x) 5 a ln(2x)? Using technology, test your predictions for different values of a. 7. Logarithms can be constructed using any positive number except 1 as a base: log a x 5 y means that a y 5 x a. Complete the accompanying table and sketch the graph of y 5 log3 x.
x
y 5 log3 x
1 9 1 3
363
c. As the brightness B increases, does the magnitude M increase or decrease? Is a sixth-magnitude star brighter or dimmer than a first-magnitude star? d. If the brightness of a star is increased by a factor of 5, by how much does the magnitude increase or decrease? 9. [Source: H. D. Young, University Physics, Vol. 1 (Reading, MA: Addison-Wesley, 1992), p. 591] If you listen to a 120-decibel sound for about 10 minutes, your threshold of hearing will typically shift from 0 dB up to 28 dB for a while. If you are exposed to a 92-dB sound for 10 years, your threshold of hearing will be permanently shifted to 28 dB. What intensities correspond to 28 dB and 92 dB? 10. In all of the sound problems so far, we have not taken into account the distance between the sound source and the listener. Sound intensity is inversely proportional to the square of the distance from the sound source; that is, I 5 k/r2, where I is intensity, r is the distance from the sound source, and k is a constant. Suppose that you are sitting a distance R from the TV, where its sound intensity is I1. Now you move to a seat twice as far from the TV, a distance 2R away, where the sound intensity is I2. a. What is the relationship between I1 and I2? b. What is the relationship between the decibel levels associated with I1 and I2? 11. If there are a number of different sounds being produced simultaneously, the resulting intensity is the sum of the individual intensities. How many decibels louder is the sound of quintuplets crying than the sound of one baby crying?
1 3 9 27
b. Now make a small table and sketch the graph of y 5 log4 x. (Hint: To simplify computations, try using powers of 4 for values of x.) 8. The stellar magnitude M of a star is approximately 22.5 log(B/B0), where B is the brightness of the star and B0 is a constant. a. If you plotted B on the horizontal and M on the vertical axis, where would the graph cross the B axis? b. Without calculating any other coordinates, draw a rough sketch of the graph of M. What is the domain?
12. An ulcer patient has been told to avoid acidic foods. If he drinks coffee, with a pH of 5.0, it bothers him, but he can tolerate both tap water, with a pH of 5.8, and milk, with a pH of 6.9. a. Will a mixture of half coffee and half milk be at least as tolerable as tap water? b. What pH will the half coffee–half milk mixture have? c. In order to make 10 oz of a milk-coffee drink with a pH of 5.8, how many ounces of each are required? 13. Lemon juice has a pH of 2.1. If you make diet lemonade by mixing 14 cup of lemon juice with 2 cups of tap water, with a pH of 5.8, will the resulting acidity be more or less than that of orange juice, with a pH of 3?
6.5 Transforming Exponential Functions to Base e In Section 6.2 we saw that an exponential function can be written in two equivalent forms: ƒstd 5 Ca t
or
ƒstd 5 Csek d t
where e k 5 a. Given a value for k, we can evaluate e k to find a. Now we can use logarithms to do the reverse; that is, given a value for a we find a value for k such that a 5 ek.
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Converting a to ek In general, if a is the growth (or decay) factor, then we can always find a value for k such that ek 5 a ln e k 5 ln a k 5 ln a
by taking ln of both sides using Rule 5 for ln So the function f std 5 Ca t can be rewritten as
ƒ(t) 5 C(e k)t 5 C(eln a)t where k 5 ln a. Recall that k is called the instantaneous or continuous growth (or decay) rate. For exponential growth, a . 1, so ek . 1 and k is positive. For exponential decay, 0 , a , 1, so 0 , ek , 1 and k is negative. Distinguishing Different Rates For functions in the form ƒ(t) 5 C(e k)t, we call the exponent k the continuous growth rate (if k . 0) or the continuous decay rate (if k , 0). For functions in the form ƒ(t) 5 Cat, we call the base a the growth (or decay) factor.
EXAMPLE
1
SOLUTION
If a . 1,
the growth factor 5 1 1 growth rate
If 0 , a , 1,
the decay factor 5 1 2 decay rate
a. Rewrite the function ƒ(t) 5 250(1.3)t using base e. b. Identify the growth rate (based on the growth factor) and the continuous growth rate, both per time period t. a. The function ƒ(t) 5 250(1.3)t can be rewritten using base e, by substituting eln 1.3 for 1.3 evaluating ln 1.3 using Rule 3 of exponents
ƒ(t) 5 250(eln 1.3)t < 250(e0.262)t 5 250e0.262t
b. Since ƒ(t) 5 250(1.3)t, the growth rate per time period t is 0.3 or 30%. Since we also have ƒ(t) 5 250e0.262 t, the continuous growth rate per time period t is 0.262 or 26.2%. EXAMPLE
2
SOLUTION
a. Rewrite the function g(t) 5 340(0.94)t using base e. Assume t is in years. b. Identify the decay factor, the decay rate, and the continuous decay rate. a. Since the function g(t) 5 340(0.94)t represents exponential decay, the value for k will be negative. The function g can be rewritten using base e, by substituting eln 0.94 for 0.94 evaluating ln 0.94 using Rule 3 of exponents
g(t) 5 340(eln 0.94)t < 340(e20.062)t 5 340e20.062t
b. The decay factor is 0.94. So the decay rate is 1 2 0.94 5 0.06 or 6% per year. The continuous decay rate is 0.062 or 6.2% per year.
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See “E4: y 5 Ce r x Sliders” and “E5: Comparing y = C a x to y 5 Ce r x ” in Exponential & Log Functions.
365
Writing an exponential function in the form y 5 Ca t or y 5 Ce kt (where k 5 ln a) is a matter of emphasis, since the graphs and functional values are identical. When we use y 5 Ca t, we may think of the growth taking place at discrete points in time, whereas the form y 5 Ce kt emphasizes the notion of continuous growth. For example, y 5 100(1.05)t could be interpreted as giving the value of $100 invested for t years at 5% compounded annually, whereas its equivalent form y 5 100(eln 1.05)t < 100e0.049t suggests that the money is invested at 4.9% compounded continuously.
Exponential Functions Using Base e The exponential function y 5 Ca t (where a . 0 and a 2 1) can be rewritten as y 5 Csek d t
where k 5 ln a
We call k the instantaneous or continuous growth (or decay) rate. Exponential growth: Exponential decay:
EXAMPLE
a.1
and
k is positive
0,a,1
and
k is negative
3
Converting to a continuous growth rate Consider the bacterial growth we described with the equation N 5 100 ? 2 t. The bacteria don’t all double at the same time, precisely at the beginning of each time period t. A continuous growth pattern is much more likely. Rewrite this equation to reflect a continuous growth rate.
SOLUTION
The equation N 5 100 ? 2 t can be rewritten to reflect a continuous growth rate by converting the base of 2 to its equivalent form using base e. We know the growth factor a can be rewritten as a 5 eln a Substitute 2 for a evaluate ln 2
2 5 eln 2 2 < e0.693
So the equation that reflects a continuous growth rate is N 5 100 ? e0.693t EXAMPLE
4
Converting from base a to base e In Chapter 5 we saw that the function ƒ(t) 5 100(0.976)t measures the amount of radioactive strontium-90 remaining as 100 milligrams (mg) decay over time t (in years). Rewrite the function using base e.
SOLUTION
To rewrite the function ƒ(t) 5 100(0.976)t using base e, we need to convert the base 0.976 to the form ek, where k 5 ln 0.976. Using a calculator to evaluate ln 0.976, we get 20.0243. Substituting into our original function, we get ƒ(t) 5 100(0.976)t < 100se20.0243 d t 5 100e20.0243t
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Since the original base 0.976 is less than 1, the function represents decay. So, as we would expect, when the function is rewritten using base e, the value of k (in this case 20.0243) is negative.
EXAMPLE
5
SOLUTION
Use a continuous compounding model to describe the growth of Medicare expenditures. In Chapter 5 we found a best-fit function for Medicare expenditures to be C(n) 5 10.6 ? (1.11)n where n 5 years since 1970 and C(n) 5 Medicare expenditures in billions of dollars. This implies a growth rate of 11% compounded annually. To describe the same growth in terms of continuous compounding, we need to rewrite the growth factor 1.11 as e k, where k 5 ln 1.11 < 0.104. Substituting into the original function gives us C(n) 5 10.6e 0.104n So Medicare expenditures are continuously compounding at about 10.4% per year.
EXAMPLE
6
Proving the rule of 70 We now have the tools to prove the rule of 70 introduced in Chapter 5. Recall that the rule said if a quantity is growing (or decaying) at R% per time period, then the time it takes the quantity to double (or halve) is approximately 70/R time periods. For example, if a quantity increases by a rate, R, of 7% each month, the doubling time is about 70/7 5 10 months. Proof Let f be an exponential function of the form ƒ(t) 5 Ce rt where C is the initial quantity, r is the continuous growth rate, and t represents time. The rate r is in decimal form, and R is the equivalent amount expressed as a percentage. Since ƒ(t) represents exponential growth, r . 0. The doubling time for an exponential function is constant, so we need only calculate the time for any given quantity to double. In particular, we can determine the time it takes for the initial amount C (at time t 5 0) to become twice as large; that is, we can calculate the value for t such that Set divide by C take ln of both sides evaluate and use ln Rule 5
ƒ(t) 5 2C Ce rt 5 2C e rt 5 2 ln e rt 5 ln 2 rt < 0.693
(1)
R 5 100r, so r 5 R/100. If we round 0.693 up to 0.70 and substitute in Equation (1) multiply both sides by 100 divide by R we get
sR/100d ? t < 0.70 R ? t < 70 t < 70/R
So the time t it takes for the initial amount to double is approximately 70/R, which is what the rule of 70 claims. We leave the similar proof about half-lives, where r , 0 represents decay, to the exercises.
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Algebra Aerobics 6.5 A calculator that can evaluate logs and powers is required for Problems 2–4. 1. Identify each of the following exponential functions as representing growth or decay. (Hint: For parts (d)–(f) use rules for logarithms.) d. P 5 250 e(ln 1.056) t a. M 5 Ne20.029t r b. K 5 100(0.87) e. A 5 20 e (ln 0.834) t c. Q 5 375 e 0.055t f. y 5 (1.2 ? 105)(eln 0.752)t 2. Rewrite each of the following as a continuous growth or decay model using base e. a. y 5 1000(1.062)t b. y 5 50(0.985)t
3. Determine the nominal and effective rates for each of the following. Assume t is measured in years. (Hint: To find the effective rate, convert from e k to a.) c. y 5 2000 e(ln1.083) t a. P 5 25,000 e (ln1.056) t b. P 5 10e(ln 1.034) t d. y 5 (1.2 ? 105)(eln1.295) t 4. Find the growth factor or decay factor for each of the following. (Hint: Convert from ek to a.) b. y 5 30e20.125t a. P 5 50,000e0.08t
Exercises for Section 6.5 Many of these exercises require a calculator that can evaluate powers and logs, some require a graphing program, and Exercise 18 requires technology that can generate a best-fit exponential function. 1. Rewrite each of the following functions using base e. a. N 5 10(1.045) t c. P 5 500(2.10) x 27 A b. Q 5 (5 ? 10 ) ? (0.072) 2. Using technology, graph each of the following. Find, as appropriate, the doubling time or half-life. c. P 5 (3.2 ? 106 )(e20.15) t a. A 5 50 e 0.025t 20.046 t b. A 5 100 e 3. Identify each of the following functions as representing growth or decay: c. ƒ(t) 5 375e0.055 t a. Q 5 Ne20.029 t b. h(r) 5 100(0.87) r 4. Identify each function as representing growth or decay. Then determine the annual growth or decay factor, assuming t is in years. a. A 5 A0 (1.0025) 20 t d. A 5 A0 e20.063 t b. A 5 A0 (1.0006) t /360 e. A 5 A0 e 0.015t t /2 c. A 5 A0 (0.992) 5. For each of the following, find the doubling time, then rewrite each function in the form P 5 P0 e r t. Assume t is measured in years. b. P 5 P02 t/25 c. P 5 P02 2t a. P 5 P02 t/5
7. The barometric pressure, p, in millimeters of mercury, at height h, in kilometers above sea level, is given by the equation p 5 760e20.128 h. At what height is the barometric pressure 200 mm? 8. After t days, the amount of thorium-234 in a sample is A(t) 5 35e20.029 t micrograms. a. How much was there initially? b. How much is there after a week? c. When is there just 1 microgram left? d. What is the half-life of thorium-234? 9. Assume ƒ(t) 5 Ce r t is an exponential decay function (so r , 0). Prove the rule of 70 for halving times; that is, if a quantity is decreasing at R% per time period t, then the number of time periods it takes for the quantity to halve is approximately 70/R. (Hint: R 5 100r.) 10. Match the function with its graph. a. ƒ(x) 5 10e20.075x c. h(x) 5 10e20.025x 20.045x b. g(x) 5 10e
15
y
10
A B
5
6. For each of the following, find the half-life, then rewrite each function in the form P 5 P0 er t. Assume t is measured in years. a. P 5 P0 A12B
t/10
b. P 5 P0 A12B
t/215
c. P 5 P0 A12B
C x
4t
–5
0
10
20
30
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11. Match the function with its graph. c. h(x) 5 10 e 0.075 x a. ƒ(x) 5 10 e 0.025 x 0.045 x b. g(x) 5 10 e
exponential decay function through the data. Then identify two points on your curve, and generate two equivalent equations that model the data, one in the form y 5 Ca t and the other in the form y 5 Ce kt.
y 30
Species density
A B C
x –4
0
12
12. The functions y 5 50 e0.04t and y 5 50(2)t/n are two different ways to write the same function. a. What does the value 0.04 represent? b. Set the functions equal to each other and use rules of natural logarithms to solve for n. c. What does the value of n represent? t/n
13. The functions y 5 2500e20.02t and y 5 2500 A12B are two different ways to write the same function. a. What does the value 20.02 represent? b. Set the functions equal to each other and use rules of natural logarithms to solve for n. c. What does the value of n represent? 14. (Requires a graphing program.) Radioactive lead-210 decays according to the exponential formula Q 5 Q0 e20.0311 t, where Q0 is the initial quantity in milligrams and t is in years. What is the half-life of lead-210? Verify your answer by graphing using technology. 15. Radioactive thorium-230 decays according to the formula t/8000 Q 5 Q0 A12B , where Q0 is the initial quantity in milligrams and t is in years. a. What is the half-life of thorium-230? b. What is the annual decay rate? c. Translate the equation into the form Q 5 Q0 e r t. What does r represent? 16. In 1859, the Victorian landowner Thomas Austin imported 12 wild rabbits into Australia and let them loose to breed. Since they had no natural enemies, the population increased very rapidly. By 1949 there were approximately 600 million rabbits. a. Find an exponential function using a continuous growth rate to model this situation. b. If the growth had gone unchecked, what would have been the rabbit population in 2000? c. Internet search: Find out what was done to curb the population of rabbits in Australia, and find the current rabbit population. 17. Biologists believe that, in the deep sea, species density decreases exponentially with the depth. The accompanying graph shows data collected in the North Atlantic. Sketch an
2000 1800 1600 1400 1200 1000 800 600 400 200 0
0
500 1000 1500 2000 2500 Depth (m)
Source: Data collected by Ron Etter, Biology Department. University of Massachusetts–Boston.
18. (Requires technology to find a best-fit exponential.) According to another version of Moore’s Law, the computing power built into chips doubles every 18 months (see Section 5.7). The accompanying table shows the computing power of some Intel chips (measured in calculations per second) between 1993 and 2005. a. Graph the data points (if possible, on a semi-log plot). Explain why an exponential function would be an appropriate model.
Chip Computing Power (millions of calculations per second)
Year
Chip Type
1993 1997 1999 2000 2005
Pentium ® Pentium II Pentium III Pentium 4 Dual-core Itanium2
66 525 1,700 3,400 27,079
Source: Intel Corporation.
b. Construct an exponential function to model the data in the following two different ways. In each case let t 5 number of years since 1993. i. Using the doubling time given by Moore’s Law, construct an exponential function, P1(t), for chip computing power. ii. Use technology to find a best-fit exponential function, P2(t), to the data in the table. Does this model come close to verifying Moore’s Law (P1(t))? 19. The number of neutrons in a nuclear reactor can be predicted from the equation n 5 n0 e(1n 2) t / T, where n 5 number of neutrons at time t (in seconds), n0 5 the number of neutrons at time t 5 0, and T 5 the reactor period, the doubling time of the neutrons (in seconds). When t 5 2 seconds, n 5 11, and when t 5 22 seconds, n 5 30. Find the initial number of neutrons, n0, and the reactor period, T, both rounded to the nearest whole number.
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20. According to Rubin and Farber’s Pathology, “death from cancer of the lung, more than 85% of which is attributed to cigarette smoking, is today the single most common cancer death in both men and women in the United States.” The accompanying graph shows the annual death rate (per thousand) from lung cancer for smokers and nonsmokers.
Annual death rate from lung cancer (per 103)
b. By contrast, those who smoke more than one pack per day show an exponential rise in the annual death rate from lung cancer. Estimate the coordinates for two points on the graph for heavy smokers, and use the points to construct an exponential model (assume a continuous growth rate). Interpret your results. 21. If an object is put in an environment at a fixed temperature, A (the “ambient temperature”), then its temperature, T, at time t is modeled by Newton’s Law of Cooling:
45 40 35
369
Heavy smokers (>1 pack/day)
T 5 A 1 Ce2 k t
30 25 20 15
All smokers
10 Never smoked
5
35–44 45–54
55–64 Age
65–74 75–84
Source: E. Rubin and J. L. Farber, Pathology, 3rd ed. (Philadelphia: Lippincott-Raven, 1998), p. 312. Copyright © 1998 by Lippincott-Raven. Reprinted by permission.
a. The death rate for nonsmokers is roughly a linear function of age. After replacing each range of ages with a reasonable middle value (e.g., you could use 60 to approximate 55 to 64), estimate the coordinates of two points on the graph of nonsmokers and construct a linear model. Interpret your results.
where k is a positive constant. Note that T is a function of t and that as t S 1` , then e2kt S 0, so the temperature T gets closer and closer to the ambient temperature, A. a. Assume that a hot cup of tea (at 160ºF) is left to cool in a 75ºF room. If it takes 10 minutes for it to reach 100ºF, determine the constants A, C, and k in the equation for Newton’s Law of Cooling. What is Newton’s Law of Cooling in this situation? b. Sketch the graph of your function. c. What is the temperature of the tea after 20 minutes? 22. Newton’s Law of Cooling (see Exercise 21) also works for objects being heated. At time t 5 0, a potato at 70ºF (room temperature) is put in an oven at 375ºF. Thirty minutes later, the potato is at 220º. a. Determine the constants A, C, and k in Newton’s Law. Write down Newton’s Law for this case. b. When is the potato at 370ºF? c. When is the potato at 374ºF? d. According to your model, when (if ever) is the potato at 375ºF? e. Sketch a graph of your function in part (a).
6.6 Using Semi-Log Plots to Construct Exponential Models for Data In Chapter 5 we learned that an exponential function appears as a straight line on a semi-log plot (where the logarithmic scale is on the vertical axis). To decide whether an exponential function is an appropriate model for a data set, we can plot the data on a semi-log plot and see if it appears to be linear. This is one of the easiest and most reliable ways to recognize exponential growth in a data set. And, as we learned in Chapters 4 and 5, a logarithmic scale has the added advantage of being able to display clearly a wide range of values.
Why Do Semi-Log Plots of Exponential Functions Produce Straight Lines? Consider the exponential function If we take the log of both sides use Rule 1 of logs
y 5 3 ? 2x log y 5 log (3 ? 2x) 5 log 3 1 log (2x)
(1)
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use Rule 3 of logs evaluate logs and rearrange, we have If we set Y 5 log y, we have
5 log 3 1 x log 2 log y < 0.48 1 0.30x Y 5 0.48 1 0.30x
(2)
So Y (or log y) is a linear function of x. Equations (1) and (2) are equivalent to each other. The graph of Equation (2) on a semi-log plot, with Y (or log y) values on the vertical axis, is a straight line (see Figure 6.11). The slope is 0.30 or log 2, the logarithm of the growth factor 2 of Equation (1). The vertical intercept is 0.48 or log 3, the logarithm of the y-intercept 3 of Equation (1).
y
Y (or log y)
100
2
10
1
log y = 0.48 + 0.30x
3 0.48 or log 3 1
x
0 1
2
3
4
5
Figure 6.11 The graph of Y 5 0.48 1 0.30x,
showing the relationship between two equivalent logarithmic scales on the vertical axis.
Figure 6.11 shows two equivalent variations of logarithmic scales on the vertical axis. One plots the value of y on a logarithmic scale (using powers of 10), and the other plots the value of log y (using the exponents of the powers of 10). Spreadsheets and some graphing calculators have the ability to instantly switch axis scales between standard linear and the logarithmic scale using powers of 10. But one can in effect do the same thing by plotting log y instead of y. Notice that on the vertical log y scale, the units are now evenly spaced. This allows us to use the standard strategies for finding the slope and vertical intercept of a straight line. In general, we can translate an exponential function in the form y 5 Ca x (where C and a . 0 and a 2 1) into an equivalent linear function log y 5 log C 1 (log a)x Y 5 log C 1 (log a)x
or where Y 5 log y
Finding the Equation of an Exponential Function on a Semi-Log Plot The graph of an exponential function y 5 Ca x (where C and a . 0 and a 2 1) appears as a straight line on a semi-log plot. The line’s equation is Y 5 log C 1 (log a)x
where Y 5 log y
The slope of the line is log a, where a is the growth factor for y 5 Ca x. The vertical intercept is log C, where C is the vertical intercept of y 5 Ca x.
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371
Growth in the Dow Jones The Dow Jones Industrial Average is based on the stock prices of thirty companies and is commonly used to measure the health of the stock market. Figure 6.12 is a semi-log plot of the Dow between the boom years of 1982 and 2000.
Y
log of the Dow Jones
log(Dow) 4.50
×
4.00
(2000, 4) 3.50
(1985, 3.1)
3.00
×
2.50 2.00 1980
x 1985
1990
1995
2000
2005
Year
Figure 6.12 Graph of log(Dow) with two points
on the best-fit line (not from the original data). Source: Dow Jones website, www.djindexes.com.
The data points lie approximately on a straight line, so an exponential model is appropriate.
EXAMPLE
1
SOLUTION
Constructing an exponential function from a line on a semi-log plot Estimate the annual percent growth rate of the Dow Jones between 1982 and 2000. We first construct a linear function for the best-fit line on the semi-log plot, then translate it into an exponential growth function to find the percent growth rate. Estimating the coordinates of two points (1985, 3.1) and (2000, 4.0) on the best-fit line, not from the data (see Figure 6.12), the slope or average rate of change between them is slope 5
change in Y 4.0 2 3.1 0.9 5 5 5 0.06 change in x 2000 2 1985 15
If we let x 5 number of years since 1980, the slope remains the same. Reading off the graph, the vertical intercept is approximately 2.8. So the equation for the best-fit line is approximately Y 5 2.8 1 0.06x
where Y 5 log(Dow)
(1)
We can translate Equation (1) into an exponential function if we substitute log(Dow) for Y rewrite as a power of 10 use Rule 6 of logs and Rule 1 of exponents Rule 3 of exponents evaluate using calculator
log(Dow) 5 2.8 1 0.06x 10 log(Dow) 5 10(2.810.06x) Dow 5 102.8 ? 100.06x Dow 5 102.8 ? (100.06)x Dow < 631 ? 1.15x
(2)
The annual growth factor is approximately 1.15, and the annual growth rate in decimal form is 0.15. So between 1982 and 2000 the Dow Jones grew by about 15% each year.
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This strategy is equivalent to using the formula in the previous box, translating log(Dow) 5 2.8 1 0.06x into the form Dow 5 Ca x, by solving 2.8 5 log C, and 0.06 5 log a.
Algebra Aerobics 6.6 Many of the problems require a calculator that can evaluate logs and exponents. 1. a. Which of the following three functions would have a straight-line graph on a standard linear plot? On a semi-log plot? y 5 3x 1 4, y 5 4 ? 3x, log y 5 (log 3) ? x 1 log 4 b. For each straight-line graph in part (a), what is the slope of the line. The vertical intercept? (Hint: Substitute Y 5 log y and X 5 log x as needed.) 2. Change each exponential equation to a logarithmic equation. a. y 5 5(3)x c. y 5 10,000(0.9)x x b. y 5 1000(5) d. y 5 (5 ? 106) (1.06)x 3. Change each logarithmic equation to an exponential equation. a. log y 5 log 7 1 (log 2) ? x b. log y 5 log 20 1 (log 0.25) ? x c. log y 5 6 1 (log 3) ? x d. log y 5 6 1 log 5 1 (log 3) ? x 4. a. Below are linear equations of log y (or Y) in x. Identify the slope and vertical intercept for each. i. log y 5 log 2 1 (log 5) ? x ii. log y 5 (log 0.75) ? x 1 log 6 iii. log y 5 0.4 1 x log 4 iv. log y 5 3 1 log 2 1 (log 1.05) ? x b. Use the slope and intercept to create an exponential function for each equation.
5. Solve each equation using the definition of log. a. 0.301 5 log a c. log a 5 20.125 b. log C 5 2.72 d. 5 5 log C 6. Change each function to exponential form, assuming that Y 5 log y. a. Y 5 0.301 1 0.477x b. Y 5 3 1 0.602x c. Y 5 1.398 2 0.046x 7. For each of the two accompanying graphs examine the scales on the axes. Then decide whether a linear or an exponential function would be the most appropriate model for the data. y
log y 2.0
100
1.5 10
1.0 0.5
1
x 0
5
10
0.0
x 0 1 2 3 4 5 6 7 8 9 10
8. Generate an exponential function that could describe the data in the accompanying graph. log (y) 7 6 5 4 3 2 1 0 0
x 2
4
6
Exercises for Section 6.6 Some exercises require a calculator that can evaluate powers and logs. Exercises 10 and 13 require a graphing program. 1. Match each exponential function in parts (a)–(d) with its logarithmic form in parts (e)–(h). a. y 5 10,000 (2)x e. log y 5 6.477 2 0.097x f. log y 5 4 1 0.301x b. y 5 1000 (1.4)x g. log y 5 3 2 0.347x c. y 5 (3 ? 106 )(0.8)x h. log y 5 3 1 0.146x d. y 5 1000 (0.45)x
2. Form the exponential function from its logarithmic equivalent for each of the following. a. log y 5 log 1400 1 (log 1.06)x b. log y 5 log (25,000) 1 (log 0.87)x c. log y 5 2 1 (log 2.5)x d. log y 5 4.25 1 (log 0.63)x
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3. Change each exponential function to its logarithmic equivalent. Round values to the nearest thousandth. a. y 5 30,000 (2)x c. y 5 (4.5 ? 106 )(0.7)x x b. y 5 4500 (1.4) d. y 5 6000(0.57)x 4. Write the exponential equivalent for each function. Assume Y 5 log y. a. Y 5 2.342 1 0.123x c. Y 5 4.74 2 0.108x b. Y 5 3.322 1 0.544x d. Y 5 0.7 2 0.004x
x 0
1
2
3
4
4.4 4.2 4.0 3.8 3.6
x 0
1
Graph A
2
3
4
Graph C
4 log (y)
log (y)
6 4 2
3 2 1
x 0
1
2
3
4
0
Graph B
1
2
3
4
6. a. From the data in the following table, create a linear equation of Y in terms of x. log y (or Y)
0 1 2 3 4
5.00000 5.60206 6.20412 6.80618 7.40824
b. Find the equivalent exponential function of y in terms of x. 7. Determine which data sets (if any) describe y as an exponential function of x, then construct the exponential function. (Hint: Find the average rate of change of Y with respect to x.) a.
x
logy (or Y)
0 10 20 30 40
2.30103 4.30103 4.90309 5.25527 5.50515
b.
9 8 7 6 5 4 3 2 1 0
10
15
20
25
30
Bacteria Population
Y
Y = 0.1185x + 4.5887
0
5
10
15
20
25
x (days)
x
Graph D
x
x 5
9. The accompanying graph shows the growth of a bacteria population over a 25-day period.
log (population)
6.8 6.6 6.4 6.2 6.0
log (y)
log (y)
5. Match each exponential function with its semi-log plot. c. y 5 (4.5 ? 106 )(0.7)x a. y 5 30,000 (2)x x b. y 5 4500 (1.4) d. y 5 6000(0.57)x
log y (or Y ) 26 24 22 20 18 16 14 12 10 8 6 4 2 0 0
373
a. Do the bacteria appear to be growing exponentially? Explain. b. Translate the best-fit line shown on the graph into an exponential function and determine the daily growth rate. 10. (Requires a graphing program.) The current population of a city is 1.5 million. Over the next 40 years, the population is expected to decrease by 12% each decade. a. Create a function that models the population decline. b. Create a table of values at 10-year intervals for the next 40 years. c. Graph the population of the city on a semi-log plot. 11. An experiment by a pharmaceutical company tracked the amount of a certain drug (measured in micrograms/liter) in the body over a 30-hour period. The accompanying graph shows the results. Y log (drug) 1.20
Drug in the Body
1.00 0.80 0.60
x
logy (or Y)
0 10 20 30 40
4.77815 3.52876 2.27938 1.02999 20.21945
8. Construct two functions that are equivalent descriptions of the data in the accompanying graph. Describe the change in log y (or Y) with respect to x.
Y = – 0.0181x + 1.0873 0.40 0.20
x 0
10
20
30
Hours
a. What was the initial amount of the drug? b. Is the amount of drug in the body decaying exponentially? If yes, state the decay rate. c. If the amount in the body is decaying exponentially, construct an exponential model.
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12. The accompanying graph shows the decay of strontium-90.
Strontium-90 (grams)
100
13. (Requires a graphing program.) The data in the following table are from the 2006 Statistical Abstract and show the rise of health care costs in the United States since 1960.
10
1
0.1 0
25
50
75
100
125
150
175
200
Time (years)
a. Is the graph a semi-log plot? b. What is the initial amount of strontium-90 for this graph? c. According to the graph, what is the half-life of strontium-90? d. What would be the yearly decay rate for strontium-90? e. Find the exponential model for the decay of strontium-90.
Year
U.S. Health Care Expenditures (billions of dollars)
1960 1970 1980 1985 1990 1995 2000 2005
28 75 225 442 717 1020 1359 2016
a. It is clear that the United States is spending more on health care as time goes on. Does this mean that we as individuals are paying more? How would you find out? b. Is it fair to say that the expenses shown are growing exponentially? Use graphing techniques to find out; measure time in years since 1960. c. Around 1985 the high cost of health care began to become an increasingly political issue, and insurance companies began to introduce “managed care” in an attempt to cut costs. Find a mathematical model that assumes a continuous growth for health costs from 1960 to 2005. What does your model predict for health costs in 2005? How closely does this reflect the actual cost in 2005?
C H A P T E R S U M M A RY Logarithms
Rules for Logarithms
The logarithm base 10 of x is the exponent of 10 needed to produce x.
If A and B are positive real numbers and p is any real number, then:
log x 5 c
means that
10 c 5 x
Logarithms base 10 are called common logarithms. The logarithm base e of x is the exponent of e needed to produce x. The number e is an irrational natural constant < 2.71828. Logarithms base e are called natural logarithms and are written as ln x: ln x 5 c If x . 1, If x 5 1, If 0 , x , 1, If x # 0 and
means that
ec 5 x
log x and ln x are both positive log 1 5 0 and ln 1 5 0 log x and ln x are both negative neither logarithm is defined log 10 5 1 and ln e 5 1
The rules for logarithms follow directly from the definition of logarithms and from the rules for exponents.
For Common Logarithms 1. 2. 3. 4. 5. 6.
log(A ? B) 5 log A 1 log B log(A/B) 5 log A 2 log B log Ap 5 p log A log 1 5 0 (since 100 5 1) log 10 x 5 x 10log x 5 x (x . 0)
For Natural Logarithms 1. 2. 3. 4. 5. 6.
ln(A ? B) 5 ln A 1 ln B ln(A/B) 5 ln A 2 ln B ln Ap 5 p ln A ln 1 5 0 (since e0 5 1) ln e x 5 x eln x 5 x (x . 0)
Solving Exponential Functions Using Logarithms Logarithms can be used to solve equations such as 10 5 2 t by taking the log of each side to get log 10 5 log (2t) ➯ 1 5 t ? log2 ➯ t < 3.32.
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Compounding
375
The graphs of these functions have similar shapes, and both
The value of P0 dollars invested at a nominal interest rate r (in decimal form) compounded n times a year for t years is r nt P 5 P0 a1 1 b n The value for r is called the annual percentage rate (APR). The actual interest rate per year is called the effective interest rate or the annual percentage yield (APY). Compounding Continuously The number e is used to describe continuous compounding. For example, the value of P0 dollars invested at a nominal interest rate r (expressed in decimal form) compounded continuously for t years is P 5 P0ert
lie to the right of the y-axis have a horizontal intercept of (1, 0) are concave down increase throughout with no maximum or minimum are asymptotic to the y-axis
Two functions are inverses of each other if what one function “does” the other “undoes.” The graphs of two inverse functions are mirror images across the line y 5 x. The logarithmic and exponential functions are inverse functions.
Continuous Growth or Decay Any exponential function ƒ(t) 5 Ca t can be rewritten as ƒ(t) 5 Ce kt
Logarithmic Functions We can define two functions: y 5 log x and y 5 ln x (where x . 0). 3
• • • • •
y y = ln x
We call k the instantaneous or continuous growth (or decay) rate. Exponential growth: Exponential decay:
a.1 0,a,1
and and
k is positive k is negative
Using Semi-Log Plots The graph of an exponential function y 5 Ca x appears as a straight line on a semi-log plot. The line’s equation is
y = log x x –1
where k 5 ln a
Y 5 log C 1 (log a)x
10
–2
Graphs of y 5 log x and y 5 ln x
where Y 5 log y
The slope of the line is log a, where a is the growth (or decay) factor for y 5 Ca x. The vertical intercept is log C, where C is the vertical intercept of y 5 Ca x.
C H E C K Y O U R U N D E R S TA N D I N G I. Is each of the statements in Problems 1–30 true or false? If false, give an explanation for your answer. 1. For the function y 5 log x, if x is increased by a factor of 10, then y increases by 1. 2. If log x 5 c, then x is always positive but c can be positive, negative, or zero. 3. ln(1.08/2) 5 ln(1.08)/ln(2) 4. Because the function y 5 log x is always increasing, it has no vertical asymptotes. 5. Both log 1 and ln 1 equal 0. 6. ln a 5 b means be 5 a. 7. 2 , e , 3 8. log(10 2) 5 (log 10)2 9. log(10 3 ? 10 5) 5 15 107 log (107 ) 10. log a 2 b 5 10 log (102 )
11. log a 5
"AB b (C 1 2) 3
1 (log A 1 log B) 2 3 log (C 1 2) 2
12. 5 ln A 1 2 ln B 5 ln(A5 1 B2) 13. If (2.3)x 5 64, then x 5
log 64 . log 2.3
14. If log 20 5 t log 1.065, then t 5 loga
20 b. 1.065
15. The amount of dollars D compounded continuously for 1 year at a nominal interest rate of 8% yields D ? e0.08. 16. ln t exists for any value of t. 17. If ƒ(t) 5 100 ? at and g(t) 5 100 ? bt are exponential decay functions and a , b, then the half-life of f is longer than the half-life of g.
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18. The graphs of the functions y 5 ln x and y 5 log x increase indefinitely. 19. The graphs of the functions y 5 ln x and y 5 log x have both vertical and horizontal asymptotes. 20. The graph of the function y 5 ln x lies above the graph of the function y 5 log x for all x . 0. 21. The graph of the function y 5 log x is always positive. 22. The graph of the function y 5 ln x is always decreasing. 23. The amount of $100 invested at 8% compounded continously has a doubling time of about 8.7 years. (Hint: e 0.08 < 1.083) 24. Of the functions of the form P 5 P0 e r t graphed in the accompanying figure, graph B has the smallest growth rate, r.
100
10
1 2001
Exports Imports
2002
2003
2004
2005
2006
Year
26. Between 2001 and 2006 America’s trade deficit (5 exports 2 imports) with China remained roughly constant. 27. We can think of both U.S. exports to China and imports from China (between 2001 and 2006) as functions of the year. The plots of both data sets appear to be approximately linear on a semi-log plot, so both functions are roughly exponential.
C
B
P
U.S. Trade with China 1000 Trade (in billions of $)
376
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A
28. If log(imports) 5 2.015 0.086t (where t 5 years since 2001) is the equation of the best-fit line for imports on a semi-log plot, then the annual growth rate for imports is about 8.6%. t
25. Of the two functions ƒ(x) 5 ln x and g(x) 5 log x graphed in the accompanying figure, graph A is the graph of ƒ(x) 5 ln x.
29. If we describe the growth in U.S. imports with the equation U.S. imports 5 103.59(1.22)t, then the annual growth rate for imports is about 22%. 30. If equations in Problems 28 and 29 are correct, they should be equivalent. II. In Problems 31–35, give an example of a function or functions with the specified properties. Express your answers using equations, and specify the independent and dependent variables.
y 3
A B x
0
10
31. A function whose graph is identical to the graph of the x function y 5 log . Å3 32. A function (that doesn’t use e) whose graph is identical to the graph of the function y 5 50.3e0.06t. (Hint: e0.06 <1.062)
–2
Problems 26–30 refer to the following table and semi-log plot. U.S. Trade with China (in billions of dollars) Year
Exports
Imports
2001 2002 2003 2004 2005 2006
19.2 22.1 28.4 34.7 41.9 50.0
102.3 125.2 152.4 196.7 243.5 263.6
Source: U.S. Bureau of the Census, Foreign Trade Statistics, www.census.gov/foreigntrade/balance.
33. A function (that uses e) whose graph is identical to the graph of the function y 5 100(0.974)t. (Hint: 0.974 < e20.0263.) 34. An exponential model describing the number of farms in the United States over time if there were 3.3 million farms in 1966 and 2.1 million farms in 2006. (Hint: (1/40) A2.1 < 0.989) 3.3 B 35. An exponential function y 5 ƒ(x) whose graph on a semi-log plot (x, log y) is the function log y 5 2 1 0.08x. (Hint: 100.08 < 1.20) III. Are the statements in Problems 36–40 true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample. 36. The number e used in the natural logarithm is a variable.
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37. Compounding at 3% quarterly is the same as compounding at 12% annually.
39. The graphs of the functions y 5 Ca x and y 5 Ce r x, where r 5 ln a, are identical.
38. Assuming your investment is growing, the effective interest rate is always greater than or equal to the nominal interest rate.
40. The interest rate r for continuous compounding in y 5 Ce rx is called the nominal interest rate or the instantaneous interest rate, depending on the application.
CHAPTER 6 REVIEW: PUTTING IT ALL TOGETHER
1. Solve each equation for the time t (in months) when the quantity Q 100. Then interpret each result. b. Q 200 ? 0.92t a. Q 50 ? 1.16t 2. a. According to the following chart, approximately how many deaths from natural disasters in the 1990s were due to floods? To severe storms? To bushfires? b. Why was this bar chart constructed with deaths displayed on a logarithmic scale?
Impact of Lead Polsoning Prevention Policy on Reducing Children’s Blood Lead Levels in the United States, 1971–2001 20 Leadbased Paint 16 Poisoning Prevention 14 Act (1971)
18 Blood lead levels (g/dl)
You will need a scientific calculator or its equivalent to evaluate logs and powers of e.
Blood lead levels Lead in plumbing banned (1986)
12 10
Residential Lead Paint Ban (1978)
8 6 4
Deaths from Natural Disasters during the 1990s
2 2002
2000
1998
1996
1994
1992
1990
1988
1986
1984
1982
1980
1978
1976
1974
1972
1970
0
1000
Year
Source: Blood lead levels: National Health and Nutrition Examination Survey. National Center for Health Statistics, Centers for Disease Control and Prevention.
Deaths (thousands)
100
10
1
Bushfire
Tsunami
Heat-wave
Severe storm
Other/geophysical
Flood
Tropical cyclone
Drought/famine
0
The best-fit exponential decay function for the data points on this graph is L(t) 18(0.88)t, where L(t) is the average blood lead level (in micrograms per deciliter) of children and t 5 years since 1976 (the first year data are available). a. What is the annual decay rate? b. Estimate from the graph when the initial blood lead level (in 1976) would be reduced by 50%. c. According to your model, when would the initial blood lead level be reduced by 50%? 5. The equation A(t) 325 ? (0.5)t and accompanying graph describe the amount of aspirin in your bloodstream at time t (measured in 20-minute periods) after you have ingested a standard aspirin of 325 milligrams (mg).
Source: Australian government, Department of Amount of Aspirin in Bloodstream
Meteorology.
4. The Lead-based Paint Poisoning Prevention Act (1971) was passed to reduce the toxic blood levels of lead in young children who might eat pieces of peeling lead-based paint. The following chart shows the subsequent drop in the average blood lead level of children, along with additional, related legislative acts.
300 250 Milligrams
3. Which of the following expressions are equivalent? d. 2 log x log y a. log(xy2) b. 2 log(xy) e. log x 2 log y f. 2 log x 2 log y c. log(xy)2
350
200 150 100 50 0 0
1
2
3
Time (in 20-minute periods)
4
5
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a. Using the equation, how much aspirin will remain after one time period? After two time periods? b. Use the graph to estimate the number of time periods it will take for the aspirin level to reach 100 mg. Now calculate the time using your equation and compare your results. c. Many doctors suggest that their adult patients take a daily dose of “baby aspirin” (81 mg) for long-term heart protection. Construct a new function B(t) to describe the amount of aspirin in your bloodstream after taking an 81-mg aspirin. (Assume the decay rate is the same as for a 325-mg aspirin.) How do the functions A(t) and B(t) differ, and how are they alike? 6. The Japanese government is concerned about its decreasing population. The Japanese people joke about when Japan will disappear entirely. According to a UN report, the Japanese population is believed to have peaked at 127.5 million in 2005 and, if the current rate continues, will contract to about 105 million people in 2050. a. What is the 45-year decay factor? The annual decay factor? b. Use the annual decay factor to construct an exponential decay function to model the Japanese population J(x), where x 5 years since 2005. c. Using the rule of 70, estimate the half-life of the population. According to your model, what is the half-life? 7. India and China have the opposite problem from that of Japan (see Problem 6): a huge population that is growing. In 2005 India had an estimated population of 1.08 billion on a land area of 1.2 million square miles, and China had a population of 1.30 billion on 3.7 million square miles. a. In 2005 did India or China have the larger population density (people per square mile)? b. China has been trying to slow population growth with its one child-per-family policy. As a result, in 2005 its annual population growth rate (0.6%) was considerably lower than that of India (1.6%). Assuming that each population continues to grow at the 2005 rate, constuct functions C(x) and I(x) for the population growth in China and India, respectively, letting x 5 years since 2005. c. Plot each function in part (b) on the same graph, for 30 years after 2005. d. Looking at the graph, is there a year when India’s population is projected to overtake China’s? If so, use your models to predict the year and population level. 8. Polonium-210 is a toxic radioactive substance named after Poland by the Curies, who discovered it. Polonium-210 poisoning is the suspected cause of death of former Soviet spy Alexander Litvinenko, in London, November 2006. a. Given that polonium has a half-life of approximately 138 days, construct a function to model the amount of polonium, P(T ), as a function of the original amount A and T, the number of half-life periods. b. Most of the world’s polonium-210 comes from Russia, which produces about 100 grams per year, which is sold commercially to the United States. How many time periods, T, would it take for 100 grams to decay until there is only 1 gram left? Translate this into days, and then into years.
c. The Soviet Union collapsed in 1991. What do you know about any pre-1991 Soviet-era stores of polonium-210? 9. At the birth of their granddaughter, the grandparents create a college fund in her name, investing $10,000 at 7% per year. a. Construct an equation to model the growth in the account. b. How much will be in the account in 18 years? c. If the annual inflation stays at 3%, how much will something that cost $10,000 today cost in 18 years? Calculate the difference between your answer in part (b) and your answer in part (c). d. Is this equivalent to the return you would have if you invested $10,000 at 4% for 18 years? If not, what does investing $10,000 at 4% represent? 10. Which of the following expressions are equivalent? a. ln ( !x/y) d. A12B ln x/ln y b. A12B ln(x/y)
e. A12B ln x A12B ln y
c. A12B ln x ln y
f. ln ( !x) /ln y
11. Solve each equation for t. a. 10 t 2.3 c. 60 30 e0.03t b. 2 log t log 4 2 d. ln(2t 5) ln(t 1)) 0 12. Simplify these expressions without using a calculator. c. 10 log(t 2) a. ln(e2) ln(3) b. e d. ln (x2 x) ln x 13. Convert each of the following expressions to a power of e. a. 1.5 b. 0.7 c. 1 14. For each of the following functions identify the growth (or decay) factor, the growth (or decay) rate, and the continuous growth (or decay) rate. a. Q 75(1.02 )t b. P(x) 50e0.3t 15. Match each function with one (or more) of the following graphs. a. y 2 x c. y 100(10)x b. y 2 log(x) d. log (y) 2 x y
y 10,000,000
8
8,000,000
6
6,000,000
4
4,000,000 2 2,000,000 0
1
2
3
4
5
x
x
0
1
2
Graph A
3
4
5
Graph C
y
y
8 7 6 5 4 3 2 1
8 6 4 2
x 0
1
2
3
Graph B
4
5
log (x) 0
1
2
3
Graph D
4
5
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16. a. In November 2006 the Bank of America offered a 12month certificate of deposit (CD) at a nominal rate (or annual percentage rate, APR) of 3.6%. What is the effective interest rate (or annual percentage yield, APY) if the bank compounds: i. Annually? ii. Monthly? iii. Continuously? b. The Bank lists the APY as 3.66%. Which of the above compounding schedules comes closest to this? 17. For each function, identify the corresponding graph. a. y1 ⫽ 2 ln x c. y3 5 ln(x ⫹ 2) d. y4 5 ln(x2) b. y2 ⫽ 2 ⫹ ln x y
–2 –1 –2 –3 –4
–2 –1 –2 –3 –4
Graph A
y
1 2 3 4 5 6
Graph C
y
4 3 2 1 –2 –1 –2 –3 –4
On October. 17, 1989, a magnitude 6.9 earthquake shook the San Francisco area. Robert Page from the U.S. Geological Survey said, “It was a wakeup call to prepare for the potentially even more devastating shocks that are inevitable in the future.” Find the ratio A/A0 for this earthquake. How many orders of magnitude larger was this earthquake’s amplitude compared with the base-level amplitude, A0?
x
1 2 3 4 5 6
22. The energy magnitude, M, radiated by an earthquake measures the potential damage to man-made structures. It can be described by the formula
4 3 2 1
x
x
1 2 3 4 5 6
–2 –1 –2 –3 –4
Graph B
M 5 A23B log E⫺2.9
1 2 3 4 5 6
where the seismic energy, E, is expressed in joules. Show that for every increase in M of one unit, the associated seismic energy E is increased by about a factor of 32.
Graph D
23. a. Given the following graph, generate a linear equation for Y(⫽ log y) in terms of x.
18. Which pair(s) of graphs show a function and its inverse? y
21. In Chapter 4 we encountered the Richter scale, which measures the amplitude of an earthquake. The Richter number R is defined as
where A is the amplitude of the shockwave caused by the earthquake and A0 is the reference amplitude, the smallest earthquake amplitude that could be measured by a seismograph at the time this definition was adopted, in 1935.
4 3 2 1
x
20. Beer has a pH of 4.5 and household lye a pH of 13.5. a. Which has the higher hydrogen ion concentration, and by how many orders of magnitude? b. What is the difference between the two pH values? How does this relate to your answer in part (a)? c. Why is it easier to use the pH number instead of [H⫹]?
R ⫽ log(A/A0)
y
4 3 2 1
379
y
y
Y (= log y) 6 5
x
x
x
4 3
Pair A
Pair B
Pair C
2 1
19. If the decibel level moves from 30 (a quiet conversation) to 80 (average street noise), by how many orders of magnitude has the intensity increased? Generate your answer in two ways: a. Using the table in the text b. Using the definition of decibels
x
0 0
2
4
6
8
10
b. Now substitute log y for Y and solve your equation for y. c. What type of function did you find in part (b)? What does this suggest about functions that appear linear on a semilog graph (with the log scale on the vertical axis)?
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E X P L O R AT I O N 6 . 1 Properties of Logarithmic Functions Objective • explore the effects of a and c on the graphs of y 5 c log(ax) and y 5 c ln(ax) Materials/Equipment • graphing calculator or computer with “E8: Logarithmic Sliders” in Exponential & Log Functions in course software or a function graphing program • graph paper Procedure Making Predictions 1. The effect of a on the equation y 5 log(ax) when a . 0 and x . 0. a. Why do we need to restrict a and x to positive values? b. Using the properties of logarithms, write the expression log(ax) as the sum of two logs. Discuss with a partner what effect you expect a to have on the graph. Now using log 2 < 0.301 and log 3 < 0.477, complete the accompanying data table and, by hand, sketch the three graphs on the same grid. Do your results confirm your predictions? What do you expect to happen to the graph of y 5 log(ax) if larger and larger positive values are substituted for a? Evaluating y 5 log(ax) when a 5 1, 2, and 3 x
y 5 log x
0.001 0.010 0.100 1.000 5.000 10.000
23.000 22.000 21.000 0.000 0.699 1.000
y 5 log(2x)
y 5 log(3x)
c. Discuss with your partner how you think the graphs of y 5 log x and y 5 log(ax) will compare if 0 , a , 1. Complete the following small data table and sketch the graphs of y 5 log x and y 5 log(x/10) on the same grid. Do your predictions and your graph agree? Predict what would happen to the graph of y 5 log(ax) if smaller and smaller positive values were substituted for a. 1 Evaluating y 5 log(ax) when a 5 1 and 10
x 0.001 0.010 0.100 1.000 5.000 10.000
y 5 log x
y 5 log(x/10)
23.000 22.000 21.000 0.000 0.699 1.000
d. How do you think your findings for y 5 log(ax) relate to the function y 5 ln(ax)?
380
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2. The effect of c on the equation y 5 c log x a. When c . 0: Talk over with your partner your predictions for what will happen as c increases. Fill in the following table and on the same grid draw a quick sketch of the three functions. Were your predictions correct? What do you expect to happen to the graph of y 5 c log x as you substitute larger and larger positive values for c? Evaluating y 5 c log x when c 5 1, 2, and 3 x
y 5 log x
0.001 0.010 0.100 1.000 5.000 10.000
23.000 22.000 21.000 0.000 0.699 1.000
y 5 2 log x
y 5 3 log x
b. When c , 0: Why can c be negative when a has to remain positive? How do you think the graphs of y 5 log x and y 5 c log x will compare if c , 0? What happens if c remains negative but k c k gets larger and larger (for example, c 5 210, 2150, 25000, etc.)? c. How do your findings on y 5 c log x relate to the function y 5 c ln x? 3. Generalizing your results. Talk over with your partner the effect of varying both a and c on the general functions y 5 c log(ax) and y 5 c ln(ax). Try predicting the shapes of the graphs of such functions as y 5 3 log(2x) and y 5 22 log(x/10). Have each partner construct a small table, and graph the results of one such function. Compare your findings. Testing Your Predictions Test your predictions by either using “E8: Logarithmic Sliders” in Exponential & Log Functions or creating your own graphs with a graphing calculator or a function graphing program. Try changing the value for a to create different functions of the form y 5 log(ax) or y 5 ln(ax). Be sure to try values of a . 1 and values such that 0 , a , 1. Then try changing the value for c in functions of the form y 5 c log x and y 5 c ln x. Let c assume both positive and negative values. Summarizing Your Results Write a 60-second summary describing the effect of varying a and c in functions of the form y 5 c log(ax) and y 5 c ln(ax). Exploration-Linked Homework 1. Use your knowledge of logarithmic functions to predict the shapes of the graphs of: a. The decibel scale: given by the function N 5 10 log II0, where I is the intensity of a sound and I0 is the intensity of sound at the threshold of human hearing. (See Section 6.4, pages 359–361.) b. Stellar magnitude: approximated by the function M 5 22.5 log BB0, where B is the brightness of a star and B0 is a constant. (See Section 6.4, Exercise 8, p.363.) 2. Explore the effect of changing the base, that is, the effect of changing b in y 5 c log b(ax). (See Section 6.5, Exercise 7, p. 365.)
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POWER FUNCTIONS OVERVIEW Power functions help us answer questions such as “Why is the most important rule of scuba diving ‘never hold your breath’?” and “Why do small animals have faster heartbeats and higher metabolic rates than large ones?” After reading this chapter you should be able to • recognize the properties of power functions • construct and interpret graphs of power functions • understand direct and inverse proportionality • use logarithmic scales to determine whether a linear, power, or exponential function is the best function model for a set of data • develop a sense about the relationship between size and shape
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7.1 The Tension between Surface Area and Volume Why do small animals have faster heartbeats and higher metabolic rates than large ones? Why are the shapes of the bodies and organs of large animals often quite different from those of small ones? To find answers to these questions, we examine how the relationship between surface area and volume changes as objects increase in size.
Scaling Up a Cube Let’s look at what happens to the surface area and volume of a simple geometric figure, the cube, as we increase its size. In Figure 7.1 we have drawn a series of cubes where the lengths of the edges are 1, 2, 3, and 4 units.1
Figure 7.1 Four cubes for which the lengths of the
edges are 1, 2, 3, and 4 units, respectively.
Surface area of a cube If we were painting a cube, the surface area would tell us how much area we would have to cover. Each cube has six identical faces, so surface area 5 6 ? sarea of one faced If the edge length of one face of the cube is x, then the surface area of that face is x 2. So the total surface area S(x) of the cube is Ssxd 5 6x 2 S(x) is called a power function of degree 2. The second column of Table 7.1 lists the surface areas of cubes for various edge lengths. Figure 7.2 shows a graph of the function S(x).
Ratio of Surface Area to Volume of Cube with Length of Edge x Edge Length x
Surface Area Ssxd 5 6x 2
Volume Vsxd 5 x 3
1 2 3 4 6 8 10
6 24 54 96 216 384 600
1 8 27 64 216 512 1000
Surface Area Volume Ssxd 6x 2 5 5 6x 5 Vsxd x3
Rsxd
6.00 3.00 2.00 1.50 1.00 0.75 0.60
Table 7.1
1
For this discussion it doesn’t matter which unit we use, but if you prefer, you may think of “unit” as being “centimeter” or “foot.”
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Since the surface area of a cube with edge length x can be represented as surface area 5 constant ? x 2 600 500 Surface area of cube
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400 300 200 100
x 0
2
4
6
8
10
Edge length of cube
Figure 7.2 Graph of the surface
area of a cube, Ssxd 5 6x 2.
we say that the surface area is directly proportional to the square (or second power) of the length of its edge. In Table 7.1, observe what happens to the surface area when we double the length of an edge. If we double the length from 1 to 2 units, the surface area becomes four times larger, increasing from 6 to 24 square units. If we double the length from 2 to 4 units, the surface area is again four times larger, increasing this time from 24 to 96 square units. In general, if we double the length of the edge from x to 2x, the surface area will increase by a factor of 22, or 4: Ssxd 5 6x 2 Ss2xd 5 6s2xd 2 5 6 ? 22 ? x 2 5 4s6x 2d Ss2xd 5 4 ? Ssxd
Surface area of a cube with edge length x: Surface area of a cube with edge length 2x: If we apply rules of exponents simplify and rearrange terms substitute S(x) for 6x 2, we get
Volume of a cube If the edge length of one face of the cube is x, then the volume V(x) of the cube is given by Vsxd 5 x 3 V(x) is an example of a power function of degree 3. The third column of Table 7.1 lists values for V(x), and Figure 7.3 is the graph of V(x). 600 500 Volume of cube
Kime07_C07pg383-452.qxd
400 300 200 100
x 0
2
4
6
8
10
Edge length
Figure 7.3 Graph of the
volume of a cube, Vsxd 5 x 3.
Since volume 5 constant ? x 3
(the constant in this case is 1)
we say that the volume is directly proportional to the cube (or the third power) of the length of its edge.
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What happens to the volume when we double the length? In Table 7.1, if we double the length from 1 to 2 units, the volume increases by a factor of 23 or 8, increasing from 1 to 8 cubic units. If we double the length from 2 to 4 units, the volume again becomes eight times larger, increasing from 8 to 64 cubic units. In general, if we double the edge length from x to 2x, the volume will increase by a factor of 8: Vsxd 5 x 3 Vs2xd 5 s2xd 3 5 23 ? x3 5 8x 3 Vs2xd 5 8 ? Vsxd
Volume of a cube with edge length x: Volume of a cube with edge length 2x: If we apply rules of exponents simplify substitute V(x) for x3, we get
Ratio of the surface area to the volume The ratio of (surface area)/volume generates a new function R(x), where S(x) 6x2 6 5 3 5 5 6x21 V(x) x x R(x) is called a power function of degree –1. As we increase the value of x, the edge length of the cube, the value of R(x) decreases, as we can see in Table 7.1 and Figure 7.4 R(x) 5
6
(Surface area)/volume
5 4 3 2 1
0
2
4 6 Edge length
8
10
Figure 7.4 Graph of R(x) =
(surface area)/volume.
As the size of the cube increases, the volume increases faster than the surface area. For example, if we double the edge length, the surface area increases by a factor of 4, but the volume increases by a factor of 8. So the ratio surface area volume decreases as the edge length increases.
Size and Shape What we learned about the cube is true for any three-dimensional object, no matter what the shape. In general, In Exploration 7.1 you can study further the effects of scaling up an object.
For any shape, as an object becomes larger while keeping the same shape, the ratio of its surface area to its volume decreases. Thus a larger object has relatively less surface area than a smaller one. This fact allows us to understand some basic principles of biology and to answer the questions we asked at the beginning of this section.
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7.1 The Tension between Surface Area and Volume
Stephen Jay Gould’s essay “Size and Shape” in Ever Since Darwin: Reflections in Natural History offers an interesting perspective on the relationship between the size and shape of objects.
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Biological functions such as respiration and digestion depend upon surface area but must service the body’s entire volume.2 The biologist J. B. S. Haldane wrote that “comparative anatomy is largely the story of the struggle to increase surface in proportion to volume.” This is why the shapes of the bodies and organs of large animals are often quite different from those of small ones. Many large species have adapted by developing complex organs with convoluted exteriors, thus greatly increasing the organs’ surface areas. Human lungs, for instance, are heavily convoluted to increase the amount of surface area, thereby increasing the rate of exchange of gases. Stephen Jay Gould wrote that “the villi of our small intestine increase the surface area available for absorption of food.” Body temperature also depends upon the ratio of surface area to volume. Animals generate the heat needed for their volume by metabolic activity and lose heat through their skin surface. Small animals have more surface area in proportion to their volume than do large animals. Since heat is exchanged through the skin, small animals lose heat proportionately faster than large animals and have to work harder to stay warm. Hence their heartbeats and metabolic rates are faster. As a result, smaller animals burn more energy per unit mass than larger animals.
Algebra Aerobics 7.1 The inside back cover of the text contains geometric formulas. 1. The function S(r) 5 4pr2 gives the surface area of a sphere with radius r. a. Compare S(r), S(2r), and S(3r). b. What happens to the surface area when the radius is doubled? When the radius is tripled? 4 2. The function V(r) 5 pr3 gives the volume of a sphere 3 with radius r. a. Compare V(r), V(2r), and V(3r). b. What happens to the volume when the radius is doubled? When the radius is tripled? 3. As the radius of a sphere increases, which grows faster, the surface area or the volume? What happens to the ratio S(r)/V(r) 5 (surface area)/volume as the radius increases? 4. The two-dimensional analog to the ratio of (surface area)/volume for a sphere is the ratio of circumference/ area for a circle. a. The circumference C(r) of a circle with radius r is 2pr. Compare C(r), C(2r), and C(4r). What happens to the circumference when the radius is doubled? Quadrupled (multiplied by 4)? b. The area A(r) of a circle with radius r is pr2. Compare A(r), A(2r), and A(4r). What happens to the area when the radius is doubled? Quadrupled (multiplied by 4)? c. Which grows faster, the circumference or the area? As a result, what happens to the ratio C(r)/A(r) 5 circumference/area as the radius increases? 2
5. Compare a sphere with radius r and a cube with edge length r. a. For what value(s) of r, if any, is the volume of the cube equal to the volume of the sphere? b. For what value(s) of r, if any, is the volume of the cube greater than the volume of the sphere? Less than? 6. Two different cylinders have the same height of 25 ft, but the base of one has a radius of 5 feet and the base of the other has a radius of 10 feet. Is the volume of one cylinder double the volume of the other? Justify your answer. 7. A cylindrical silo has a radius of 12 feet. If a certain amount of feed fills the silo to a depth of 5 feet, will twice the amount of feed fill the silo to a depth of 10 feet? Justify your answer. 8. The volume V of some regular figures can be defined as the area B of the base times the height h, or V 5 Bh. a. Use this definition to find the formulas for the volumes in Figures 7.5 and 7.6. b. In each case, if the height is doubled, by what factor is the volume increased? c. If all dimensions are doubled, by what factor is the volume increased? a r
b
h h
Cylinder
Triangular prism
Figure 7.5
Figure 7.6
For those who want to investigate how species have adapted and evolved over time, see D.W. Thompson, On Growth and Form (New York: Dover, 1992), and T. McMahon and J. Bonner, On Size and Life (New York: Scientific American Books, 1983).
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Exercises for Section 7.1 A calculator that can evaluate powers is required throughout. Formulas for geometric figures are on the inside back cover of the text. 1. Solve the following formulas for the indicated variable. a. V 5 lwh, solve for l c. P 5 2l 1 2w, solve for w hb b. A 5 , solve for b d. S 5 2x2 1 4xh, solve for h 2 2. In parts (a)–(c), evaluate the functions at R and 3R. a. Csrd 5 2pr, the circumference of a circle with radius r b. Asrd 5 pr2, the area of a circle with radius r c. Vsrd 5 43pr 3, the volume of a sphere with radius r d. Describe what happens to C(r), A(r), and V(r) when the radius triples from R to 3R. 3. Using the formulas in Exercise 2, simplify the following ratios and identify which ones are increasing as r increases. Describe each result in geometric terms. a.
Asrd Csrd
b.
Vsrd Asrd
c.
Vsrd Csrd
4. Consider the formulas V1 5 43pr3 (volume of a sphere) and V2 5 pr2h (volume of a cylinder). Simplify the following ratios. Which one(s) increase as r increases (assume h is held constant)? Describe your result in geometric terms. V V a. 2 b. 1 V1 V2 5. A tiny sphere has a radius of 0.000 000 000 1 5 10210 meter, which is roughly equivalent to the radius of a protein molecule. Answer the following questions. Express your answers in scientific notation. a. Find its surface area, S, in square meters, where S 5 4pr2. b. Find its volume, V, in cubic meters, where V 5 43pr3. c. Find the ratio of the surface area to the volume. d. As r increases, does the ratio in part (c) increase or decrease? 6. The radius of Earth is about 6400 kilometers. Assume that Earth is spherical. Express your answers to the questions below in scientific notation. a. Find the surface area of Earth in square meters. b. Find the volume of Earth in cubic meters. c. Find the ratio of the surface area to the volume. 7. If the radius of a sphere is x meters, what happens to the surface area and to the volume of a sphere when you: a. Quadruple the radius? b. Multiply the radius by n? c. Divide the radius by 3? d. Divide the radius by n? 8. Assume a box has a square base and the length of a side of the base is equal to twice the height of the box. a. If the height is 4 inches, what are the dimensions of the base?
b. Write functions for the surface area and the volume that are dependent on the height, h. c. If the volume has increased by a factor of 27, what has happened to the height? d. As the height increases, what will happen to the ratio of (surface area)/volume? 9. A box has volume V 5 length ? width ? height. a. Find the volume of a cereal box with dimensions of length 5 19.5 cm, width 5 5 cm, and height 5 27 cm. (Be sure to specify the unit.) b. If the length and width are doubled, by what factor is the volume increased? c. What are two ways you could increase the volume by a factor of 4 and keep the height the same? 10. A box with a lid has a square base with each base side x inches and a height of h inches. a. Write the formula for the volume of the box. b. Write the formula for the surface area. (Hint: The surface area of the box is the sum of the areas of each side of the box.) c. If the side of the base is tripled and the height remains constant, by what factor is the volume increased? d. If the side of the base triples (from x to 3x) with h held constant, what is the change in the surface area? 11. Consider a cylinder with volume V 5 pr 2h. What happens to its volume when you double its height, h? When you double its radius, r? 12. Consider two solid figures, a sphere and cylinder, where each has radius r. The volume of a sphere is Vs 5 43pr 3 and the volume of a cylinder is Vc 5 pr 2 h.
r
h r
a. If h 5 1, when is the volume of the sphere greater than the volume of the cylinder? b. What value of h would make the volumes the same? c. Complete this statement: “The volume of the cylinder with radius r is greater than the volume of a sphere of radius r, when h . ____.” 13. The volume, V, of a cylindrical can is V 5 pr 2h and the total surface area, S, of the can is S 5 area of curved surface 1 2 ? sarea of based 5 2prh 1 2pr 2 where r is the radius of the base and h is the height. a. Assume the height is three times the radius. Write the volume and the surface area as functions of the radius.
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b. As r increases, which grows faster, the volume or the surface area? Explain.
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5 4
14. (Hint: See Exploration 7.1.) To celebrate Groundhog Day, a gourmet candy company makes a solid, quarter-pound chocolate groundhog that is 2 inches tall. a. If the company wants to introduce a solid chocolate groundhog in the same shape but twice as tall (4 inches high), how much chocolate is required? b. How tall would a solid chocolate groundhog be if made with twice the original quarter-pound amount (for a total of 8 ounces)? 15. A circle has a radius of x units. Which graph could represent the ratio of area to circumference? Which could represent the ratio of circumference to area? Explain your answers.
3
j(x)
2
h(x)
1
x 0
1
2
3
4
5
16. A cylinder (with no top or bottom) has a radius of x feet and a height of 3 feet. a. Construct a function for the volume and another for the surface area of the cylinder. volume b. Construct the ratio surface area . What would happen to the ratio as x increases?
7.2 Direct Proportionality: Power Functions with Positive Powers In the last section we encountered three functions of the edge length x of a cube: Surface area Volume surface area Ratio of volume
Ssxd 5 6x 2 Vsxd 5 x 3 6 Rsxd 5 or 6x21 x
All three are called power functions since they are of the form dependent variable 5 constant ? sindependent variabled power or output 5 constant ? inputpower
Power Functions A power function y 5 ƒ(x) can be represented by an equation of the form y 5 kx p where k and p are constants.
For the functions: Ssxd 5 6x 2, we have k 5 6 and p 5 2. Vsxd 5 x 3, we have k 5 1 and p 5 3. Rsxd 5 6x21, we have k 5 6 and p 5 21. EXAMPLE
1
Decide whether or not each of the following functions is a power function of the form y 5 kx p, and if it is, identify the values of k and p. a. y 5 3x4 b. y 5 3x4 1 1 c. y 5 2x24 d. y 5 26 "x e. y 5 4 ? 2 x
SOLUTION
a. y 5 3x4 is a power function where k 5 3 and p 5 4. b. y 5 3x4 1 1 is not a power function since it cannot be written in the form y 5 kx p. c. y 5 2x24 is a power function where k 5 2 and p 5 24.
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d. y 5 26 "x 5 26x 1/2 is a power function where k 5 26 and p 5 12. e. y 5 4 ? 2 x is not a power function, since the variable x is in the exponent. It’s an exponential function.
Direct Proportionality In Chapter 2, for linear functions, we said that y is directly proportional to x if y equals a constant times x. For example, if y 5 4x, then y is directly proportional to x. We can extend the same concept to any power function with positive exponents. If y 5 kx p and p is positive, we say that y is directly proportional to x p. In Section 7.1, we saw that the surface area of a cube is directly proportional to the square of its edge length and that the volume is directly proportional to the cube of its edge length. The symbol ~ is used to indicate direct proportionality. For example, if y 5 5x 3, then y is directly proportional to x3, which we write as y ~ x 3.
Direct Proportionality If y 5 kx p
(k 2 0 and p . 0)
we say that y is directly proportional to x p. We write this as y ~ xp The coefficient k is called the constant of proportionality.
EXAMPLE
2
SOLUTION
Write formulas to represent the following relationships. a. The circumference, C, of a circle is directly proportional to its radius, r. b. The area, A, of a circle is directly proportional to its radius, r, squared. c. The volume, V, of a liquid flowing through a tube is directly proportional to the fourth power of the radius, r, of the tube. a. C 5 kr, where k 5 2p. b. A 5 kr 2, where k 5 p.
c. V 5 kr4 for some constant k.
Properties of Direct Proportionality Consider a general power function ƒsxd 5 kx p where p . 0. If we double the input from x to 2x, we have evaluating ƒ at 2x using rules of exponents and rearranging terms substituting ƒ(x) for kx p
?
SOMETHING TO THINK ABOUT
Weight is directly proportional to length cubed, but the ability to support the weight, as measured by the cross-sectional area of bones, is proportional to length squared. Why does this mean that Godzilla or King Kong could exist only in the movies?
ƒs2xd 5 5 5 5
ks2xd p ks2px pd 2p skx pd 2pƒsxd
Doubling the input multiplies the output by 2p. We saw in Section 7.1 that if we double the input for the function Ssxd 5 6x 2 (the surface area of a cube), the output is multiplied by 22 or 4. If we multiply the input by m, changing the input from x to mx, then evaluating ƒ at mx using rules of exponents and rearranging terms substituting ƒ(x) for kx p
ƒsmxd 5 ksmxd p 5 ksm px p d 5 m pskx p d 5 m pƒsxd
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So multiplying the input by m multiplies the output by m p. For example, if we triple the input for Ssxd 5 6x 2, the output is multiplied by 32 or 9. Note that the value for k (the constant of proportionality) is irrelevant in these calculations. In general,
If y is directly proportional to x p (where p . 0), then y 5 kx p for some nonzero constant k. Multiplying the input by m multiplies the output by m p. For example, tripling the input multiplies the output by 3 p.
EXAMPLE
3
Given each function, what happens if the input is doubled? Increased by a factor of 10? Cut in half? a. y 5 2x4 b. hszd 5 22x5
SOLUTION
a. If the input is doubled, the output is multiplied by 2 4 or 16. If the input is multiplied by 10, the output is multiplied by 10 4 or 10,000. If the input is cut in half, the output 1 would be multiplied by A12B 4 5 16 . b. If the input is doubled, the output is multiplied by 25 or 32. If the input is multiplied by 10, the output is multiplied by 105 or 100,000. If the input is cut in half, the 1 output would be multiplied by A12B 5 5 32 .
EXAMPLE
4
SOLUTION
a. What is the difference among 2ƒ(x), ƒ(2x), and ƒ(x) 1 2? b. If ƒsxd 5 x 2, evaluate the three expressions in part (a) when x 5 4. a. 2ƒ(x) means to multiply the value of ƒ(x) by 2. ƒ(2x) means to use 2x as the input for the function ƒ. ƒ(x) 1 2 means to evaluate ƒ(x) and then add 2. b. If ƒsxd 5 x 2 and x 5 4, then 2ƒs4d 5 2 ? 42 5 2 ? 16 5 32 ƒs2 ? 4d 5 ƒs8d 5 82 5 64, or equivalently, 5 s2 ? 4d 2 5 22 ? 42 5 4 ? 16 5 64 ƒs4d 1 2 5 4 2 1 2 5 16 1 2 5 18
EXAMPLE
5
Pedaling into the wind When you pedal a bicycle, it’s much easier to pedal with the wind than into the wind. That’s because as the wind blows against an object, it exerts a force upon it. This force is directly proportional to the wind velocity squared. a. Construct an equation to describe the relationship between the wind force and the wind velocity. b. If the wind velocity doubles, by how much does the wind force go up?
SOLUTION
a. Fsvd 5 kv2 for some constant k, where v is the wind velocity and F(v) is the wind force.
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b. If the velocity doubles, then the wind force is multiplied by 2 2 or 4. So pedaling into a 20-mph wind requires four times as much effort as pedaling into a 10-mph wind. (See Figure 7.7.) Wind force
F(2v) = 4F(v)
F(v) 0
v
Wind velocity
2v
Figure 7.7 Doubling the wind velocity
quadruples the wind force. EXAMPLE
6
Earth’s core The radius of the core of Earth is slightly over half the radius of Earth as a whole, yet the core is only about 16% of the total volume of Earth. How is this possible?
SOLUTION
We can think of both Earth and its core as approximately spherical in shape (see Figure 7.8). Earth Core
r
Figure 7.8 A sketch of a
spherical Earth with radius r and its core.
If we let r 5 radius of Earth, then Earth’s volume is VEarth 5 43 pr 3 If the radius of the core were exactly half that of Earth A2r B , then the core’s volume would be Vcore 5 43p A2r B 3 3
use rules of exponents
5 43pQ2r 3R
multiply and regroup terms
5 Q213RQ43pr 3R
substitute 18 for Q213R and VEarth for 43pr 3
5 18VEarth 5 12.5% of the volume of Earth
Since the radius of the core is slightly more than half the radius of Earth, 16% of Earth’s volume is a reasonable estimate for the volume of the core.
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EXAMPLE
7
393
In Chapter 4 we encountered the following formula used by police. It estimates the speed, S, at which a car must have been traveling given the distance, d, the car skidded on a dry tar road after the brakes were applied: S 5 "30d < 5.48d 1/2 Speed, S, is in miles per hour and distance, d, is in feet. a. Use the language of proportionality to describe the relationship between S and d. b. If the skid marks were 50 feet long, approximately how fast was the driver going when the brakes were applied? c. What happens to S if d doubles? Quadruples? d. Is d directly proportional to S?
SOLUTION
a. The speed, S, is directly proportional to d 1/2 and the constant of proportionality is approximately 5.48. b. If the length of the skid marks, d, is 50 feet, then the estimated speed of the car S < 5.48 ? s50d 1/2 < 5.48s7.07d < 39 mph. c. If the skid marks double in length from d to 2d, then the estimate for the speed of the car increases from "30d to "30 ? s2dd 5 "2 ? "30d. So the estimated speed goes up by a factor of "2 < 1.414 or, equivalently, by about 41.4%. If the skid marks quadruple in length from d to 4d, then the estimated speed of the car goes from "30d to "30 ? s4dd 5 "4 ? "30d 5 2 "30d. So the speed estimate goes up by a factor of 2, or by 100%. For example, if the skid marks doubled from 50 to 100 feet, the speed estimate would go up from 39 mph to 1.414 ? 39 < 55 mph. If the skid marks quadrupled from 50 to 200 feet, then the speed estimate would double from about 39 to almost 78 mph. d. To determine if d is directly proportional to S, we need to solve our original equation for d. S 5 "30d
Given square both sides of the equation divide both sides by 30
S 2 5 30d S2 5d 30 d5
or
S2 30
So d is directly proportional to S 2 but not to S.
Direct Proportionality with More Than One Variable When a quantity depends directly on more than one other quantity, we no longer have a simple power function. For example, the volume, V, of a cylindrical can depends on both the radius, r, of the base and the height, h. The equation describing this relationship is V 5 area of base ? height V 5 pr2h We say V is directly proportional to both r 2 and h.
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EXAMPLE
8
Deflection A person stands at the center of a fir plank (with a cross section of 2s by 12s ) that is anchored at both ends (see Figure 7.9). Architects estimate the downward deflection of the plank using the formula D 5 s6.6 ? 1026 d ? P ? L3 where D is the deflection in inches, P is the weight of the person in pounds, and L is the length of the plank in feet.3
Figure 7.9 A person
standing at the center of a plank.
a. Describe the relationship among D, P, and L using the language of proportionality. b. What is the downward deflection of the plank if the person weighs 200 lb and the plank is 6r long? c. What happens to D if we increase P by 50%? If we increase L by 50%? SOLUTION
a. D is directly proportional to P and to L3. b. If P 5 200 lb and L 5 6r, then D 5 s6.6 ? 1026 d ? 200 ? 63 rewrite 200 and rearrange terms
5 s6.6 ? 63 ? 2d ? 102 ? 1026
evaluate
5 2851.2 ? 1024 < 0.3 inches
So the plank is about a third of an inch lower in the middle, under the 200-lb person. c. Increasing the person’s weight by 50% means multiplying the weight by 1.5. Since the deflection is directly proportional to the person’s weight, that means that the deflection is multiplied by 1.5 or, equivalently, increased by 50%. So if we replace a 200-lb person (who causes a deflection of 0.3 inches) with someone 300 lb (or 50% heavier), the deflection would become (1.5) ? (0.3) 5 0.45, or almost half an inch. Increasing the plank’s length by 50% means multiplying the length by 1.5. Since the deflection is directly proportional to the cube of the plank length, the deflection is multiplied by s1.5d 3 < 3.4 or, equivalently, increased by 240%. So for a 200-lb person, if the plank length of 6r is increased by 50% [to (1.5) ? 6 5 9r ], the projected deflection would be s3.4d ? s0.3 inchesd 5 1.02 inches, which is at the edge of reliability for this model.
3
The deflection formula for the plank is quite accurate for deflections up to about an inch. After that it starts to produce unrealistic D values very quickly since L is cubed.
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Algebra Aerobics 7.2 1. In each case, indicate whether or not the function is a power function. If it is, identify the independent and dependent variables, the constant of proportionality, and the power. a. A 5 pr 2 c. z 5 w 5 1 10 e. y 5 3x 5 b. y 5 z 5 d. y 5 5 x 2. Identify which (if any) of the following equations represent direct proportionality: a. y 5 5.3x 2 c. y 5 5.3 ? 2 x 2 b. y 5 5.3x 1 10 3. Given g(x) 5 5x3 a. Calculate g(2) and compare this value with g(4). b. Calculate g(5) and compare this value with g(10). c. What happens to the value of g(x) if x doubles in value? d. What happens to g(x) if x is divided by 2? 4. Given h(x) 5 0.5x2 a. Calculate h(2) and compare this value with h(6). b. Calculate h(5) and compare this value with h(15). c. What happens to the value of h(x) if x triples in value? d. What happens to h(x) if x is divided by 3?
5. Express each of the following relationships with an equation. a. The volume of a sphere is directly proportional to the cube of its radius. b. The volume of a prism is directly proportional to its length, width, and height. c. The electrostatic force, f, is directly proportional to the particle’s charge, c. 6. Express in your own words the relationship between y and x in the following functions. x5 a. y 5 3x 5 b. y 5 2.5x 3 c. y 5 4 7. Express each of the relationships in Problem 6 in terms of direct proportionality. 8. a. Solve P 5 aR 2 for R. b. Solve V 5 A13B pr 2h for h. 9. If ƒsxd 5 0.1x 3, evaluate and describe the difference between the following functions. a. ƒ(2x) and 2ƒ(x) b. ƒ(3x) and 3ƒ(x) c. Evaluate each function in part (a) for x 5 5.
Exercises for Section 7.2 A calculator that can evaluate powers is useful here. A graphing program is recommended for Exercises 15 and 16. 1. Evaluate the following functions when x 5 2 and x 5 22: a. ƒsxd 5 5x2 c. hsxd 5 25x2 3 b. gsxd 5 5x d. ksxd 5 25x3 2. Identify which of the following are power functions. For each power function, identify the value of k, the constant of proportionality, and the value of p, the power. a. y 5 23x2 b. y 5 3x10
c. y 5 x2 1 3 d. y 5 3x
3. Each of the following represents direct proportionality with powers of both x and y. Identify the constant of proportionality, k, and then evaluate each function for x 5 4 and y 5 3. a. P 5 12x3y2 c. T 5 4px2y3 b. M 5 14x1/2y d. N 5 s18xyd 1/3 4. Find the constant of proportionality, k, for the given conditions. a. y 5 kx3, and y 5 64 when x 5 2. b. y 5 kx3/2, and y 5 96 when x 5 16. c. A 5 kr2, and A 5 4p when r 5 2. d. v 5 kt2, and v 5 2256 when t 5 4.
5. Write the general formula to describe each variation and then solve the variation problem. a. y is directly proportional to x, and y 5 2 when x 5 10. Find y when x is 20. b. p is directly proportional to the square root of s, and p 5 12 when s 5 4. Find p when s 5 16. c. A is directly proportional to the cube of r, and A 5 8p 3 3 when r 5 " 6. Find A when r 5 2 ? " 6. d. P is directly proportional to the square of m, and P 5 32 when m 5 8. If m is halved, what happens to P? 6. The data in the table satisfy the equation y 5 kx n, where n is a positive integer. Find k and n. x y
2 1
3 2.25
4 4
5 6.25
7. Assume Y is directly proportional to X 3. a. Express this relationship as a function where Y is the dependent variable. b. If Y 5 10 when X 5 2, then find the value of the constant of proportionality in part (a). c. If X is increased by a factor of 5, what happens to the value of Y?
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d. If X is divided by 2, what happens to the value of Y? e. Rewrite your equation from part (a), solving for X. Is X directly proportional to Y? 8. The distance, d, a ball travels down an inclined plane is directly proportional to the square of the total time, t, of the motion. a. Express this relationship as a function where d is the dependent variable. b. If a ball starting at rest travels a total of 4 feet in 0.5 second, find the value of the constant of proportionality in part (a). c. Complete the equation and solve for t. Is t directly proportional to d? 9. a. Assume L is directly proportional to x5. What is the effect of doubling x? b. Assume M is directly proportional to x p, where p is a positive integer. What is the effect of doubling x? 10. In “Love That Dirty Water” (Chicago Reader, April 5, 1996), Scott Berinato interviewed Ernie Vanier, captain of the towboat Debris Control. The Captain said, “We’ve found a lot of bowling balls. You wouldn’t think they’d float, but they do.” When will a bowling ball float in water? The bowling rule book specifies that a regulation ball must have a circumference of exactly 27 inches. Recall that the circumference of a circle with radius r is 2pr. The volume of a sphere with radius r is 43pr3. a. What is a regulation bowling ball’s radius in inches? b. What is the volume of a regulation bowling ball in cubic inches? (Retain at least two decimal places in your answer.) c. What is the weight in pounds of a volume of water equivalent in size to a regulation bowling ball? (Water weighs 0.03612 lb/in3). d. A bowling ball will float when its weight is less than or equal to the weight of an equivalent volume of water. What is the heaviest weight of a regulation bowling ball that will float in water? e. Typical men’s bowling balls are 15 or 16 pounds. Women commonly use 12-pound bowling balls. What will happen to the men’s and to the women’s bowling balls when dropped into the water? Will they sink or float? 11. When a variable is directly proportional to the product of two or more variables, we say that the variable is jointly proportional to those variables. Express each of the following as an equation. a. x is jointly proportional to y and the square of z. b. V is jointly proportional to l, w, and h. c. w is jointly proportional to the square of x and the cube root of y. d. The volume of a cylinder is jointly proportional to its height and the square of its radius. 12. Write a general formula to describe each variation. Use the information given to find the constant of proportionality. a. Q is directly proportional to both the cube root of t and the square of d, and Q 5 18 when t 5 8 and d 5 3. b. A is directly proportional to both h and the square of the radius, r, and A 5 100p when r 5 5 and h 5 2.
c. V is directly proportional to B and h, and V 5 192 when B 5 48 and h 5 4. d. T is directly proportional to both the square root of p and the square of u, and T 5 18 when p 5 4 and u 5 6. 13. The cost, C, in dollars, of insulating a wall is directly proportional to the area, A, of the wall (measured in ft2) and the thickness, t, of the insulation (measured in inches). a. Write a cost equation for insulating a wall. b. If the insulation costs are $12 when the area is 50 ft2 and the thickness is 4 inches, find the constant of proportionality. c. A storage room is 15 ft by 20 ft with 8-ft-high ceilings. Assuming no windows and one uninsulated door that is 3 feet by 7 feet, what is the total area of the insulated walls for this room? d. What is the cost of insulating this room with 4-inch insulation? With 6-inch insulation? 14. Suppose you are traveling in your car at speed S and you suddenly brake hard, leaving skid marks on the road. A “rule of thumb” for the distance, D, that the car will skid is given by D5
S2 30ƒ
where D 5 distance the car skids (in feet), S 5 speed of the car (in miles per hour), and f is a number called the coefficient of friction that depends on the road surface and its condition. For a dry tar road, ƒ < 1.0. For a wet tar road, ƒ < 0.5. (We saw a variation of this problem in Example 7, Section 7.2.) a. What is the equation giving distance skidded as a function of speed for a dry tar road? For a wet tar road? b. Generate a small table of values for both functions in part (a), including speeds between 0 and 100 miles per hour. c. Plot both functions on the same grid. d. Why do you think the coefficient of friction is less for a wet road than for a dry road? What effect does this have on the graph in part (c)? e. In the accompanying table, estimate the speed given the distances skidded on dry and on wet tar roads. Describe the method you used to find these numbers. Distance Skidded (ft)
Estimated Speed (mph) Dry Tar
Wet Tar
25 50 100 200 300 f. If one car is going twice as fast as another when they both jam on the brakes, how much farther will the faster car skid? Explain. Does your answer depend on whether the road is dry or wet? 15. (Graphing program recommended.) Assume a person weighing P pounds is standing at the center of a 4s 3 12s fir plank that spans a distance of L feet. The downward deflection Ddeflection (in inches) of the plank can be described by Ddeflection 5 (5.25 ? 10]7) ? P ? L3
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a. Graph the deflection formula, Ddeflection, assuming P 5 200 pounds. Put values of L (from 0 to 25 feet) on the horizontal axis and deflection (in inches) on the vertical axis. b. A rule used by architects for estimating acceptable deflection, Dsafe, in inches, of a beam L feet long bent downward as a result of carrying a load is
c. The area, A, of a triangle is directly proportional to its base, b, and height, h. d. The reaction rate, R, is directly proportional to the concentration of oxygen, [O2], and the square of the concentration of nitric oxide, [NO]. e. When you drop a small sphere into a dense fluid such as oil, it eventually acquires a constant velocity, v, that is directly proportional to the square of its radius, r.
Dsafe 5 0.05L Add to your graph from part (a) a plot of Dsafe.. c. Is it safe for a 200-pound person to sit in the middle of a 4s 3 12s fir plank that spans 20 feet? What maximum span is safe for a 200-pound person on a 4s 3 12s plank? 16. (Graphing program recommended.) In Exercise 15 we looked at how much a single load (a person’s weight) could bend a plank downward by being placed at its midpoint. Now we look at what a continuous load, such as a solid row of books spread evenly along a shelf, can do. A long row of paperback fiction weighs about 10 pounds for each foot of the row. Typical hardbound books weigh about 20 lb/ft, and oversize hardbounds such as atlases, encyclopedias, and dictionaries weigh around 36 lb/ft. The following function is used to model the deflection D (in inches) of a 1s 3 12s common pine board spanning a length of L feet carrying a continuous row of books: D 5 s4.87 ? 1024 d ? W ? L4 where W is the weight per foot of the type of books along the shelf. This deflection model is quite good for deflections up to 1 inch; beyond that the fourth power causes the deflection value to increase very rapidly into unrealistic numbers. a. How much deflection does the formula predict for a shelf span of 30 inches with oversize books? Would you recommend a stronger, thicker shelf? b. Plot Dhardbound, Dpaperback, and Doversize on the same graph. Put L on the horizontal axis with values up to 4 feet. c. For each kind of book identify what length, L, will cause a deflection of 0.5 inch in a 1s 3 12s pine shelf. 17. Construct formulas to represent the following relationships. a. The distance, d, traveled by a falling object is directly proportional to the square of the time, t, traveled. b. The energy, E, released is directly proportional to the mass, m, of the object and the speed of light, c, squared.
397
18. a. An insulation blanket for a cylindrical hot water heater is sold in a roll 48 in 3 75 in 3 2 in. Assuming a hot water heater 48 inches high, for what diameter water heater is this insulation blanket made? (Round to the nearest 12 inch.) b. What is the volume of the hot water heater in part (a)? (Note: Ignore the thickness of the heater walls.) c. If 1 gallon 5 231 in3, what is the maximum number of gallons of water that a cylinder the size of this water heater could hold? 19. The volume, V, of a cylinder, with radius r and height h, is given by the formula V 5 pr 2h . Describe what happens to V under the following conditions. a. The radius is doubled; the radius is tripled. b. The height is doubled; the height is tripled. c. The radius r is multiplied by n, where n is a positive integer. d. The height is multiplied by n, where n is a positive integer. 20. If you were designing cylinders, what are two different ways you could quadruple the volume? (See Exercise 19.) 21. Given hsxd 5 0.5x2: a. Calculate h(2) and compare this value with h(6). b. Calculate h(5) and compare this value with h(15). c. What happens to the value of h(x) if x triples in value? d. What happens to the value of h(x) if x is divided by 3? 22. a. Let ƒsxd 5 2x2. Describe the difference between ƒ(3x) and 3ƒ(x). b. Let ƒsxd 5 5x3. Describe the difference between ƒ(4x) and 4ƒ(x). c. In general, given ƒsxd 5 kx p, describe the difference between ƒ(nx) and nƒ(x), where p and n are positive integers.
7.3 Visualizing Positive Integer Powers What do the graphs of power functions look like when the input values are infinitely large or assume negative values? Let’s examine two basic power functions, ƒsxd 5 x 2 and gsxd 5 x 3.
The Graphs of ƒ(x) = x 2 and g(x) = x3 What happens when x ≥ 0? For ƒsxd 5 x 2 and gsxd 5 x 3, as x S 1`, both x 2 and x 3 S 1`. If x 5 0, then x 2 and x 3 are both equal to 0, and if x 5 1, both x2 and x3 equal 1. So ƒ(x) and g(x) intersect at (0, 0) and (1, 1).
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We can see from the graph in Figure 7.10 that when 0 , x , 1, then x 3 , x 2, but when x . 1, then x3 . x2. So the graph of gsxd 5 x 3 eventually dominates the graph of ƒsxd 5 x 2. y 10
g (x) = x 3 5
f (x) = x 2 x
0 0
1
2
Figure 7.10 Comparing the graphs of
ƒsxd 5 x 2 and gsxd 5 x 3 when x $ 0.
What happens when x < 0? When x is negative, the functions ƒsxd 5 x 2 and gsxd 5 x 3 exhibit quite different behaviors (see Table 7.2 and Figure 7.11). When x is negative, x2 is positive but x 3 is negative. As x S 2`, x 2 S 1` but x 3 S 2` . So the domain for both functions is (2`, 1` ), but the range for ƒsxd 5 x 2 is [0, 1 ` ). The range for gsxd 5 x 3 is (2`, 1` ). x
ƒsxd 5 x 2
gsxd 5 x 3
24 23 22 21 0 1 2 3 4
16 9 4 1 0 1 4 9 16
264 227 28 21 0 1 8 27 64
Table 7.2 y
y
25
25
x –5
5
–25
f (x) = x 2
x –5
5
–25
g (x) = x 3
Figure 7.11 The graphs of ƒsxd 5 x 2 and gsxd 5 x 3.
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Exploration 7.2 can help reinforce your understanding of power functions.
399
Odd vs. Even Powers As we saw with the graphs of ƒsxd 5 x 2 and gsxd 5 x 3, the exponent of the power function affects the shape of the graph. The graphs of power functions with positive even powers are <-shaped and the graphs of positive odd powers greater than 1 are -shaped. Why does it matter if the exponent is even or odd? If n is a positive integer and if x is positive, then x n is positive. But if x is negative, we have to consider whether n is even or odd. If n is even, then x n is positive, since (negative number)even power 5 positive number If n is odd, then x n is negative, since (negative number)odd power 5 negative number So whether the exponent of a power function is odd or even will affect the shape of the graph. If we graph the simplest power functions, y 5 x, y 5 x 2, y 5 x 3, y 5 x4, y 5 x 5, y 5 x 6, and so on, we see quickly that the graphs fall into two groups: the odd powers and the even powers (Figure 7.12).
Odd Powers
Even Powers
y
y
10
10
x 2
–2
x 2
–2
y = x2
y=x –10
–10
y
y
10
10
x –2
x –2
2
2
y = x3
y = x4
–10
–10
y
y
10
10
x –2
2
x –2
2
y = x5 –10
y = x6 –10
Figure 7.12 Graphs of odd and even power functions.
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All the power functions graphed in Figure 7.12 have a positive coefficient k (51). So the graphs of the even powers in Figure 7.12 are all concave up. Graphs of the odd powers (except for the power 1) change concavity at x 5 0. When x , 0, the graphs are concave down. When x . 0, the graphs are concave up. Symmetry of the graphs For the odd positive powers, the graphs are rotationally symmetric about the origin; that is, if you hold the graph fixed at the origin and then rotate it 1808, you end up with the same graph. As can be seen in Figure 7.13(a), the net result is the same as a double reflection, first across the y-axis and then across the x-axis. The graphs of even powers are symmetric across the y-axis. You can think of the y-axis as a dividing line: the “left” side of the graph is a mirror image, or reflection, of the “right” side, as shown in Figure 7.13(b).
y
y
x
x
(a)
(b)
Figure 7.13 (a) Power functions with positive odd powers are
symmetric about the origin. (b) Power functions with positive even powers are symmetric across the y-axis.
The Effect of the Coefficient k The program “P1: k & p Sliders” in Power Functions can help you visualize the graphs of y = kx p for different values of k and p.
What is the effect of different values for the coefficient k on graphs of power functions in the form y 5 kx p
where p is a positive integer and k 2 0?
What happens when k > 0? We know from our work with power functions of degree 1 (that is, linear functions of the form y 5 kx) that k affects the steepness of the line. For all power functions, where k . 1, the larger the value for k, the more vertical the graph of y 5 kx p becomes compared with y 5 x p. As k increases, the steepness of the graph increases. We say the graph is stretched vertically. When 0 , k , 1, the graph of y 5 kx p is flatter than the graph of y 5 x p and lies closer to the x-axis. We say the graph is compressed vertically. The graphs in Figure 7.14 illustrate this effect for power functions of degrees 3 and 4.
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10
y y = 10x3
y
y = 5x3
401
y = 10x4
18
y = 5x4
y = x3 y = x4
y = 0.5x3 3
y = 0.25x
–2 2
x
y = 0.5x4
y = 0.25x4 x –2
2
–10
–2
Figure 7.14 When k . 0, the larger the value of k, the more closely the power
function y 5 kx p “hugs” the y-axis for both odd and even powers.
What happens when k < 0? We know from our work with linear functions that the graphs of y 5 kx and y 5 2kx are mirror images across the x-axis. Similarly, the graphs of y 5 kx p and y 5 2kx p are mirror images of each other across the x-axis. For example, y 5 27x 3 is the mirror image of y 5 7x3. Figure 7.15 shows various pairs of power functions of the type y 5 kx p and y 5 2kx p.
10
y
10
y y = 7x 3
y = 3x x –4
x
4 y = –3x
–4
y = –7x 3 –10
–10
10
4
y
10
y y = 0.4x6
y = 0.5x2 x –4
4
y –10
x = –0.5x 2
–4
4
y = –0.4x6 –10
Figure 7.15 In each case, the graphs of y 5 kx p
and y 5 2kx p (shown in blue and black, respectively) are mirror images of each other across the x-axis.
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In sum: the value of k can stretch, compress, or reflect the graph The graph of a power function in the form y 5 kx p is the graph of y 5 x p • vertically stretched by a factor of k, if k . 1 • vertically compressed by a factor of k, if 0 , k , 1 • vertically stretched or compressed by a factor of k k k and reflected across the x-axis, if k , 0
Graphs of Power Functions with Positive Integer Powers For functions of the form y 5 kx p (where p . 0 and k 2 0) The graph goes through the origin. If the power p is even, the graph is y-axis.
< -shaped
and symmetric across the
If the power p is odd, and . 1, the graph is shaped roughly like symmetric about the origin.
and is
The coefficient k compresses or stretches the graph of y 5 x p. The graphs of y 5 kx p and y 5 2kx p are mirror images of each other across the x-axis.
EXAMPLE
1
Match each set of functions with the appropriate graph in Figure 7.16. a. isxd 5 2x 2 jsxd 5 23x 2 ksxd 5 20.5x 2
b. lsxd 5 x 3 msxd 5 3x 3 nsxd 5 0.5x 3
y
y
20
20
D E
F
x –10
10
–20
A
B
x –5
C
5
–20
Figure 7.16 Two families of power functions.
SOLUTION
a. A: jsxd 5 23x 2 B: isxd 5 2x 2 C: ksxd 5 20.5x 2
b. D: msxd 5 3x 3 E: lsxd 5 x 3 F: nsxd 5 0.5x 3
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Algebra Aerobics 7.3 A graphing program is useful for Problems 5, 6, 9, and 10. 1. If ƒsxd 5 4x3, evaluate the following. a. ƒ(2) b. ƒ(22) c. ƒ(s) d. ƒ(3s) 2. If gstd 5 24t3, evaluate the following. a. g(2) b. g(22) c. g A12tB d. g(5t) 2 3. If ƒsxd 5 3x , evaluate: a. ƒ(4)
c. ƒ(s)
e. ƒ(2s)
g. ƒ A2s B
b. ƒ(24)
d. 2ƒ(s)
f. ƒ(3s)
h. ƒ A4s B
7. Without graphing the functions, describe the differences in the graphs of each pair of functions. a. y1 5 0.2x3 and y2 5 0.3x3 b. y1 5 20.15x4 and y2 5 20.2x 4 8. Determine whether p in each power function of the form y 5 ax p in Figures 7.17 and 7.18 is even or odd and whether a . 0 or a , 0. Then find the value of a. a.
4. a. For each of the following functions, generate a small table of values, including positive and negative values for x, and sketch the functions on the same grid: ƒsxd 5 4x 2
b.
y 4
y 4
(1,3)
x –2
and gsxd 5 4x 3
x
2
–2
2 (1,–2)
b. What happens to ƒ(x) and to g(x) as x S 1`? c. What happens to ƒ(x) and to g(x) as x S 2`? d. Specify the domain and range of each function. e. Where do the graphs of these functions intersect? f. For what values of x is g(x) . ƒ(x)? 5. Graph both functions ƒsxd 5 2x 2 and gsxd 5 2x 3 over the interval [22, 2] and then determine values for x where: a. ƒ(x) 5 g(x) b. ƒ(x) . g(x) c. ƒ(x) , g(x) 6. Plot each pair of functions on the same grid. a. y1 5 22x 2 and y2 5 22x 3 b. y1 5 20.1x3 and y2 5 20.1x4
–4
–4
Figure 7.18
Figure 7.17
9. Draw a quick sketch by hand of the power functions. a. y 5 x9 b. y 5 x10 If possible, check your graphs using technology. 10. Draw a rough sketch of: a. y 5 x, y 5 4x, and y 5 24x (all on the same grid) b. y 5 x4, y 5 0.5x4, and y 5 20.5x4 (all on the same grid) If possible, check your graphs using technology.
Exercises for Section 7.3 A graphing program is useful for many of the exercises in this section. 1. Sketch by hand and compare the graphs of the following: y1 5 x2
y2 5 2x2
y4 5 22x
2
y5 5
1 2 2x
y3 5 2x2 y6 5
y 5 x3
y 5 x5
b. y 5 x2 y 5 x4
y 5 x6
c. y 5 x3
y 5 2x3
y 5 22x3
d. y 5 x2 y 5 4x2
y 5 24x2
b. gsxd 5 x 6
If possible, check your results using a graphing program.
d. j(x) 5 22x2
y 5
5
212x2
2. (Graphing program option.) For each part sketch by hand the three graphs on the same grid and clearly label each function. Describe how the three graphs are alike and not alike. a. y 5 x1
3. Match the function with its graph. Explain why you have chosen each graph. a. ƒsxd 5 x 2 c. hsxd 5 2x 2 y
x
x –5
0
5
–5
Graph A
5
Graph C
0 –5
0
y
x
5
5
x
–5 –5 Graph B
0 Graph D
5
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4. Match the function with the appropriate graph. Explain why you have chosen each graph. a. ƒsxd 5 x3 b. gsxd 5 13x3 c. hsxd 5 3x3 y 5
AB C
10. Consider the accompanying graphs of four functions of the form y 5 kx n. Which (if any) individual functions: a. Are symmetric across the x-axis? b. Are symmetric across the y-axis? c. Are symmetric about the origin? d. Have a positive even power for n? e. Have a positive odd power for n? f. Have k , 0?
x –5
5
5
y
y 5
x –5
–5
5. Begin with the function ƒsxd 5 x4. a. Create a new function g(x) by vertically stretching ƒ(x) by a factor of 6. b. Create a new function h(x) by vertically compressing ƒ(x) by a factor of 12 . c. Create a new function j(x) by first vertically stretching ƒ(x) by a factor of 2 and then reflecting it across the x-axis. 6. (Graphing program optional.) Plot the functions ƒsxd 5 x 2, gsxd 5 3x 2, and hsxd 5 14x2 on the same grid. Insert the symbol . or , to make the relation true. a. For x . 0, g(x) ___ ƒ(x) ___ h(x) b. For x , 0, g(x) ___ ƒ(x) ___ h(x) 7. (Graphing program recommended.) Plot the functions ƒsxd 5 x2 and gsxd 5 x5 on the same grid. Insert the symbol ., ,, or 5 to make the relation true. a. For x 5 0, ƒ(x) ___ g(x) b. For 0 , x , 1, ƒ(x) ___ g(x) c. For x 5 1, ƒ(x) ___ g(x) d. For x . 1, ƒ(x) ___ g(x) e. For x , 0, ƒ(x) ___ g(x) 8. Consider the accompanying graph of ƒsxd 5 kx n, where n is a positive integer. y 10 a. Is n even or odd? b. Is k . 0 or is k , 0? c. Does ƒ(22) 5 2ƒ(2)? x –5 5 d. Does ƒ(2x) 5 2ƒ(x)? e. As x S 1`, ƒsxd S __ –10 f. As x S 2`, ƒsxd S __ 9. Consider the accompanying graph of ƒsxd 5 kx n, where n is a positive integer. a. Is n even or odd? b. Is k . 0 or is k , 0? c. Does ƒ(22) 5 ƒ(2)? d. Is ƒ(2x) 5 ƒ(x)? e. As x S 1`, ƒsxd S __ f. As x S 2`, ƒsxd S __
5
x –5
5
–5
–5
Graph A
Graph C
y
y 5
5
x
x –5
5
–5
5
–5
–5 Graph B
Graph D
11. Sketch by hand the graph of each function: ƒsxd 5 x3, gsxd 5 2x3, hsxd 5 12x3, j(x) 5 22x3 a. Identify the k value, the constant of proportionality, for each function. b. Which graph is a reflection of ƒ(x) across the x-axis? c. Which graph is both a stretch and a reflection of ƒ(x) across the x-axis? d. Which graph is a compression of ƒ(x)? 12. a. Complete the partial graph in the accompanying figure in three different ways by: i. Reflecting the graph across the y-axis to create Graph A. ii. Reflecting the graph across the x-axis to create Graph B. iii. Reflecting the graph first across the y-axis and then across the x-axis to create Graph C. y
5
5 x
–5
5 x –5 –10
5
b. Which of the finished graphs is symmetric across the y-axis?
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c. Which of the finished graphs is symmetric across the x-axis? d. Which of the finished graphs is symmetric about the origin? 13. (Graphing program optional.) Given ƒsxd 5 4x2, construct a function that is a reflection of ƒsxd across the horizontal axis. Graph the functions and confirm your answer. 14. (Graphing program optional.) Are ƒsxd 5 x3 and g(x) 5 2x 3 reflections of each other across the vertical axis? Graph the functions and confirm your answer.
7.4
405
15. (Graphing program optional.) Evaluate each of the following functions at 0, 0.5, and 1. Then, on the same grid, graph each over the interval [0, 1]. Compare the graphs. a. y1 5 x
y2 5 x1/2
b. y5 5 x
y6 5 x
2
3
y3 5 x1/3
y4 5 x1/4
y7 5 x
4
16. (Graphing program required). Now graph the functions in Exercise 15 over the interval [1, 4] and compare the graphs.
Comparing Power and Exponential Functions Which Eventually Grows Faster, a Power Function or an Exponential Function? Although power and exponential functions may appear to be similar in construction, in each function type the independent variable or input assumes a very different role. For power functions the input, x, is the base, which is raised to a fixed power. Power functions have the form output 5 k ? sinputd power y 5 kx p If k and p are both positive, the power function describes growth. For exponential functions the input, x, is the exponent applied to a fixed base. Exponential functions have the form output 5 C ? a input y 5 Ca x If C . 0 and a . 1, the exponential function describes growth. Consider the functions y 5 x 3, a power function; y 5 3x, an exponential function; and y 5 3x, a linear function. Table 7.3 compares the role of the independent variable, x, in the three functions.
x
Linear Function y 5 3x (x is multiplied by 3)
0 1 2 3 4 5
3 3 3 3 3 3
? ? ? ? ? ?
0 1 2 3 4 5
5 5 5 5 5 5
0 3 6 9 12 15
Power Function y 5 x3 (x is the base raised to the third power) 0 1 2 3 4 5
? ? ? ? ? ?
0 1 2 3 4 5
? ? ? ? ? ?
0 1 2 3 4 5
5 5 5 5 5 5
0 1 8 27 64 125
Exponential Function y 5 3x (x is the exponent for base 3) 30 5 1 31 5 3 32 5 9 33 5 27 34 5 81 35 5 243
Table 7.3
Visualizing the difference Table 7.3 and Figure 7.19 show that the power function y 5 x3 and the exponential function y 5 3x both grow very quickly relative to the linear function y 5 3x.
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Yet there is a vast difference between the growth of an exponential function and the growth of a power function. In Figure 7.20 we zoom out on the graphs. Notice that now the scale on the x-axis goes from 0 to 10 (rather than just 0 to 4 as in Figure 7.19) and the y-axis extends to 8000, which is still not large enough to show the value of 3x once x is slightly greater than 8. The linear function y 5 3x is not shown, since at this scale it would appear to lie flat on the x-axis.
y
y
80
8000
y = 3x Exponential function
60
y = 3x Exponential function
6000
y = x3 Power function
40
4000
y = 3x
20
y = x3 Power function
2000
Linear function
x 0
1
2
3
x
4
0
2
4
6
8
Figure 7.19 A comparison of
Figure 7.20 “Zooming out” on
y 5 3x, y 5 x3, and y 5 3x.
Figure 7.19.
10
The exponential function y 5 3x clearly dominates the power function y 5 x3. The exponential function continues to grow so rapidly that its graph appears almost vertical relative to the graph of the power function. What if we had picked a larger exponent for the power function? Would the exponential function still overtake the power function? The answer is yes. Let’s compare, for instance, the graphs of y 5 3x and y 5 x10. If we could zoom in on the graph in Figure 7.21, we would see that for a while the graph of y 5 3x lies below the graph of y 5 x10. For example, when x 5 2, then 32 , 210. But eventually 3x . x10. Figure 7.21 shows that somewhere after x 5 30, the values for 3x become substantially larger than the values for x10.
8 • 1017
y y = 3x Exponential function
y = x10 Power function 0
x 0
10
20
30
40
50
Figure 7.21 Graph of y 5 x10
and y 5 3x.
Any exponential growth function will eventually dominate any power function.
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EXAMPLE
1
407
Comparing power and exponential functions a. Construct a table using values of x $ 0 for each of the following functions: y 5 x2
and
y 5 2x
Then plot the functions on the same grid. b. If x . 0, for what value(s) of x is x 2 5 2x? c. Does one function eventually dominate? If so, which one and after what value of x? SOLUTION
a. Table 7.4 and Figure 7.22 compare values for y 5 x 2 and y 5 2 x. y 60
x
y 5 x2
y 5 2x
0 1 2 3 4 5 6
0 1 4 9 16 25 36
1 2 4 8 16 32 64
y = 2x
30
y = x2
0
Table 7.4
3
6
x
Figure 7.22 Comparison of
y 5 x2 and y 5 2x.
b. For positive values of x, if x 5 2 or x 5 4, then x 2 5 2 x. c. In Table 7.4 and Figure 7.22, we can see that if x . 4, then 2 x . x 2, so the function y 5 2 x will dominate the function y 5 x 2 for x . 4.
Algebra Aerobics 7.4 y
1. a. For each of the following functions construct a table using integer values of x between 0 and 5. Then sketch the functions on the same grid. y 5 4x and
y
4
5
y 5 x3 x
b. Does one function eventually dominate? If so, which one and after approximately what value of x? 2. By inspection, determine if each of the graphs in Figure 7.23 is more likely to be the graph of a power function or of an exponential function. 3. Which function eventually dominates? a. y 5 x10 or y 5 2x b. y 5 s1.000 005d x or y 5 x1,000,000
–4
0
x –5
4
0
5
Graph C
Graph A
y
y
4
5
x –4
4
x –5
5
–4
–5
Graph B
Graph D
Figure 7.23 Graphs of power and exponential functions.
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Exercises for Section 7.4 A graphing program is useful for several of the exercises.
e. When x lies in the interval (1, 6), then ƒ(x) _____ g(x). f. When x lies in the interval (7, 1` ), then ƒ(x) _____ g(x).
1. (Graphing program recommended.) Graph y 5 x and y 5 4x on the same grid. a. For positive values of x, where do your graphs intersect? Do they intersect more than once? b. For positive values of x, describe what happens to the right and left of any intersection points. You may need to change the scales on the axes or change the windows on a graphing calculator in order to see what is happening. c. Which eventually dominates, y 5 x4 or y 5 4x ?
5. (Graphing program optional.) Use a table or graphing utility to determine where 2x . x2 and where 3x . x3 for nonnegative values of x.
2. Examine the following two different versions of the graphs of both ƒsxd 5 3x3 and gsxd 5 3x. Notice that each version uses different scales on both axes. Make sure to check each version when answering the following questions.
6. Match the following data tables with the appropriate function. a. ƒsxd 5 2/x b. ƒsxd 5 x2 c. ƒsxd 5 2x
4
y
2000
5
g (x) 1000
f (x)
g (x) x –5
5
–5
f (x)
–5
a. b. c. d. e.
When x , 0, how many times do ƒ(x) and g(x) intersect? When x . 0, how many times do ƒ(x) and g(x) intersect? Estimate the points of intersection. As x S 1` , does ƒsxd 5 3x3 or gsxd 5 3x dominate? Which function, 10 ? ƒsxd or g(x), will dominate as x S 1` ?
3. The following figures show two different versions of the graphs of ƒsxd 5 2x2 and gsxd 5 2x . Notice that each version uses different scales on the axes. Make sure to check each version when answering the following questions. 5
f (x)
120
g (x)
x
y
x
y
x
y
1 2 3 4
2 1 2/3 1/2
1 2 3 4
2 4 8 16
1 2 3 4
1 4 9 16
Table 1
7
–400
4. a. Which eventually dominates, y 5 s1.001d x or y 5 x 1000 ? b. As the independent variable approaches 1` , which function eventually approaches zero faster, an exponential decay function or a power function with negative integer exponent?
Table 3
7. (Graphing program recommended.) If x is positive, for what values of x is 3 ? 2x , 3 ? x2? For what values of x is 3 ? 2x . 3 ? x2? 8. For x . 0, match each function with its graph. a. y 5 x2 c. y 5 4x3 x b. y 5 2 d. y 5 4(3x )
20
y
20
10
y
10
x
g (x)
100
Table 2
0
1
2
3
x 0
4
1
3
4
3
4
Graph C
Graph A
80
2
60 40
f (x) 20
y
20
y
20
x –5
5
–5
7 10
Supply the appropriate inequality symbol. a. As x S 2` , ƒ(x) ___ g(x). b. As x S 1` , ƒ(x) ___ g(x). c. When x lies in the interval (2`, 21), then ƒ(x) _____ g(x). d. When x lies in the interval (20.5, 1) then ƒ(x) _____ g(x).
10
x 0
1
2 Graph B
3
4
x 0
1
2 Graph D
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9. Match the function with its graph. a. hsxd 5 2x2 b. isxd 5 2s3d x
c. jsxd 5 2x3
409
b. Find an exponential function of the form y 5 ab x that goes through the two points. c. Find a power function of the form y 5 ax p that goes through the two points.
y 20
13. Find power, exponential, and linear equations that go through the two points (1, 0.5) and (4, 32). A
B
14. Think of the graphs of ƒsxd 5 ax3 and gsxd 5 as3d x, where a . 0, and then decide whether each of the following statements is true or false. a. ƒ(x) and g(x) have the same vertical intercept. b. ƒ(x) intersects g(x) only once. c. As x S 1`, gsxd . ƒsxd. d. As x S 2`, both ƒ(x) and g(x) approach 0.
C
10
x 0
1
2
3
10. Given the functions ƒsxd 5 x4 and gsxd 5 s4d x: a. Find ƒ(3x) and 3ƒ(x). Summarize the difference between these functions and ƒ(x). b. Find g(3x) and 3g(x). Summarize the difference between these functions and g(x).
15. If you know that one of the graphs in the accompanying figure is a power function of the form y 5 ax p and the other is an exponential function of the form y 5 Cb x, then determine if each of the following is true or false. a. C $ 1 b. p is an even integer. c. 0 , b , 1 d. The graphs of the functions will only intersect twice. (Hint: Think of the long-term behavior of these functions.) y
11. Consider a power function ƒsxd 5 x n where n is a positive integer, and an exponential function gsxd 5 b x where b . 1. Describe how ƒ(ax) and g(ax) differ from ƒ(x) and g(x) (see Exercise 10).
5
12. Given the points (1, 2) and (6, 72): a. Find a linear function of the form y 5 ax 1 b that goes through the two points.
x –5
5
7.5 Inverse Proportionality: Power Functions with Negative Integer Powers Recall that the general form of an equation for a power function is output 5 constant ? sinputd power In Sections 7.1 through 7.4 we focused on functions where the power was a positive integer. We now consider power functions where the power is a negative integer. Using the rules for negative exponents, we can rewrite power functions in the form y 5 kx negative power where k is a constant, as y5
k x
positive power
For example, y 5 3x22 can be rewritten y 5 x32 . In this form (with a positive power) it is easier to make calculations and to see what happens to y as x increases or decreases in value. In Section 7.1 we constructed a function Rsxd 5 6x 5 6x21 from the ratio of surface area to volume of a cube with edge length x.We described Rsxd 5 6x21 as a power function of degree 21. Repeating the table and graph from Section 7.1 here as Table 7.5 and Figure 7.24 reminds us that as x increases, R(x) decreases. So the shape
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of this power function with a negative exponent is quite different from the shape of those with a positive exponent. 6 5
Edge Length x
Surface Area Volume Rsxd 5 6x
1 2 3 4 6 8 10
6.00 3.00 2.00 1.50 1.00 0.75 0.60
(Surface area)/volume
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4 3 2 1
0
2
4 6 Edge length
8
10
Figure 7.24 The graph of
Rsxd 5 6x shows that as the edge length, x, increases, the ratio of (surface area)/volume decreases.
Table 7.5
Inverse Proportionality The headline for a March 8, 2004, article in The Economic Times read “Grace is inversely proportional to the crisis an individual faces.” The author was arguing that the greater the crisis, the less gracious an individual is likely to be. Mathematics has a formal definition for the same concept. For power functions in the form y5
constant xp
where p is positive, we say that y is inversely proportional to x p. For example, if y 5
8 x3
we say that y is inversely proportional to x3. If y is inversely proportional to x p, then as x increases, y decreases.
Direct and Inverse Proportionality Let p be a positive number and k 2 0. If y 5 kx p, then y is directly proportional to x p. k 5 kx2p, then y is inversely proportional to x p. xp In both cases k is called the constant of proportionality. If y 5
EXAMPLE
1
Examples of direct and inverse proportionality Write formulas to represent the following relationships. a. Boyle’s Law says that the volume, V, of a fixed quantity of gas is inversely proportional to the pressure, P, applied to it. b. The force, F, keeping an electron in orbit is inversely proportional to the square of the distance, d, between the electron and the nucleus. c. The acceleration, a, of an object is directly proportional to the force, F, applied upon the object and inversely proportional to the object’s mass, m.
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SOLUTION
a. V 5 b. F 5
EXAMPLE
2
k P for some constant k k for some constant k d2
c. a 5
kF m
411
for some constant k
In each formula, identify which variables are directly or inversely proportional to each other, and specify the constant of proportionality. a. a 5 vt, where a 5 acceleration, v 5 velocity, and t 5 time. b. F 5
GM1M2 , d2
Newton’s Law of Universal Gravitation, where F is the force of gravity,
G is a gravitational constant, M1 and M2 are the masses of two bodies, and d is the distance between the two bodies.
EXAMPLE
3
SOLUTION
a. Acceleration, a, is directly proportional to velocity, v, and inversely proportional to time, t. The constant of proportionality is 1. b. The force of gravity, F, is directly proportional to the masses of the two bodies, M1 and M2, and inversely proportional to d 2, the square of the distance between them. The constant of proportionality is G, a gravitational constant.
You’ve just found a great house to rent for $2,400 a month. You would need to share the rent with several friends. a. Construct an equation for the function R(n) that shows the rental cost per person for n people. What kind of function is this? b. Graph the function for a reasonable domain. c. Evaluate R(4) and R(6), and describe what they mean in this context. a. R(n) 5 2400/n is a power function where R(n) is inversely proportional to n. b. $3000
Cost/person
SOLUTION
$2000
$1000
$0 0
2
4
6
8
10
Number of people
c. R(4) 5 600, which means that the cost per person would be $600 if there are four renters. R(6) 5 400, which means that the cost per person would be $400 if there are six renters.
Properties of Inverse Proportionality If y is inversely proportional to x p, then y 5
k x p (where p . 0) for some nonzero constant k.
For an inversely proportional relationship, when we multiply the input by m, the output is multiplied by m1 p (where p . 0). Similar to our argument for direct proportionality, if ƒsxd 5
k x p,
then
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evaluate ƒ at mx
k smxd p k 5 sm px pd
ƒsmxd 5
use rules of exponents 1 mp
5 a
1 k b ? a pb mp x
k substitute ƒ(x) for a pb x
5 a
1 b ? ƒsxd mp
factor out
Note that k, the constant of proportionality, does not play a role in these calculations. For example, if ƒsxd 5 x32 , then doubling the input would multiply the output by 212 or 14 . So if we double the input, the output would be reduced to one-fourth or 25% of the original amount. These doubling calculations do not depend on 3, the constant of proportionality.
If y is inversely proportional to x p, then y 5
k x p,
where p . 0.
Multiplying the input by m multiplies the output by m1p. For example, doubling the input multiplies the output by 21p.
EXAMPLE
4
SOLUTION
For each of the following functions, what happens if the input is doubled? Multiplied by 10? Cut in half? a. y 5 x14 b. y 5 2x23 a. If the input is doubled, the output is multiplied by 214 5
1 16
or 0.0625. If the input is
1 multiplied by 10, the output is multiplied by 101 4 5 10,000 5 1 1 in half, the output is multiplied by s0.5d 4 5 0.0625 5 16.
0.0001. If the input is cut
b. The function y 5 2x23 can be rewritten as y 5
2 x 3 . In this form it’s easier to see that if the input is doubled, the output is multiplied by 213 5 18 5 0.125. If the input is 1 multiplied by 10, the output is multiplied by 101 3 5 1000 5 0.001. If the input is cut 1 1 in half, the output is multiplied by s0.5d 3 5 0.125 5 8.
EXAMPLE
5
SOLUTION
The cardinal rule of scuba diving The most important rule in scuba diving is “Never, ever, hold your breath.” Why? Think of your lungs as balloons filled with air. As a balloon descends underwater, the surrounding water applies pressure, compressing the balloon. As the balloon ascends, the pressure is lessened and the balloon expands until it attains its original size when it reaches the surface. The volume V of the balloon is inversely proportional to the
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413
pressure P; that is, as the pressure increases, the volume decreases. The relationship is given by a special case of Boyle’s Law for Gases, V5
1 P
V 5 P21
or equivalently
where V is the volume in cubic feet and P the pressure measured in atmospheres (atm). (One atm is 15 lb/in2, the atmospheric pressure at Earth’s surface.) Each additional 33 feet of water depth increases the pressure by 1 atm. Table 7.6 and Figure 7.25 describe the relationship between pressure and volume.
Pressure vs. Volume Depth (ft)
Pressure (atm)
Volume (ft3)
0 33 66 99
1 2 3 4
1 1/2 1/3 1/4
Table 7.6
Volume (cubic feet)
1.00
0.50
0
1
2 3 Pressure (atm)
4
Figure 7.25 Graph of volume vs. pressure for a balloon descending underwater.
We can see in Table 7.6 that if we start with 1 cubic foot of air in the balloon at the surface and descend 33 feet, the surrounding pressure has doubled, from 1 to 2 atm, and the volume of air is cut in half. If the balloon descends to 99 feet, doubling the pressure again from 2 to 4 atm, the volume of air is cut in half again—leaving only one-fourth of the original volume. Why does this matter to divers? Suppose you are swimming in a pool, take a lung full of air at the surface, and then dive down to the bottom. As you descend, the build-up of pressure will decrease the volume of air in your lungs. When you ascend back to the surface, the volume of air in your lungs will expand back to its original size, and everything is fine. But when you are scuba diving, you are constantly breathing air that has been pressurized at the surrounding water pressure. If you are scuba diving 33 feet below the surface of the water, the surrounding water pressure is at 2 atm, twice that at the surface. What will happen then if you fill your lungs from your tank, hold your breath, and ascend to the surface? When you reach the surface, the pressure will drop in half, from 2 atm down to 1 atm, so the volume of air in your lungs will double, rupturing your lungs! Hence the first rule of scuba diving is “Never, ever, hold your breath.” EXAMPLE
6
SOLUTION
Designing a soda can A designer is asked to redesign a 12-ounce soda can. The volume must remain constant at 22 cubic inches (just enough to hold the 12 ounces and a little air) and the shape must remain cylindrical. What are her options? If r 5 radius of the can (in inches) and h 5 height (in inches), then volume of can 5 area of base ? height of can 5 pr 2h
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Since the volume of the can must be 22 cubic inches, 22 5 pr 2h Solving for h gives h5
22 pr 2
So h is inversely proportional to r 2. If we substitute an approximation of 3.14 for p, we get h<
22 7 < 3.14r 2 r 2
Figure 7.26 shows a graph of the relationship between the height of the can and its radius.
h 10
Height (inches)
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10/11/07
(1, 7)
(1.5, 3.1)
0
3
r
Radius (inches)
Figure 7.26 The relationship between height, h, and radius, r, of a can holding 22 cubic inches.
The designer can pick any point on the curve to determine the potential dimensions of the can. For example, if r 5 1, then h 5 7. So the point (1, 7) on the curve represents a radius of 1s (hence a diameter of 2s ) and a height of 7s . If r 5 1.5, then 7 7 h 5 s1.5d 2 5 2.25 < 3.1. The point (1.5, 3.1) on the curve represents a radius of 1.5s (diameter of 3s ) and a height of 3.1s . The two points are labeled on the graph (Figure 7.26), and the corresponding can sizes are drawn in Figure 7.27.
7
3.1
1
1.5
Figure 7.27 Two possible sizes
for the soda can.
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Inverse Square Laws When the output is inversely proportional to the square of the input, the functional relationship is called an inverse square law. In the preceding example, where we held the volume of the soda can fixed at 22, the resulting function h 5 7/r 2 is an inverse square law. Inverse square laws are quite common in the sciences.
EXAMPLE
7
Seeing the light The intensity of light is inversely proportional to the square of the distance between the light source and the viewer. a. Describe light intensity as a function of the distance from the light source. b. What happens to the intensity of the light if the distance doubles? Triples?
SOLUTION
a. If d is the distance from the light source, then I(d ), the intensity of the light, is inversely proportional to d, so we can write Isdd 5 dk2 for some constant k. b. If the distance doubles, then the light intensity output is multiplied by 212 5 14 . So the intensity drops to one-fourth of the original intensity. If the distance triples, then the light intensity is multiplied by 312 5 19 . So the intensity drops to one-ninth of the original intensity. For example, if you are reading a book that is 3 feet away from a lamp, and you move the book to 6 feet away (doubling the distance between the book and the light), the light will be one-fourth as intense. If you move the book from 3 to 9 feet away (tripling the distance), the light will be only one-ninth as intense. The reverse is also true; for example, if the book is 6 feet away and the light seems too dim for reading, by cutting the distance in half (to 3 feet), the illumination will be four times as intense.
EXAMPLE
8
SOLUTION
Gravitational force between objects The gravitational force between you and Earth is inversely proportional to the square of the distance between you and the center of Earth. a. Express this relationship as a power function. b. What happens to the gravitational force as the distance between you and the center of Earth increases by a factor of 10? c. Why do astronauts appear to be weightless in space? a. The power function k Fsdd 5 d 2 5 kd 22 describes the gravitational force F(d) between you and Earth in terms of a constant k times d 22, where d is the distance between you and the center of Earth. b. If the distance between you and Earth’s center increases by a factor of 10, then the 1 gravitational force is multiplied by 101 2 5 100 ; that is, multiplying the distance by 10 decreases the gravitational force to one-hundredth of its original size. c. Suppose an astronaut starts at the surface of Earth (roughly 4000 miles from Earth’s center) and travels to 40,000 miles above Earth’s center. She will have increased her distance from Earth’s center by a factor or 10. The pull of Earth’s gravity there would be one-hundredth that on Earth’s surface. So she would appear to be weightless.
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Why many inverse square laws work Inverse square laws in physics often depend on a power source and simple geometry. Imagine a single point as a source of power, emitting perhaps heat, sound, or light. We can think of the power radiating out from the point as passing through an infinite number of concentric spheres. The farther away you are from the point source, the lower the intensity of the power, since it is spread out over the surface area of a sphere that increases in size as you move away from the point source. Therefore, the intensity, I, of the power you receive at any point on the sphere will be a function of the distance you are from the point source. The distance can be thought of as the radius, r, of a sphere with the point source at its center: intensity 5 I5
power from a source surface area of sphere power 4pr2 a
1 factor out 2 r
5
power b 4p r2
If the power from the source is constant, we can simplify the expression by power substituting a constant k 5 4p and rewrite our equation as I5
k r2
The intensity you receive, I, is inversely proportional to the square of your distance from the source if the power from the point source is constant. If you double the distance, the intensity you receive is one-fourth of the original intensity. If you triple the distance, the intensity you receive is one-ninth of the original intensity. In particular, if r 5 1, then I 5 k; if r doubles to 2, then I 5 4k 5 0.25k. The graph of I 5 rk2 is sketched in Figure 7.28. I
Intensity
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k 0.25k 0
1 2 Distance from point source
r
Figure 7.28 Graph of I 5 rk2 , the
relationship between intensity and distance from a point source.
EXAMPLE
9
Direct and inverse proportionality in the same equation: The relative effects of the moon and the sun on tides The force that creates ocean tides on the surface of Earth varies inversely with the cube of the distance from Earth to any other large body in space and varies directly with the mass of the other body. a. Construct an equation that describes this relationship.
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417
b. The sun has a mass 2.7 ? 107 times larger than the moon’s, but the sun is about 390 times farther away from Earth than the moon. Would you expect the sun or the moon to have a greater effect on Earth’s tides? SOLUTION
a. Using T for the tide-generating force, d for the distance between Earth and another body, and m for the mass of the other body, we have km or T 5 kmd 23 d3 for some constant of proportionality k. b. If Mmoon is the mass of the moon and Msun is the mass of the sun, then T5
Msun 5 2.7 ? 107 ? Mmoon If dmoon is the distance from the moon to Earth and dsun is the distance from the sun to Earth, then dsun 5 390dmoon If Tsun is the tidal force of the sun, then using the formula from part (a), we have Tsun 5
k ? M sun sdsun d 3
substitute for Msun and dsun
5
k ? s2.7 ? 107 ? M moon d s390dmoon d 3
regroup terms
5
2.7 ? 107 k ? Mmoon ? 3903 sdmoon d 3
Tsun <
substitute in Tmoon and simplify
1 ? Tmoon 2
Thus, despite the fact that the sun is more massive than the moon, because it is much farther away from Earth than the moon, its effect on Earth’s tides is about one-half that of the moon.
Algebra Aerobics 7.5 1. Describe each proportionality with a sentence. (Assume k is a constant.) k k "y a. x 5 kyz b. y 5 x 2 c. D 5 z 3 2. Which of the following represent quantities that vary inversely? (Assume k is a constant.) a. y 5 "x c. y 5 kx23 e. y 5 8 ? 22x d. V 5 13pr2h
b. xy 5 k
3. You inflate a balloon with 1 ft3 of compressed air from your scuba tank while diving 99 feet underwater. The pressure at 99 ft 3 is equal to 4 atm. When you have ascended to the surface, by how much will the volume have increased (assuming the balloon doesn’t burst)? 4. a. Rewrite each of the following expressions using positive exponents. i. 15x23 iii. 3.6x21 24
ii. 210x
22 23
iv. 2x y
b. Rewrite each of the following expressions using negative exponents. 1.5 2 i. 2 iii. 2 2 3x x 6 6 ii. 2 3 iv. 3 4 x yz x 5. The time in seconds, t, needed to fill a tank with water is inversely proportional to the square of the diameter, d, of the pipe delivering the water. Write an equation describing this relationship. 6. A light is 4 feet above the book you are reading. The light seems too dim, so you move the light 2 feet closer to the book. What is the change in light intensity? 7. If gsxd 5 x34, what happens to g(x) when: a. x doubles? 2x23,
8. If hsxd 5 a. Triple x?
b. x is divided by 2? what happens to h(x) if you: b. Divide x by 3?
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9. Write an equation of variation where: a. x is directly proportional to both y and the square of z. b. a is directly proportional to both b and the cube of c and inversely proportional to d. c. a is directly proportional to both the square of b and the square root of c and inversely proportional to both d and e. 10. Construct an equation of variation and find the constant of proportionality for each of the following situations. (Hint: Substitute values for x, y, and z to
find the constant of proportionality, k. Then solve for the indicated value.) a. x is inversely proportional to the square root of y. When x 5 3, y 5 4. Find x when y 5 400. b. x is directly proportional to both y and z. When x 5 3 and y 5 4, z 5 0.5. Find y when x 5 6 and z 5 2. c. x is directly proportional to the square of y and inversely proportional to the cube root of z. When x 5 6, y 5 6 and z 5 8. Find x when y 5 9 and z 5 0.027.
Exercises for Section 7.5 Exercises 21 and 22 require technology that can generate a bestfit power function. 4 answer the following: x3 As x changes from 1 to 4, y changes from —— to ——. When x . 0, as x increases, y ——. As x changes from 2 to 12, y changes from —— to ——. When x . 0, as x decreases, y —— . We say that y is ———— proportional to ——, and —— is the constant of proportionality.
1. For y 5 a. b. c. d. e.
2. For ƒsxd 5 kx21 and gsxd 5 kx22, find: a. ƒ(2x), ƒ(3x), and ƒ A2x B b. g(2x), g(3x), and g A2x B 3. For h(x) 5 kx23 and jsxd 5 2kx 24, find: a. h(2x), hs23xd , and h A13xB b. j(2x), js23xd , and j A13xB 4. Find the value for k, the constant of proportionality, if: a. y 5 xk and y 5 3 when x 5 2. 1 b. y 5 xk2 and y 5 4 when x 5 8. 1 c. y 5 xk2 and y 5 16 when x 5 2. k d. y 5 and y 5 1 when x 5 9. "x
5. Create an equation that meets the given specifications and then solve for the indicated variable. a. If P is inversely proportional to f, and P 5 0.16 when f 5 0.1, then what is the value for P when f 5 10? b. If Q is inversely proportional to the square of r, and Q 5 6 when r 5 3, then what is the value for Q when r 5 9? c. If S is inversely proportional to w and p, and S 5 8 when w 5 4 and p 5 12, what is the value for S when w 5 8 and p 5 1? d. W is inversely proportional to the square root of u and W 5 13 when u 5 4. Find W when u 5 16. 6. Assume y is inversely proportional to the cube of x. a. If x doubles, what happens to y? b. If x triples, what happens to y?
c. If x is halved, what happens to y? d. If x is reduced to one-third of its value, what happens to y? 7. a. B is inversely proportional to x4. What is the effect on B of doubling x? b. Z is inversely proportional to x p, where p is a positive integer. What is the effect on Z of doubling x? 8. The time, t, required to empty a tank is inversely proportional to r, the rate of pumping. If a pump can empty the tank in 30 minutes at a pumping rate of 50 gallons per minute, how long will it take to empty the tank if the pumping rate is doubled? 9. The intensity of light from a point source is inversely proportional to the square of the distance from the light source. If the intensity is 4 watts per square meter at a distance of 6 m from the source, find the intensity at a distance of 8 m from the source. Find the intensity at a distance of 100 m from the source. 10. In Exercise 9, it you wanted to quadruple the intensity, what would need to be done to the distance? 11. A light fixture is mounted flush on a 10-foot-high ceiling over a 3-foot-high counter. How much will the illumination (the light intensity) increase if the light fixture is lowered to 4 feet above the counter? 12. The frequency, F (the number of oscillations per unit of time), of an object of mass m attached to a spring is inversely proportional to the square root of m. a. Write an equation describing the relationship. b. If a mass of 0.25 kg attached to a spring makes three oscillations per second, find the constant of proportionality. c. Find the number of oscillations per second made by a mass of 0.01 kg that is attached to the spring discussed in part (b). 13. Boyle’s Law says that if the temperature is held constant, then the volume, V, of a fixed quantity of gas is inversely proportional to the pressure, P. That is, V 5 Pk for some constant k. What happens to the volume if: a. The pressure triples? b. The pressure is multiplied by n?
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the distance from one wave crest to the next. [See D. W. Thompson, On Growth and Form (New York: Dover Publications, 1992).]
c. The pressure is halved? d. The pressure is divided by n? 14. The pressure of the atmosphere around us is relatively constant at 15 lb/in2 at sea level, or 1 atmosphere of pressure (1 atm). In other words, the column of air above 1 square inch of Earth’s surface is exerting 15 pounds of force on that square inch of Earth. Water is considerably more dense. As we saw in Section 7.5, pressure increases at a rate of 1 atm for each additional 33 feet of water. The accompanying table shows a few corresponding values for water depth and pressure. Water Depth (ft)
Pressure (atm)
0 33 66 99
1 2 3 4
a. What type of relationship does the table describe? b. Construct an equation that describes pressure, P, as a function of depth, D. c. In Section 7.5 we looked at a special case of Boyle’s Law for the behavior of gases, P 5 V1 (where V is in cubic feet, P is in atms). i. Use Boyle’s Law and the equation you found in part (b) to construct an equation for volume, V, as a function of depth, D. ii. When D 5 0 feet, what is V? iii. When D 5 66 feet, what is V? iv. If a snorkeler takes a lung full of air at the surface, dives down to 10 feet, and returns to the surface, describe what happens to the volume of air in her lungs. v. A large, flexible balloon is filled with a cubic foot of compressed air from a scuba tank at 132 feet below water level, sealed tight, and allowed to ascend to the surface. Use your equation to predict the change in the volume of its air. 15. a. Construct an equation to represent a relationship where x is directly proportional to y and inversely proportional to z. b. Assume that x 5 4 when y 5 16 and z 5 32. Find k, the constant of proportionality. c. Using your equation from part (b), find x when y 5 25 and z 5 5. 16. a. Construct an equation to represent a relationship where w is directly proportional to both y and z and inversely proportional to the square of x. b. Assume that w 5 10 when y 5 12, z 5 15, and x 5 6. Find k, the constant of proportionality. c. Using your equation from part (b), find x when w 5 2, y 5 5, and z 5 6. 17. Waves on the open ocean travel with a velocity that is directly proportional to the square root of their wavelength,
Wavelength
a. If you are in a fixed spot in the ocean, the time interval wave length between successive waves equals wave velocity . Show that this time interval is directly proportional to the square root of the wavelength. b. On one day waves crash on the beach every 3 seconds. On the next day, the waves crash every 6 seconds. On the open ocean, how much farther apart do you expect the wave crests to be on the second day than on the first? 18. When installing Christmas lights on the outside of your house, you read the warning “Do not string more than four sets of lights together.” This is because the electrical resistance, R, of wire varies directly with the length of the wire, l, and inversely with the square of the diameter of the wire, d. a. Construct an equation for electrical wire resistance. b. If you double the wire diameter, what happens to the resistance? c. If you increase the length by 25% (say, going from four to five strings of lights), what happens to the resistance? 19. The rate of vibration of a string under constant tension is inversely proportional to the length of the string. a. Write an equation for the vibration rate of a string, v, as a function of its length, l. b. If a 48-inch string vibrates 256 times per second, then how long is a string that vibrates 512 times per second? c. In general, it can be said that if the length of the string increases, the vibration rate will _______. d. If you want the vibration rate of a string to increase, then you must _________ the length of the string. e. Playing a stringed instrument, such as a guitar, dulcimer, banjo, or fiddle, requires placing your finger on a fret, effectively shortening the string. Doubling the vibration produces a note pitched one octave higher, and halving the vibration produces a note pitched one octave lower. If the number of vibrations decreased from 440 to 220 vibrations per second, what happened to the length of the string to cause the change in vibration? 20. The weight of a body is inversely proportional to the square of the distance from the body to the center of Earth. Assuming an Earth radius of 4000 miles, a man who weighs 200 pounds on Earth’s surface would weigh how much 20 miles above Earth? 21. (Requires technology to generate a best-fit power function.) Oil is forced into a closed tube containing air. The height, H, of the air column is inversely proportional to the pressure, P. a. Construct a general equation to describe the relationship between H and P.
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b. Using technology, construct a best-fit power function for the accompanying data collected on height and pressure. Height, H (in) 13.3 10.7 9.2 7.1 6.3 5.1 4.1 3.5 3.0
Pressure, P (lb/in2)
proportional to the pressure, P, exerted on the gas; that is, V 5 Pk for some constant k. Use technology to determine a best-fit function model for the accompanying measurements collected on the volume of air as the pressure on it was increased.
6.7 8.1 10.2 12.9 14.4 17.6 21.4 25.1 29.4
22. (Requires technology to generate a best-fit power function.) Recall that Boyle’s Law states that for a fixed mass of gas at a constant temperature, the volume, V, of the gas is inversely
Pressure, P (atm)
Volume of Air, V (cm3)
1.0098 1.1610 1.3776 1.6350 1.9660 2.3828 2.9834 3.9396 5.0428 6.2687
20 18 16 14 12 10 8 6 4 2
7.6 Visualizing Negative Integer Power Functions The Graphs of f(x) = x21 and g(x) = x22 Let’s take a close look at two power functions with negative exponents, ƒsxd 5 x21 and gsxd 5 x22. Two important questions are “What happens to these functions when x is close to 0?” and “What happens as x approaches 1` or 2` ?” What happens when x = 0? If we rewrite ƒsxd 5 x21 as ƒsxd 5 1x and gsxd 5 x22 as gsxd 5 12 , it is clear that x both functions are undefined when x 5 0. So neither domain includes 0. The graph of each function is split into two pieces: where x . 0 and x , 0. What happens when x S + ` ? Table 7.7 and Figure 7.29 show that when x is positive and increasing, both ƒ(x) and g(x) are positive and decreasing. As x S 1` , both 1x and x12 grow smaller and smaller, approaching, but never reaching, zero. This means that the range does not include 0. Graphically, as x gets larger and larger, both curves get closer and closer to the x-axis but never touch it. So both graphs are asymptotic to the x-axis. y 10
Values for ƒ(x) and g(x) when x > 0 x 0 1/100 1/4 1/3 1/2 1 2 3 4 100 Table 7.7
ƒsxd 5 1x
gsxd 5 x12
Undefined 100 4 3 2 1 1/2 1/3 1/4 1/100
Undefined 10,000 16 9 4 1 1/4 1/9 1/16 1/10,000
g(x ) = 1/x 2
f (x) = 1/x
(1, 1)
x 0
4
Figure 7.29 Graphs of ƒsxd 5 1x
and gsxd 5
1 x2
when x . 0.
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What happens when x approaches 0? We can also see in Table 7.7 and Figure 7.29 that when x is positive and grows smaller and smaller, approaching zero, the values for both ƒsxd 5 1x and gsxd 5 x12 get larger and larger. Let’s examine some of the underlying calculations in Table 7.7 for fractional values of x between 0 and 1. ƒ A14B 5
1 1/4
514
1 4
51?
4 1
54
1 ƒ A100 B5
1 1/100
514
1 100
51?
100 1
g A14B 5
1 2 s1/4d
514
1 16
51?
16 1
1 g A100 B5
1 2 s1/100d
514
1 10,000
5 100
5 16 10,000 1
51?
5 10,000
If x is positive and approaches zero, then ƒsxd 5 1x and gsxd 5 x12 approach positive infinity. Graphically, as x gets closer and closer to 0, both curves get closer and closer to the y-axis but never touch it. So both graphs are asymptotic to the y-axis. What happens when x < 0? When x is negative, the graphs of ƒ(x) and g(x) have very different shapes. If x , 0, then 1/x is negative but 1/x 2 is positive (see Table 7.8 and Figure 7.30). As x S 0 through negative values, 1/x S 2` and 1/x 2 S 1` . So both ƒ(x) and g(x) are again asymptotic to the y-axis, but in different directions. Values for ƒ(x) and g(x) when x < 0 x
ƒsxd 5 1x
gsxd 5 x12
2100 24 23 22 21 21/2 21/3 21/4 21/100
21/100 21/4 21/3 21/2 21 22 23 24 2100
1/10,000 1/16 1/9 1/4 1 4 9 16 10,000
Table 7.8
5
y
5
y
x
x –5
5
–5
–5 Graph of f (x) = 1/x
Figure 7.30 The graphs of ƒsxd 5 1x and gsxd 5 x12 .
5
–5 Graph of g (x) = 1/x2
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As x S 2` , both 1/x and 1/x 2 S 0. So both functions are again asymptotic to the x-axis, but on different sides of the x-axis. The graphs of ƒ(x) and g(x), for both positive and negative values of x, are in Figure 7.30. Each graph consists of two pieces, split by the y-axis. When x . 0, the graphs are somewhat similar, both positive and concave up. However, when x , 0, the graphs lie on opposite sides of the x-axis. One graph is negative and concave down, the other positive and concave up. The graphs of ƒ(x) and g(x) are asymptotic to both the x- and y-axes.
Odd vs. Even Powers The graphs of power functions with negative powers are all similar in that they are composed of two non-intersecting curves that are each asymptotic to both the x- and y-axes. However, like the positive power functions, the graphs of negative power functions fall into two categories, even and odd powers, that are distinctly different as shown in Figure 7.31.
Odd Powers
Even Powers
y
y
10
10
x –2
2
x –2
2
y = 1/x 2
y = 1/x
–10
–10
y
y
10
10
x –2
2
x –2
2
y = 1/x3
y = 1/x4
–10
–10
y
y
10
10
x –2
2
x –2
2
y = 1/x5
–10
y = 1/x6 –10
Figure 7.31 Graphs of power functions with even
and odd powers.
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All of the graphs in Figure 7.31 are split into two pieces and have a positive coefficient k (5 1). So both pieces of the even-power graphs are concave up. However, the two pieces of the odd-power graphs differ in concavity. Here when x , 0, the graphs are concave down. When x . 0, the graphs are concave up. Asymptotes Figure 7.31 confirms that all odd and even negative power functions are asymptotic to both axes. (Note that positive power functions—odd or even—have no asymptotes.) All negative power functions “blow up” at the point 0; that is, the function is not defined at 0, but as x gets closer and closer to 0, the values of y 5 k/x p (where p . 0) approach either 1∞ or 2∞. We call 0 a singularity. This singularity forces the negative power functions to “explode” at x 5 0, causing them to be asymptotic to the y-axis. Similarly, as x approaches either 1∞ or 2∞ (at either end of the x-axis), the values of y approach, but never reach, 0. This means that the functions are asymptotic to the x-axis. Symmetry of the graphs All of the odd negative powers exhibit rotational symmetry about the origin. If you hold the graph fixed at the origin and rotate it 180º, you end up with the same image. The net result is equivalent to a double reflection, first about the y-axis and then about the x-axis. All of the even negative power functions are symmetric about the y-axis. If you think of the y-axis as the dividing line, the “left” side of the graph is a mirror image, or reflection, of the “right” side.
The Effect of the Coefficient k In Exploration 7.3 you can study more about the effect of k on the graphs of negative power functions.
We saw in Section 7.3 that the value of the coefficient k compresses or stretches the graphs of power functions with positive integer powers. We also saw that y 5 kx p and y 5 2kx p are mirror images of each other across the x-axis. These properties hold true for all power functions. The graph of power functions in the form y 5 xkp (where p . 0 and k 2 0) is the graph of y 5 x1p • vertically stretched by a factor of k, if k . 1 • vertically compressed by a factor of k, if 0 , k , 1 • vertically stretched or compressed by a factor k k k, and reflected across the x-axis, if k , 0
EXAMPLE
1
a. Describe how the graph of gsxd 5 x32 is related to the graph of ƒsxd 5 x12 . Support your answer by graphing the functions on the same grid using technology. b. Describe how the graph of gsxd 5 24x1 3 is related to the graph of ƒsxd 5 x13 . Support your answer by graphing the functions on the same grid using technology.
SOLUTION
a. g(x) 5
3 x2
5 3Qx12R. Since the coefficient 3 is greater than 1, the graph of g(x) is the
graph of ƒsxd 5
1 x2
vertically stretched by a factor of 3. See Graph A in Figure 7.32.
b. g(x) 5 24x1 3 5 214 Qx13R. Since the coefficient 214 is negative and 0 , k 214 k , 1, gsxd is the graph of ƒsxd 5
1 x 3,
compressed by a factor of
across the x-axis. See Graph B in Figure 7.32.
1 4
and then reflected
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CHAPTER 7 POWER FUNCTIONS
y
y
5
5
g(x) = – 13 4x
3
g(x) = x2
1
f (x) = x3
x 1
–5
x
5
f (x) = x2
–5
5
–5
–5
Graph A
Graph B
Figure 7.32 Graph A: The black graph is ƒsxd 5 x12 and the blue graph
is gsxd 5 x32 . Graph B: The black graph is ƒsxd 5
1 x3
and the blue graph is
gsxd 5 2 41x3 .
Graphs of Power Functions with Negative Integer Powers The graph of y 5 xkp (where p . 0 and k 2 0): never goes through the origin, since the function is not defined when x 5 0, is composed of two non-intersecting curves that are each asymptotic to both the x- and y-axes. If the power p is even, the graph is symmetric across the y-axis. If the power p is odd, the graph is symmetric about the origin. The value of the coefficient k compresses or stretches the graph of y 5 The graphs of y 5
EXAMPLE
2
k xp
1 x p.
and y 5 2xkp are mirror images of each other across the x-axis.
For ƒsxd 5 x22 and gsxd 5 x23, construct the following functions and their graphs. a. ƒs2xd and 2ƒsxd b. 3ƒ(x) and its reflection across the x-axis c. gs2xd and 2gsxd d. 4g(x) and its reflection across the x-axis
SOLUTION
a. We can rewrite ƒsxd 5 x22 as ƒsxd 5 ƒs2xd 5
1 2 s2xd
5
1 . Then x2
1 x2
and 2ƒsxd 5 2 12 x
See Figure 7.33 Graph A. b. 3ƒsxd 5 3 A x12B 5 x32 The function y 5 2x32 is the reflection of 3ƒ(x) across the x-axis. See Figure 7.33 Graph B. c. We can rewrite gsxd 5 x23 as gsxd 5 gs2xd 5
1 3 s2xd
1 x 3 . Then
5 2 13 x
and 2gsxd 5 2 13 x
So gs2xd and 2gsxd are equivalent functions. See Figure 7.33 Graph C.
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7.6 Visualizing Negative Integer Power Functions
d. 4gsxd 5 4 A x13B 5
425
4 x3
The function y 5 2 x43 is the reflection of 4g(x) across the x-axis. See Figure 7.33 Graph D. y 10
y
g(–x) = –g(x)
10
f (–x) = f (x) x –4
4
x –4
4
–f (x) –10
–10 Graph C
Graph A
y 10
4g(x)
y 10 3f(x)
x –4
4
x –4
4 –4g(x) –3f(x) –10
–10 Graph D
Graph B
Graph of two equivalent functions gs2xd and 2gsxd . Figure 7.33 Four graphs of pairs of functions.
EXAMPLE
3
SOLUTION
a. Graph the power function ƒsxd 5 x23. b. Evaluate ƒ(0.5) and ƒ(22). c. Use part (b) and the fact that ƒ(x) is a power function to predict the values for ƒ(10) and ƒ(20.5). d. Now evaluate ƒ(10) and ƒ(20.5) to confirm your predictions in part (c). a. See Figure 7.34 on next page. b. ƒs0.5d 5
2 s0.5d3
5
2 s0.125d
5 16 and ƒs22d 5
2 s22d 3
5
2 s28d
5 20.25. So the points
(0.5, 16) and (22, 20.25) lie on the graph. c. Changing the input from 0.5 to 10 is equivalent to multiplying the original input by 1 20. So the output will be multiplied by 201 3 5 8000 . The value of ƒ(10) should be 1 1 A8000B ? ƒs0.5d 5 A8000B ? 16 5 0.002. Changing the input from 0.5 to 20.5 is equivalent to multiplying the input of 0.5 by 21. 1 The output will be multiplied by s21d 3 5 21. So ƒ(20.5) 5 2ƒ(0.5) 5 216.
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CHAPTER 7 POWER FUNCTIONS
d. Evaluating the functions directly, we have ƒs10d 5 ƒs20.5d 5
2 s20.5d 3
2 s20.125d
5
2 103
5
2 1000
5 0.002 and
5 216, confirming our predictions in part (c). y 20 (0.5, 16)
–2
x 2
(–2, –0.25)
–20
Figure 7.34 The graph of ƒsxd 5 x23 .
Algebra Aerobics 7.6 A graphing program is useful for Problems 3 and 4. 1. Rewrite each of the following expressions with positive exponents and then evaluate each expression for x 5 22. a. x22 c. 4x23 e. 221x24 g. 22x24 23 23 23 b. x d. 24x f. 22x h. 2x24 2. a. Select the most likely function type for each graph in Figure 7.35. a y 5 ax y5 y 5 ax 2 x a a y5 2 y 5 ax 3 y5 3 x x y
y
10
10
3. Sketch the graph for each equation: a. y 5 x210 b. y 5 x211 4. On the same grid, graph the following equations. a. y 5 x22 and y 5 x23. Where do these graphs intersect? How are the graphs similar and how are they different? b. y 5 4x22 and y 5 4x23. Where do these graphs intersect? How are the graphs similar and how are they different? 5. Match the appropriate function with each of the graphs in Figure 7.36 5 1 25 ƒsxd 5 2 gsxd 5 2 hsxd 5 2 x x x y
x –4
4
x –4
4
A B x –10 Graph A
–10 Graph B
C
Figure 7.35 Two unidentified graphs.
b. After you have decided on the function type, determine the value of a for each graph.
Figure 7.36 Three
unidentified graphs.
Exercises for Section 7.6 For many of the exercises in this section the use of a graphing program is required or recommended. 1. A cube of edge length x has a surface area Ssxd 5 6x 2 and a volume Vsxd 5 x 3 . We constructed the function 2 S(x) R(x) 5 V(x) 5 6x 5 6x . Consider R(x) as an abstract x3
function. What is the domain? Construct a small table of values, including negative values of x, and plot the graph. Describe what happens to R(x) when x is positive and x S 0 . What happens to R(x) when x is negative and x S 0?
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2. a. (Graphing program recommended.) Make a table of values and sketch a graph for each of the following functions. Be sure to include negative and positive values for x, as well as values for x that lie close to zero. 1 y1 5 x
1 y2 5 2 x
1 y3 5 3 x
3. a. (Graphing program recommended.) Generate a table of values and a graph for each of the following functions: gsxd 5 5x
hsxd 5
x 5
tsxd 5
7. Match each function with its graph. Explain your choices.
1 y4 5 4 x
b. Describe the domain and range of each function. c. Describe the behavior of each function as x approaches positive infinity and as x approaches negative infinity. d. Describe the behavior of each function when x is near 0.
1 x
ƒsxd 5
c. hsxd 5
b. gsxd 5 x 23
d. j(x) 5 2
y
Graph A
Graph C
5
y2 5 x 23
y3 5 x 25
b. y1 5 x 0
y2 5 x 22
y3 5 x 24
c. y1 5 2x 21
y2 5 4x 21
y3 5 22x 21
Graph B
5
y Graph D
5
y
x 5
x –5
5
–5
8. Match each function (where x . 0) with the appropriate graph. Explain your choices. 1 1 a. ƒsxd 5 x23 b. gsxd 5 5 c. hsxd 5 x x
5
y B
d. y 5 x22
C
y Graph A
x 5
–5
–5
6. Match each of the following functions with its graph: a. y 5 3s2 x d c. y 5 x23
Graph C
y
–5
5
–5
y
5
–5
5. (Graphing program required.) Using graphing technology, on the same grid graph y 5 x 2 and y 5 x22. a. Over what interval does each function increase? Decrease? b. Where do the graphs intersect? c. What happens to each function as x approaches positive infinity? Negative infinity?
b. y 5 x 3
2 x2
x –5
4. (Graphing program optional.) In each part, sketch the three graphs on the same grid and label each function. Describe how the three graphs are similar and how they are different. a. y1 5 x 21
2 x2
a. ƒsxd 5 x24
5 x
b. Describe the ways in which the graphs in part (a) are alike and the ways in which they are not alike.
A
20
5
x
0 10
5
x –5
5
x –5
–5
y Graph B
427
Graph D
5
0
5
y 5
x
x –5
5
–5
–5
5
–5
9. Begin with the function ƒsxd 5 x23. Then: a. Create a new function g(x) by vertically stretching ƒ(x) by a factor of 4. b. Create a new function h(x) by vertically compressing ƒ(x) by a factor of 12 . c. Create a new function j(x) by first vertically stretching ƒ(x) by a factor of 3 and then by reflecting the result across the x-axis. 10. (Graphing program optional.) Graph the functions ƒsxd 5 x12 , gsxd 5 x42 , and hsxd 5 4x1 2.
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For parts (a)–(c), insert the symbol . or , to make the relation true. a. For x . 0, ƒ(x) _______ g(x) b. For x . 0, g(x) _______ h(x) c. For x . 0, h(x) _______ ƒ(x) d. Describe how the graphs are related to each other when x , 0. Generalize your findings to any value for the coefficient k . 0. 11. (Graphing program required.) Graph the functions ƒsxd 5 x15 and gsxd 5 x14 and then insert the symbol . or , to make the relation true. a. For x . 1, ƒ(x) _______ g(x) b. For 0 , x , 1, ƒ(x) _______ g(x) c. For x , 0, ƒ(x) _______ g(x) 12. Consider the accompanying graph of ƒ(x) 5 k ? x1n , where n is a positive integer. y 5 a. Is n even or odd? b. Is k . 0 or is k , 0? c. Is ƒs21d . 0 or is ƒs21d , 0? x d. Does ƒs2xd 5 2ƒsxd ? –5 5 e. As x S 1`, ƒsxd S ___ f. As x S 2`, ƒsxd S ___ –5 13. Consider the accompanying graph of f(x) 5 k ? x1n , where n is a positive integer. a. Is n even or odd? y 1 b. Is k . 0 or is k , 0? x 5 c. Is ƒs21d . 0 or is ƒs21d , 0? –5 d. Does ƒs2xd 5 ƒsxd ? –5 e. As x S 1`, ƒsxd S ___ f. As x S 2`, ƒsxd S ___
Which functions: a. Are symmetric across the y-axis? b. Are symmetric about the origin? c. Have an even power n? d. Have an odd power n? e. Have k , 0? 15. (Graphing program optional.) Sketch the graph of the function ƒsxd 5 x23. Then consider ƒ(x) in relation to the following functions: g(x) 5 2x23,h(x) 5
1 2x3
and
j(x) 5 2
2
x3
a. Identify k, the constant of proportionality, for each function. b. Which function has a graph that is a reflection of ƒ(x) across the x-axis? c. Which function has a graph that is both a stretch and a reflection of ƒ(x) across the x-axis? d. Which function has a graph that is a compression of ƒ(x)? 16. a. Complete the partial graph in the accompanying figure three different ways by adding: i. The reflection of the graph across the y-axis to create Graph A. ii. The reflection of the graph across the x-axis to create Graph B. iii. The result of reflecting the graph first across the y-axis and then across the x-axis to create Graph C. b. Which of the finished graphs is symmetric across the y-axis? c. Which of the finished graphs is symmetric about the origin? d. Which of the finished graphs is symmetric across the x-axis?
14. Consider the following graphs of four functions of the form y 5 xkn, where n . 0.
y 5
Graph A
5
Graph C
y
y 5
x 5
–5
x
–5 Graph B
5
5
–5
y
y
Graph D
–5
5
5
–5
5
–5
x –5
x –5
x –5
5
–5
17. (Graphing program required.) Given ƒsxd 5 4x 22, construct a function g(x) that is a reflection of ƒ(x) across the horizontal axis. Graph the function and confirm your answer. 18. (Graphing program required.) Are ƒsxd 5 x23 and gsxd 5 2x23 reflections of each other across the vertical axis? Graph the functions and confirm your answer.
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19. If ƒsxd 5 x14 and gsxd 5 x15 , construct the following functions. a. ƒ(2x) and 2ƒ(x) c. g(2x) and 2g(x) b. 2ƒ(x) and ƒ(2x) d. 2g(x) and g(2x) e. The function whose graph is the reflection of ƒ(x) across the x-axis f. The function whose graph is the reflection of g(x) across the x-axis 20. (Graphing program required.) Begin with a function ƒsxd 5 x13 . In each part describe how the graph of g(x) is
429
related to the graph of ƒ(x). Support your answer by graphing each function with technology. a. g(x) 5 x23 b. g(x) 5 2 x13 c. g(x) 5 2 12x23 21. (Graphing program required.) Determine which of the graphs of the following pairs of functions intersect. If the graphs intersect, find the point or points of intersection. a. y 5 2x y 5 4x2 d. y 5 x21 y 5 x22 y 5 4x3 b. y 5 4x2 e. y 5 4x22 y 5 4x23 22 2 c. y 5 x y 5 4x
7.7 Using Logarithmic Scales to Find the Best Functional Model Looking for Lines Throughout this course, we have examined several different families of functions: linear, exponential, logarithmic, and power. In this section we address the question “How can we determine a reasonable functional model for a given set of data?” The simple answer is “Look for a way to plot the data as a straight line.” We look for straight-line representations not because the world is intrinsically linear, but because straight lines are easy to recognize and manipulate. In Figure 7.37 we plot the functions (for x . 0) of y 5 3 1 2x y 5 3 ? 2x
linear function exponential function
y 5 3x2
power function
first with a linear scale on both axes (a standard plot), then with a linear scale on the x-axis and a logarithmic scale on the y-axis (a semi-log plot), and finally with a logarithmic scale on both axes (a log-log plot).
Standard Plot
Semi-Log Plot
100 Exponential function
80
40
1000
Power function
100
Linear function
10
Linear function
20 1
2
3 4
5 (a)
6
7
8 9 10
100
Power function
10
Linear function
1
1 0
Exponential function
Exponential function
Power function
60
0
Log-Log Plot
1000
0
1 2 3 4
5 6 7 8 9 10 (b)
1
2
3
4 5 6 7 8 10
(c)
Figure 7.37 (a) Standard plot. The graph of a linear function appears as a straight line. (b) Semi-log plot. The
graph of an exponential function appears as a straight line. (c) Log-log plot. The graph of a power function appears as a straight line.
The course software “E11: Semi-Log Plots of y 5 Ca x ” in Exponential & Log Functions and “P2: Log-Log Plots of Power Functions” in Power Functions can help you visualize the ideas in this section.
On the standard plot in Figure 7.37(a), only the linear function appears as a straight line; the power and exponential functions curve steeply upward. On the semi-log plot in Figure 7.37(b), the exponential function now is a straight line, and the power and linear functions curve downward. As we saw in Chapter 5, exponential growth will always appear as a straight line on a semi-log plot, where the independent variable (input) is plotted on a linear scale and the dependent variable (output) on a logarithmic scale. On the log-log plot in Figure 7.37(c), only the power function appears as a straight line.
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Why Is a Log-Log Plot of a Power Function a Straight Line? As we did with exponential functions and semi-log plots in Chapter 6, we can use the properties of logarithms to understand why power functions appear as straight lines on a log-log plot. Let’s analyze our power function: y 5 3x2
(x . 0)
(1)
log y 5 logs3x d 2
Take the logarithm of both sides use Rule 1 of logs
5 log 3 1 log x2
use Rule 3 of logs
5 log 3 1 2 log x log y < 0.48 1 2 log x
evaluating log 3, we get
(2)
If we let X 5 log x and Y 5 log y, we can rewrite the equation as Y < 0.48 1 2X
(3)
So Y (or log y) is a linear function of X (or log x). Equations (1), (2), and (3) are all equivalent. The graph of Equation (3) on a log-log plot, with log x values on the horizontal and log y values on the vertical axis, is a straight line (see Figure 7.38). The slope of the line is 2, the power of x in Equation (1). The vertical intercept is 0.48 or log 3, the logarithm of the constant of proportionality in Equation (1).
Y (or log y) 4 3 2 1
X (or log x) 0
0.5
1
Figure 7.38 The graph of
Y < 0.48 1 2X.
?
SOMETHING TO THINK ABOUT
On what type of plot would a logarithmic function appear as a straight line?
The slope, 2, described by Equation (3) means that each time X (or log x) increases by one unit, Y (or log y) increases by two units. Remember that a log scale is multiplicative. So, adding 1 to log x is equivalent to multiplying x by 10. Adding 2 to log y is equivalent to multiplying y by 10 ? 10 or 10 2. We can confirm this by examining Equation (1), where we also see that each time x increases by a factor of 10, y increases by a factor of 10 2.
Translating Power Functions into Equivalent Logarithmic Functions In general, we can translate a power function in the form y 5 kx
p
where k and x . 0
into an equivalent linear function log y 5 log k 1 p log x Y 5 log k 1 pX
or where X 5 log x and Y 5 log y
The graph of the power function will appear as a straight line on a log-log plot with log x on the horizontal and log y on the vertical axis. The slope of the line is the power p of x and the vertical intercept is log k, where k is the constant of proportionality.
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Finding the Equation of a Power Function on a Log-Log Plot The graph of a power function y 5 kx p, where k and x . 0, appears as a straight line on a log-log plot and the equation of the line is Y 5 log k 1 pX
where X 5 log x and Y 5 log y
The slope of the line is p, where p is the power of x for y 5 kx p. The vertical intercept is log k, where k is the constant of proportionality for y 5 kx p.
Analyzing Weight and Height Data In 1938, Katherine Simmons and T. Wingate Todd measured the average weight and height of children in Ohio between the ages of 3 months and 13 years. Table 7.9 shows their data. Using a standard plot Let’s examine the relationship between height and weight. Figure 7.39 shows height versus weight for boys and for girls using a standard linear scale on both axes. Simmons and Todd's Ohio Data
Measuring Children
1 4 DATA
See Excel or graph link file CHILDSTA.
1 2 3 4 5 6 7 8 9 10 11 12 13
Weight (kg)
Height (cm)
Boys
Girls
Boys
Girls
6.5 10.8 13.2 15.2 17.4 19.6 22.0 24.8 28.2 31.5 35.6 39.2 42.0 46.6
5.9 9.9 12.5 14.7 16.8 19.2 22.0 24.5 27.9 32.1 35.2 39.5 46.6 52.0
61.3 76.1 87.4 96.2 103.9 110.9 117.2 123.9 130.1 136.0 141.4 146.5 151.1 156.7
59.3 74.2 86.2 95.5 103.2 110.3 117.4 123.2 129.3 135.7 140.8 147.8 155.3 159.9
150 140 130 Height (cm)
Age (yr)
160
120 110 100 90 Boys Girls
80 70 60 0
10
20 30 40 Weight (kg)
50
60
Table 7.9
Figure 7.39 Standard linear plot
Source: Data adapted from D. W. Thompson, On Growth and Form (New York: Dover Paperback, 1992), p. 105.
of height vs. weight data for boys and girls.
The filled circles are the data for boys, the open squares the data for girls. Table 7.9 also gives their respective ages, and in general, baby girls are smaller than baby boys and teenage girls are larger than teenage boys. But in the graph in Figure 7.39, we see that at the same weight the heights of boys and girls are roughly the same. The line in Figure 7.39 shows the linear model that best approximates the combined data. The line does not describe the data very well, and it doesn’t seem reasonable that height is a linear function of weight for growing children. Other models may describe the data better.
?
SOMETHING TO THINK ABOUT
Why is there no zero on an axis with a logarithmic scale?
Using a semi-log plot Figure 7.40 shows the same data, this time using a linear scale for weight and a logarithmic scale for height. Figure 7.40 also shows a best-fit exponential function, which appears as a straight line on the semi-log plot. This model is not a good fit to the data, and it doesn’t seem reasonable to suppose that height is an exponential function of weight.
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Simmons and Todd's Ohio Data
Simmons and Todd's Ohio Data
160 150
160
140
140
130
130
120
120
110
110
150
Height (cm)
Height (cm)
100 90 80
100 90 80
Boys Girls
70
Boys Girls 70
60
60 0
10
20 30 40 Weight (kg)
50
60
6 7 8 910
Figure 7.40 Semi-log plot of height vs. weight data for boys and girls.
20 30 40 50 Weight (kg)
Figure 7.41 Log-log plot of height vs. weight data.
Using a log-log plot The same data are plotted again on a log-log plot (a plot with logarithmic scales on both axes) in Figure 7.41. In this case, the line is a reasonably good approximation to the data and certainly the best fit of the three models considered. Since we know that a straight line on a log-log plot represents a power function, we can find the equation for the bestfit straight line on the log-log plot and transform the equation into a power function. Finding the equation of a power function on a log-log plot To construct a linear function for the best-fit line on the log-log plot in Figure 7.41, we can regraph the line using H(5 log h) and W(5 log w) scales on the axes, where h 5 height (cm) and w 5 weight (kg) (see Figure 7.42). 3.0 2.5
H (log h)
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2.0 1.5 1.0 0.5 0
0.5
1
1.5
2
W (log w)
Figure 7.42 Best-fit line for height vs. weight
data graphed on a log-log plot.
The equation will then be of the form H 5 b 1 mW
(1)
where H 5 log h and W 5 log w. To find values for the slope m and the vertical intercept b, we can estimate two points on the line, say (0.5, 1.5) and (1.5, 2.0). So slope 5
2.0 2 1.5 5 0.5 1.5 2 0.5
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433
From the graph, the vertical intercept appears to be approximately 1.3. So the linear equation would be H 5 1.3 1 0.5W
(2)
where H 5 log h and W 5 log w. We can transform this equation into an equivalent power function in the form h 5 kw p
(3)
where p is the slope of the best-fit line on the log-log plot and log k is the vertical intercept (see box on p. 433). The slope of the best-fit line is 0.5, so p 5 0.5. The vertical intercept is approximately 1.3, so definition of log evaluate 101.3
log k 5 1.3 k 5 101.3 k < 20
Substituting p 5 0.5 and k 5 20 in Equation (3), we get h 5 20w0.5
(or 20w(1/2))
(4)
We now have a power function that models the relationship between height, h, and weight, w. Interpreting our results Is it plausible to argue that height is a power law function of weight for growing children? Equation (4) says that h ~ w 1/2
or equivalently
w ~ h2
where the symbol ~ means “is directly proportional to.” Let’s think about whether this is a reasonable exponent. Suppose children grew self-similarly; that is, they kept the same shape as they grew from the age of 3 months to 13 years. Then, as we have discussed in Section 7.1, their volume, and hence their weight, would be proportional to the cube of their height: w 5 kh3
(self-similar growth)
But of course children do not grow self-similarly; they become proportionately more slender as they grow from babies to young adults (see Figure 7.43). Their weight therefore grows less rapidly than would be predicted by self-similar growth, and we expect an exponent less than 3 for weight as a function of height.
0.42
0.75 2.75 6.75 12.75 25.75 Years
Figure 7.43 The change in human body shape with increasing age. Source: Medawar, P. B. 1945. “Size and Shape in Age” in Essays on Growth and Form Presented to D’Arcy Thompson (W. E. LeGros Clark and P. B. Medawar, eds.), pp. 1572187. Oxford: Clarendon Press.
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On the other hand, suppose that children’s bodies grew no wider as their height increased. Since weight is proportional to height times cross-sectional area, with constant cross-sectional area we would expect weight to increase linearly with height. So w ~ h1. But certainly, children do become wider as they grow taller, so we expect the exponent of h to be greater than 1. Therefore, the experimentally determined exponent of approximately 2 seems reasonable. Other considerations There is a fourth possibility for plotting our height-weight data: We could use a linear scale for the height axis and a logarithmic scale for the weight axis. On such a plot, a straight line will represent a model in which weight is an exponential function of height. If we generate this plot, the data fall along what is very close to a straight line. Based only on the data, we might argue that an exponential function is a reasonable model. However, it is very difficult to construct a plausible physical explanation of such a model, and it would not help us understand how children grow. It may be, of course, that this relationship reveals some unknown physical or biological law. Perhaps you, the reader, will be the one to find some previously unsuspected explanation.
Allometry: The Effect of Scale Allometry is the study of how the relationships among different physical attributes of an object change with scale. In biology, allometric studies naturally focus on relationships in living organisms. Biologists look for general “laws” that can describe the relationship, for instance, between height and weight (as in this section) or between surface area and volume (as in Section 7.1) as the organism size increases. The relationships may occur within one species or across species. These laws are characteristically power functions. Sometimes we can predict the exponent by simple reasoning about physical properties and verify our prediction by examining the data; other times we can measure the exponent from data but cannot give a simple explanation for the observed value. Surface area vs. body mass In Section 7.1 we made the argument that larger animals have relatively less surface area than smaller ones. Let’s see if we can describe this relationship as a power law and then look at some real data to see if our conclusion was reasonable. Making Predictions. Since all animals have roughly the same mass per unit volume, their mass should be directly proportional to their volume. We will substitute mass for volume in our discussion since we can determine mass easily by weighing an animal. Since volume is measured in cubic units of some length, if animals of different sizes have roughly the same shape, then we would expect mass, M, to be proportional to the cube of the length, L, of the animal, M ~ L3
(1)
that is, M 5 k1 ? L3 for some constant k1. Also, we would expect surface area, S, to be proportional to the square of the length: S ~ L2 That is, S 5 k 2 ? L2 for some constant k2.
(2)
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By taking the cube root and square root, respectively, in relationships (1) and (2), we may rewrite them in the form M 1/3 ~ L
and S 1/2 ~ L
Since M 1/3 ~ L and L ~ S 1/2, we can now eliminate L and combine these two statements into the prediction that S 1/2 ~ M 1/3 By squaring both sides, we have the equivalent relationship S ~ M 2/3
(3)
It is useful to eliminate length as a measure of the size of an animal, since it is a little more ambiguous than mass. Should a tail, for instance, be included in the length measurement? Testing Our Predictions. This prediction, that surface area is directly proportional to the two-thirds power of body mass, can be tested experimentally. Figure 7.44 shows data collected for a wide range of mammals, from mice, with a body mass on the order of 1 gram, to elephants, whose body mass is more than a million grams (1 metric ton). Here surface area, S, measured in square centimeters, is plotted on the vertical axis and body mass, M, in grams, is on the horizontal. The scales on both axes are logarithmic.
106
S
105
Surface area (cm2)
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104
103
102
10 1
M 10 102 103 104 105 106 107 Body mass (g)
Figure 7.44 A log-log plot of body surface vs. body mass for a wide range of mammals. Source: McMahon, T. A. 1973. “Size and Shape in Biology,” Science, 179: 1201–1204.
The data lie pretty much in a straight line. Since this is a log-log plot, that implies that S is directly proportional to M p for some power p, where p is the slope of the best-fit line. To estimate the slope we can convert the scales on the axes in Figure 7.44 to log(surface area) and log(body mass); that is, we use only the exponents of 10 on both the
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vertical and the horizontal axes. Then the best-fit line passes through (3, 3) and (6, 5), 3 2 so the slope is 56 2 2 3 5 3 . This implies that S ~ M 2/3 confirming our simple prediction of a two-thirds power law. This means that every time the horizontal coordinate increases by a factor of 10 3, the vertical coordinate increases by a factor of 10 2 . We have approximate verification of the fact that mammals of different body masses have roughly the same proportion of surface area to the two-thirds power of body mass. This power function is just another way of describing our finding in Section 7.1 on the relationship between surface area and volume. Metabolic rate vs. body mass Another example of scaling among animals of widely different sizes is metabolic rate as a function of body mass. This relationship is very important in understanding the mechanisms of energy production in biology. Figure 7.45 shows a log-log plot of metabolic heat production, H, in kilocalories per day, versus body mass, M, in kilograms, for a range of land mammals. The data fall in a straight line, so we would expect H to be directly proportional to a power of M. The slope of this line, and hence the exponent of M, is 34 5 0.75. As body mass increases by four factors of 10, say from 1021 to 103 kg, metabolic rate increases by three factors of 10, from 10 to 104 kcal/day. Thus H ~ M 3/4, where M is mass in kilograms and H is metabolic heat production in kilocalories (thousands of calories) per day.
105 Elephant
H, Heat production (kcal/day)
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104
Steer
Cow
Chimpanzee
Sheep Goat
103
Rabbit
Macaque Cat Guinea Pig Rat
102
10 Mouse 1 10–3
10–2
10–1
1
10
102
103
104
M, Body mass (kg)
Figure 7.45 A log-log plot of heat production vs. body
mass. Source: Kleiber, M. 1951. “Body Size and Metabolism,” Hilgardia, 6: 315–353.
Kleiber’s Law. This scaling relationship is called “Kleiber’s Law,” after the American veterinary scientist who first observed it in 1932. It has been verified by many series of subsequent measurements, though its cause is not fully understood. Animals have evolved biological modifications partially to avoid the consequences of scaling laws. For example, our argument for the two-thirds power law of surface area vs. body mass was based on the assumption that the shapes of animals stay roughly the same as their size increases. But elephants have relatively thicker legs than gazelles in order to provide the extra strength needed to support their much larger weight. Such
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437
changes in shape are very important biologically but are too subtle to be seen in the overall trend of the data we have shown. Despite these adaptations, inexorable scaling laws give an upper limit to the size of land animals, since eventually the animal’s weight would become too heavy to be supported by its body. That’s why large animals such as whales must live in the ocean and why giant creatures in science fiction movies couldn’t exist in real life. Allometric Laws. Allometric laws provide an overview of the effects of scale, valid over several orders of magnitude. They help us compare important traits of elephants and mice, or of children and adults, without getting lost in the details. They help us to understand the limitations imposed by living in three dimensions.
Algebra Aerobics 7.7 Older ages, slope = 1.0 100
Younger ages, slope = 1.2
70 Arm length (cm)
A calculator that can evaluate powers and logs is required for Problems 2, 3, and 7. 1. Rewrite each of the following as an equivalent equation by taking the log of both sides. a. y 5 3 ? 2 x c. y 5 12 ? 10 x b. y 5 4x3 d. y 5 0.15x22 2. Rewrite each of the following so that y is a power or exponential function of x. a. log y 5 0.067 1 1.63 log x b. log y 5 2.135 1 1.954x c. log y 5 21.963 1 0.865x d. log y 5 0.247 2 0.871 log x 3. By inspection, determine whether y is an exponential or a power function of x. Then rewrite the expression so that y is a (power or exponential) function of x. a. log y 5 log 2 1 3 log x b. log y 5 2 1 x log 3 c. log y 5 0.031 1 1.25x d. log y 5 2.457 2 0.732 log x e. log y 5 20.289 2 0.983x f. log y 5 21.47 1 0.654 log x 4. a. Identify the type of function. i. y 5 4x3 ii. y 5 3x 1 4 iii. y 5 4 ? 3x b. Which of the functions in part (a) would have a straight-line graph on a standard linear plot? On a semi-log plot? On a log-log plot? If possible, use technology to check your answers. c. For each straight-line graph in part (b), use the original equation in part (a) to predict the slope of the line. 5. Interpret the slopes of 1.2 and 1.0 on the accompanying graph in terms of arm length and body height. [Note: Slopes refer to the slopes of the lines, where the units of the axes are translated to log(body height) and log(arm length).]
50 30 20
10 10
20
40
100 300 Body height (cm)
900
Source: T. A. McMahon and J. T. Bonner, On Size and Life (New York: Scientific American Books, 1983), p. 32.
6. For each of the three graphs below, examine the scales on the axes. Then decide whether a linear, exponential, or power function would be the most appropriate model for the data. y 100
log y 2.0 1.5
10
1.0 0.5
1 0
x
5 Graph A
10
0.0
x
0 1 2 3 4 5 6 7 8 9 10 Graph B
y 1000 100 10 1
1
10 100 Graph C
x
1000
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0.01
0.1
1
Elephant
Horse
a. Would a linear, power, or exponential function be the best model for the relationship between heart rate and body mass? b. The slope of the best-fit line shown on the graph is 20.23. Construct a function for heart rate in terms of body mass. c. Interpret the slope of 20.23 in terms of heart rate and body mass. 9. Using the graph in Figure 7.45 on page 438, what rate of heat production does this model predict for a 70-kg human being? Considering that each kilocalorie of heat production requires consumption of one food Calorie, does this seem about right?
Goat
10
Rabbit
100
Pig
Dog
Rat
Mouse
1000
Chipmunk
Heart rate (beats per min)
7. a. Using Figure 7.44 on page 437, estimate the surface area for a human being whose mass is 70 kg (or 7 ? 104 g). b. Construct a power function for the plot in Figure 7.44 and use it to calculate the value for part (a). c. How do your answers in parts (a) and (b) compare? d. Translate your answers into units of pounds and square inches. 8. The accompanying graph shows the relationship between heart rate (in beats per minute) and body mass (in kilograms) for mammals ranging over many orders of magnitude in size.
10 100 Body mass (kg)
1000
Source: Robert E. Ricklefs, Ecology (San Francisco: W. H. Freeman, 1990), p. 66.
Exercises for Section 7.7 Several exercises require a graphing program that can convert axes from a linear to a logarithmic scale. 1. Match each function with its standard linear-scale graph and its logarithmic-scale graph. (Hint: Evaluate each function for x 5 4.) b. y 5 100x5 a. y 5 100 (5)x y
y 350,000
350,000
300,000
300,000
250,000
250,000
200,000
200,000
150,000
150,000
100,000
100,000 50,000
50,000
x
x 0
1
2
3
4
0
5
1
2
4
5
3
4
5
Graph C
Graph A log y 6
y 1,000,000
5
100,000
4
10,000
3
1000
2
100
1
10
log x 0
3
0.1
0.2
0.3
0.4
Graph B
0.5
0.6
0.7
x
1 0
1
2 Graph D
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2. The accompanying graph shows data on U.S. annual death rate from cancer of the large intestine as a function of age. Would an exponential or power function be a better model for the data? Justify your answer.
Annual U.S. Death Rate from Cancer of the Large Intestine in Relation to Age, 1968
Deaths per million people per year
1000
100
439
4. Given the following three power functions in the form y 5 kx p, y1 5 x3
y2 5 5x3
y3 5 2x4
a. Use the rules of logarithms to change each power function to the form: log y 5 log k 1 p log x. b. Substitute in each equation in part (a), Y 5 log y and X 5 log x and the value of log k to obtain a linear function in X and Y. c. Compare your functions in parts (a) and (b) to the original functions. What does the value of log k represent in the linear equation? What does the value of the slope represent in the linear equation? 5. Using rules of logarithms, convert each equation to its power function equivalent in the form y 5 kx p. a. log y 5 log 4 1 2 log x c. log y 5 log 1.25 1 4 log x b. log y 5 log 2 1 4 log x d. log y 5 log 0.5 1 3 log x
10
1
20
40
60
80 100
Age (in years)
Source: J. Cairns, Cancer and Society (San Francisco: W. H. Freeman and Company, 1978), p. 37.
6. Given linear equations in the form Y 5 B 1 mX, where Y 5 log y, X 5 log x, and B 5 log k, change each linear equation to its power function equivalent, in the form y 5 kx p. (Note: Round values to the nearest hundredth.) a. Y 5 0.34 1 4X c. Y 5 1.0 1 3X b. Y 5 20.60 1 4X d. Y 5 1.0 2 3X 7. a. Create a table of values for the function y 5 x5 for x 5 1, 2, 3, 4, 5, 6. Then plot the values on the given log-log scale. Log-log Scale y
3. Does the accompanying graph suggest that an exponential or a power function would be a better choice to model infant mortality rates in the United States from 1915 to 1977? Explain your answer.
100,000 10,000 1000 100
Infant Mortality Rates in the United States, 1915–1977
10
100 90 80 70
x
1 1
10
Rate per 1000 live births
60
b. Add to the table you created in part (a) the values for log x and log y. Then graph the results on the log y vs. log x scale provided below.
50 40 30
5 20
log y
4 3 2
10 1900
1920
1940
1960
1980
Source: Healthy People: The Surgeon General’s Report on Health Promotion and Disease Prevention, DHEW (PHS) Publication 79-55071, Department of Health, Education, and Welfare, Washington, DC, 1979.
1 0 0
0.2
0.4
0.6
0.8
log x 1.0
c. Describe the relationship between the plots in parts (a) and (b).
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8. Match each power function with its log y vs. log x plot. a. y 5 10x 2 c. y 5 12x 4 3 b. y 5 4x d. y 5 10x22
log y 3.0 2.5 2.0 1.5 1.0 0.5 0.0 log x –0.5 0 0.2 0.4 0.6 0.8
e. What egg mass does your formula predict for a 12.7kilogram turkey? f. A giant hummingbird (Patagona gigas) lays an egg of mass 2 grams. What size does your formula predict for the mass of the adult bird?
log y 3.0
10. Find the equation of the line in each of the accompanying graphs. Rewrite each equation, expressing y in terms of x.
2.0
y 5
1.0
4 0.0
log x 0
0.2
Graph A
0.4
0.6 0.8
Graph C log y 3.0
log y 1.2
2
2
1
1
1.0 0.0
3
4
log y 5
–0.4 0.6 0.8
2
5
0
1
Graph A
0.4
0.2 0.4
(0, 3)
x 1
2.0
0
3
(0, 3)
0
log x
(2, 5)
4
3
0.8
0.0
log y 5
(4, 5)
0
0.2
Graph B
0.4
0.6 0.8
log x
Graph D
(4, 5)
4
log y 7 6
(2, 7)
5 3
(0, 3)
4 3
2
9. (Technology recommended for graphing and changing axis scales.) The following data give the typical masses for some birds and their eggs:
(0, 3)
2
1
1
x 0
1
2
3
4
5
0
1
Graph B
Species
Adult Bird Mass (g)
Egg Mass (g)
Ostrich Goose Duck Pheasant Pigeon Hummingbird
113,380.0 4,536.0 3,629.0 1,020.0 283.0 3.6
1,700.0 165.4 94.5 34.0 14.0 0.6
Source: W. A. Calder III, Size, Function and Life History (Boston: Harvard University Press, 1984).
a. Create a fourth column in the table with the values for the egg mass ratio adult bird mass. Is the ratio the same for all of the birds? (Note: The first ratio value should be 0.015.) Write a sentence that describes what you discover. b. Graph egg mass (vertical axis) vs. adult bird mass (horizontal axis) three times using standard linear, semi-log, and log-log plots. c. Examine the three graphs you made and determine which looks closest to linear. Find a linear equation in the form Y 5 b 1 mX to model the line. (Hint: To find the slope of the line, you will need to use log y on the vertical axis, and possibly log x on the horizontal axis). d. Once you have a linear model, transform it to a form that gives egg mass as a function of adult bird mass.
2 3 Graph C
2
3
Graph D
11. Use the accompanying graphs to answer the following questions. log y
log y
l1
l3
l2
l4 x
log x
a. Assume l1 and l2 are straight lines that are parallel. In each case, what type of equation would describe y in terms of x? How are the equations corresponding to l1 and l2 similar? How are they different? b. Assume l3 and l4 are also parallel straight lines. For each case, what type of equation would describe y in terms of x? How are the equations corresponding to l3 and l4 similar? How are they different? 12. (Technology recommended for graphing and changing axis scales.) a. Assume that ƒ(1) 5 5 and ƒ(3) 5 45. Then find an equation for f assuming f is: i. A linear function ii. An exponential function iii. A power function
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c. Interpret the slope in terms of oxygen consumption and body mass. In particular, by how much does oxygen consumption increase when body mass increases by a factor of 10? By a factor of 104?
b. Verify that you get a straight line when you plot: i. Your linear function on a standard plot ii. Your exponential function on a semi-log plot iii. Your power function on a log-log plot
c.
x y
1 8
2 16
3 32
4 64
5 128
x y
1 9
2 12
3 15
4 18
5 21
16. a. The left-hand graph below shows the relationship between population density and length of an organism. The slope of the line for log(density) vs. log(length) is 22.25. Express the relationship between population density and length in terms of direct proportionality. b. The right-hand graph shows the relationship between population density and body mass of mammals. The slope of the line for log (density) vs. log (mass) is 20.75. Express the relationship between population density and body mass in terms of direct proportionality. c. Are your statements in parts (a) and (b) consistent with the fact that body mass is directly proportional to length3? (Hint: Calculate the cube of the length to the 20.75 power.) 1016
14. (Graphing program required.) Examine the following table. x
y
log x
log y
1 2 3 4 5 6
5 80 405 1280 3125 6480
0 0.301 03 0.477 12 0.602 06 0.698 97 0.778 15
0.698 97 1.903 09 2.607 46 3.107 21 3.494 85 3.811 58
Treating x as the independent variable: a. Create a scatter plot of y versus x. b. Create a scatter plot of log y versus x. c. Create a scatter plot of log y versus log x. d. Does the table represent a linear, power, or exponential function? Explain your reasoning. e. Create the equation that models the data.
106 Elephant Horse
105 Dog
1010
Invertebrates, small rodents, and small birds
104
Birds, reptiles, amphibians, and small mammals Large mammals 10–4
10–2 1 Length (m)
102
104
104 103 102 10 1 10–1 10–2 10–1 1 10 102 Body mass (kg)
103
104
Source: T. A. McMahon and John T. Bonner, On Size and Life (New York: Scientific American Books, 1983), p. 228.
17. Public health researchers have developed a new measure of overall population health called “health-adjusted life expectancy” (HALE). HALE represents the number of expected years of life lived in full health. The following chart shows health-adjusted life expectancy versus the health care expenditure per capita for various countries. Note that here the log scale is on the horizontal axis. HALE versus Total Health Expenditure per Capita, 2002
Sheep
80.0 70.0
Rat Squirrel Mouse
10 0.01 0.1
Large invertebrates
106
10–6
105
Small invertebrates
108
1
Cat
103 10
Amoebae 1012
Man
104
2
Bacteria
102
1 10 100 1000 Body mass (kg)
Source: R. E. Ricklefs, Ecology (San Francisco: W. H. Freeman, 1990), p. 67.
a. Would a power or an exponential function be the best model of the relationship between oxygen consumption and body mass? b. The slope of the best-fit line (for log(oxygen) vs. log(mass)) shown on the graph is approximately 34 . Construct the basic form of the functional model that you chose in part (a).
60.0 HALE
Oxygen consumption (cm3 per hour)
15. The accompanying graph shows oxygen consumption versus body mass in mammals.
Population density (number/km2)
1014
Population density (number/km2)
13. (Technology optional.) For each table, create a linear, exponential, or power function that best models the data. Support your answer with a graph. a. x 1 2 3 4 5 y 3 12 27 48 75 b.
441
50.0 40.0 30.0 20.0 10
100
1000
10000
Per-capita total health expenditure
a. Sketch a best-fit line through the data. Replace each label x on the horizontal axis with log(x) 5 X. (So 10 becomes log 10 5 1, 100 becomes log 100 5 2, etc. You end up with the new labels 1, 2, 3, and 4 on the horizontal axis.)
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b. Estimate two points on the line to construct an equation for your line for y in terms of X. (Note that the vertical axis, where x 5 0, is not on the graph, so you can’t use it to estimate the vertical intercept.) Then rewrite your equation as y in terms of log x.
c. Why is it reasonable that HALE is a log function of per capita health expenditure? d. What does your result suggest about functions that appear to be linear on a semi-log plot, where the logarithmic scale is on the horizontal axis?
C H A P T E R S U M M A RY Power Functions
Even Positive Powers
y 5 kx
y = 10x4
y
A power function has the form
18
y = 5x4
p
y = x4
where k is called the constant of proportionality. If p . 0 (and k 2 0), then
y = 0.5x4
y is directly proportional to x p if y 5 kx p y is inversely proportional to x p if y 5 kx 2p a5
y = 0.25x4
k b xp
x 0
–2
Positive Integer Powers For power functions of the form y 5 kx p (where p . 0 and k 2 0) • The graph goes through the origin. • If the power p is odd (and p . 1), the graph is shaped roughly like and is symmetric about the origin. • If the power p is even, the graph is <-shaped and symmetric across the y-axis. • The value of the coefficient k compresses or stretches the graph of y 5 x p. • The graphs of y 5 kx p and y 5 2kx p are mirror images of each other across the x-axis. Odd Positive Powers 10
y y = 10x3
2
–2
y = 5x3 y = x3
Negative Integer Powers For power functions of the form y 5 and k 2 0)
k 2p x p or kx (where p . 0
• The graph never goes through the origin, since the function is not defined when x 5 0. • The graph is composed of two non-intersecting curves each asymptotic to both the x- and the y-axes. • If the power p is even, the graph is symmetric across the y-axis. • If the power p is odd, the graph is symmetric about the origin. • The value of the coefficient k compresses or stretches the graph of y 5 x1p. • The graphs of y 5 xkp and y 5 2xkp are mirror images of each other across the x-axis.
y = 0.5x3 y = 0.25x3
–2 0
2
Odd Negative Powers y
Even Negative Powers y
10
x
10
y = 1/x = x –1
y = 1/x 2 = x –2
x –10
–2
2
–10
x –2
2
–10
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Check Your Understanding
Semi-Log and Log-Log Graphs To find the best function to model a relationship between two variables, we look for a way to plot the data so they appear as a straight line: • The graph of a linear function appears as a straight line on a standard plot that has linear scales on both axes. • The graph of an exponential function appears as a straight line on a semi-log plot that has a linear scale on the horizontal axis and a logarithmic scale on the vertical axis. The slope is the logarithm of the growth (or decay) factor. • The graph of a power function appears as a straight line on a log-log plot that has logarithmic scales on both axes. The slope is the exponent of the power function.
443
Finding the Equation of a Power Function on a Log-Log Plot The graph of a power function y 5 kx p (where k and x . 0) appears as a straight line on a log-log plot. The equation of the line is Y 5 log k 1 pX
where X 5 log x and Y 5 log y
• The slope of the line is p, where p is the power of x for y 5 kx p. • The vertical intercept is log k, where k is the constant of proportionality for y 5 kx p.
C H E C K Y O U R U N D E R S TA N D I N G I. Is each of the statements in Problems 1–21 true or false? Give an explanation for your answer.
14. Of the three functions f, g, and h in the accompanying figure, only function h could be a power function.
1. The graph of y 5 23x5 is decreasing.
2. The graph of y 5 2x21 increases when x , 0 and decreases when x . 0.
y
h g
f
3. The graphs of y 5 3.5x4 and y 5 5.1x2 intersect only at the origin. 4. If y 5 x3 and if x increases by 1, then y increases by a factor of 3. 5. If y 5 5x21 and if x doubles, then y decreases by a 1
factor of 2.
x
6. The function y 5 3 ? 2x is a power function.
7. The graphs of the functions y 5 3x2 and y 5 3x22 are reflections of each other across the x-axis. 8. The graphs of the functions y 5 4x21 and y 5 24x21 are reflections of each other across the x-axis. 22
9. If y 5 x zero.
, as x becomes very large, then y approaches
1 10. If y 5 x 2, as x approaches 1` , then y approaches 1` .
11. The function y 5 x10 eventually grows faster than the function y 5 1.5x.
12. The functions y 5 x2 and y 5 x21 intersect at the point (1, 1).
15. If F 5 km23 and k is a nonzero constant, then F is directly proportional to m. 23 16. If Ksdd 5 3d 22, then Ks22d 5 4 . 1 17. The graphs of two power functions ƒsxd 5 x p and
gsxd 5
1 x q are shown in the accompanying figure. For
these functions, q , p. y
2
13. The graph of y 5 x5 on a log-log plot is a straight line with slope of 25.
f
g
x
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18. If a quantity M is inversely proportional to the cube of a quantity Q, then M 5 Qk3 for k, a nonzero constant.
28. A power function whose graph would be similar to the graph of y 5 23x but would have a different power of x.
19. In the power function y 5 x1p graphed in the accompanying
29. A power function T(m) where doubling the value of m increases the value of T(m) by a factor of 16.
figure, p is an even positive integer.
III. Is each of the statements in Problems 30–40 true or false? If a statement is true, explain why. If a statement is false, give a counterexample.
y
30. All linear functions are power functions. y=
31. All power functions y 5 kx p, where p is a positive integer, pass through the origin.
1
xp
32. For the two power functions m(x) 5 x p and n(x) 5 x q in the accompanying figure, both p and q have negative values.
x
y
20. If ƒ(q) 5 2q , then ƒ(22) 5 232.
n
5
m
21. The graphs of the power functions A, B, and C, all of the form G 5 k ? m p, are shown in the accompanying figure. For each of these functions, k , 0.
x
G
A
n
m
C B
B
II. In Problems 22–29, construct a function or functions with the specified properties. 22. A power function y 5 ƒ(x) whose graph is steeper than the graph of y 5 4x6 when x . 1. 23. A power function y 5 g(x) whose graph is a reflection of the graph of y 5 23.2x4 across the horizontal axis. 24. A power function of even degree whose graph opens downward. 25. A function w 5 h(m) whose graph will eventually dominate the function w 5 500m 10. 26. A function that is not a power function. 27. A function whose graph on a log-log plot is a straight line.
33. The graphs of all power functions y 5 kx p, where p is a negative integer, lie in quadrants I (where x $ 0 and y $ 0) and III (where x # 0 and y # 0). 34. The graphs of all power functions with integer powers are asymptotic to the horizontal axis. 35. The graphs of all power functions y 5 kx p, where p is a positive even integer, are symmetric across the y-axis. 36. The ratio of (surface area)/volume is less for a larger sphere than for a smaller one. 37. The graphs of the functions F(m) 5 a ? m p and G(m) 5 2a ? m p, where a 2 0 and p is a positive integer, are reflections of each other across the vertical axis. 38. The graphs of power functions y 5 kx p, where p is a negative integer, never pass through the origin. 39. The graphs of power functions ƒ(x) 5 kx p, where p is a positive integer, on a log-log plot are straight lines with a positive slope. 40. The graphs of exponential functions g(x) 5 a ? b x on a log-log plot are also straight lines.
CHAPTER 7 REVIEW: PUTTING IT ALL TOGETHER A standard calculator that can do multiplication and division is required throughout. Some problems are identified as needing a calculator that can calculate fractional powers (such as square roots) or a graphing program (that for one problem can change axis scales from linear to logarithmic). 1. Suppose you have a solid rectangular box, with dimensions x by 2x by 2x.
x 2x 2x
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a. What is its volume V(x)? If you double the value of x, from X to 2X, what happens to the volume? b. What is the surface area S(x)? If you double the value of x, from X to 2X, what happens to the surface area? c. Construct the ratio R(x) 5 S(x)/V(x). If you double the value of x, from X to 2X, what happens to R(x)? Describe your result in terms of surface area and volume. d. As x increases, what happens to R(x) 5 (surface area)/volume? 2. Inside a bottle of Bayer aspirin is a small plastic cylinder (about 5/8" in height and 1/2" in diameter) whose tightly packed contents are used to absorb water to keep the aspirin dry. a. What is the volume (in cubic inches) of the cylinder (and hence a good approximation of the volume of its contents)? [Note: The volume of a cylinder with radius r and height h is pr2h. b. If you examine the contents of the cylinder under a microscope, you would see an incredibly convoluted surface. The total surface area is about the size of a football field (160' wide and 360' long, including the two end zones). What is the area in square feet? In square inches? c. Interpret your results. Why would the manufacturers want such a large surface area in such a tiny volume?
445
where W is in watts, A is in square feet , and S is in miles per hour. a. Is W directly proportional to A? To S? To S3? b. Calculate the circular area A (in square feet) swept out by wind turbine blades with a rotor radius of 10 ft. (Round off to the nearest integer.) c. Using your answer in part (b), alter the general formula for W to represent the wattage W10 generated by a wind turbine with 10-ft blades. Using the new formula, how many watts would be generated by a wind speed of 30 mph? How many kilowatts is that? If a typical suburban house needs on average about 1.8 kilowatts of power at any point in the day, how many houses could this turbine support? d. If the wind speed doubles, what happens to the wattage output? If the wind speed is cut in half, what happens to the wattage output? e. Is it better to have ten wind turbines in an area that averages 20-mph winds or two turbines in an area that averages 40-mph winds? 7. Some northern cities celebrate New York’s eve with First Night events, which include artists making large, elaborate ice sculptures. (See the accompanying photo from Fairbanks, Alaska.)
3. The website for La Riviera Gourmet in Lexington, Massachusetts, states, “A 10" cake takes up 3 quarts of volume, whereas a 12" cake takes up 6 quarts of volume. That’s why a 12" cake feeds twice as many and costs almost twice as much!” Does this seem reasonable? 4. Which of the following functions represent power functions? Which power functions represent direct and which inverse proportionality? a. Q 5 4t c. Q 5 4 1 t2 e. Q 5 4t22 b. Q 5 4t2 d. Q 5 4 ? 2t 5. A bathtub holds about 15 gallons of water. a. Construct a function to describe the water volume VF(t) in gallons at minute t if the tub is being filled at a rate of 3 gallons/minute. What would be a reasonable domain for the function? Graph VF(t) by hand. Does VF(t) represent direct proportionality? b. Assuming the tub is full to the brim, construct a new function VD(t) to describe the water volume (in gallons) at minute t if the tub has a leak that is draining water at a rate of 0.5 gallons/minute. What would be a reasonable domain for VD(t)? Graph VD(t) by hand. Does VD(t) represent inverse proportionality? 6. In some areas wind turbines offer a viable clean source of energy. The amount of power, W, (in watts) delivered by a wind turbine is a function of S, wind speed, and A, the area swept out by the wind turbine rotors. Wind engineers can estimate the wattage using the formula
W 5 0.0052AS3
An artist planned to build a sculpture 4' high to be carved out of a 4'32'32' block of ice. But she decides it would look better proportionally scaled up to 6' high. Her friend says, "You’ll have to order 50% more ice if you make it 6 ft instead of 4 ft." a. Is the friend correct? b. Give the dimensions of the ice block she should order for the 6' sculpture? What percent increase in the volume of ice would that be? c. If the temperature gets above freezing and the 6' sculpture loses one-tenth of its volume through melting, how tall will it be then? 8. (Requires a calculator that can compute square roots.) In any stringed instrument, such as a violin, guitar, or ukulele, each note is determined by the frequency F of a vibrating string. F is directly proportional to the square root of the tension T and
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inversely proportional to the length L of the string. The higher the frequency, the higher the note. a. Construct a general formula for F. b. If the string length is cut in half (by holding down a string in the middle against the neck of the instrument), is the resulting note higher or lower? By how much? c. If the tension is increased by 10%, will the resulting note be higher or lower? By how much? 9. Construct a formula for each situation. a. The time, t, taken for a journey is directly proportional to the distance traveled, d, and inversely proportional to the rate r at which you travel. b. On a map drawn to scale, the distance D1 between two points is directly proportional to the distance D2 between the corresponding two points in real life. c. The resistance R is directly proportional to the voltage, V, squared and inversely proportional to the wattage, W. 10. The following four graphs are power functions with positive exponents (and hence represent direct proportionality). Which graphs depict even and which depict odd power functions? y
y
14. (Requires a calculator that can compute square roots.) Film studios, photographers, and the lighting industry usually measure light intensity in foot-candles—historically, the amount of light one candle generates on a surface one foot away. Light intensity I (in foot-candles) is directly proportional to candlepower P (which measures how much light is generated by a light source) and inversely proportional to the square of the distance, d, from the light source. (See Section 7.5, Example 7.) The relationship is the inverse square law
I = P/d2 where I is measured in foot-candles, P in candlepower, and d in feet. a. A student’s angle-arm lamp can be positioned as close as 1 foot above the work surface and as far as 3 feet, measuring to the center of the bulb. If the 60-watt bulb has 70 candlepower, how many foot-candles illuminate the surface at the lowest and the highest positions? b. Where would you position the bulb to get 50 foot-candles of illumination, the standard level recommended for reading?
x
10
0
–5
5
x –5
5
–25
–10 Graph A 160
c. Now consider your function P(n) as an abstract function, where n can be any positive or negative real number. Graph this extended version of P(n) and identify any asymptotes.
Graph C
y
50
15. (Requires a calculator that can compute square roots.) Surgeons use a lamp that supplies 4000 foot-candles when suspended 3 feet above the operating surface. (See the preceding problem.) a. What candlepower would the source have to provide? b. Lighting engineers assume that surgery requires somewhere in the range of 2000 to 5000 foot-candles. How high above the operating surface would you hang the lamp to give 2000 foot-candles? 16. Which of the following graphs could depict power functions with a negative exponent (and hence inverse proportionality)?
y
100
5
y
5
–5
–50
–160 Graph B
Graph D
11. Describe any symmetries and asymptotes for each function graphed in the preceding problem.
x –10
y2 5 1.000001x,
0
10
x –5
Graph A
12. Of the following three functions, as x S 1∞, which would eventually dominate (have the largest values for y)?
y1 5 x1000,
100
x
x –5
y
5
0
5
Graph C
y y 10
y3 5 10100,000x
13. (Graphing program optional.) a. In Nashua, New Hampshire, you can rent an indoor skydiving facility for $450 per half hour. What is the cost per person, P(n), if there are n friends sharing the experience? b. What is the cost per person for two people? For five people? What is the trade-off between number of people and cost per person?
x –5
5
x –5
0 Graph B
5
–10 Graph D
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17. Describe any apparent symmetries and asymptotes for each of the function graphs in the preceding problem.
20. (Requires a graphing program that can convert the scale on either axis from linear to logarithmic.) These data give the sound reduction in decibels through walls of different densities, measured in pounds per square foot of the wall. (Note: This assumes a fixed depth to the wall material.)
18. (Requires a calculator that can compute fractional powers.) Because wind speed increases with height, wind turbines are placed as high as possible, and certainly above the level of any existing structures or trees that may block the wind. In order to estimate typical wind speeds at various heights, wind engineers first measure the average wind speed S33 (in mph) at a height of 33 feet. Then they can estimate the wind speed Sh at hight h using the formula
Sound Reduction through Walls
Sh 5 S33(h/33)2/7 where h is in feet, and Sh and S33 are both in miles per hour. a. For an area likely to have hurricanes, wind engineers design for a wind speed of 90 mph at a height of 33 feet. Estimate the wind speed under hurricane conditions at the top of a building 120 feet tall. b. Rearrange the formula to express h in terms of Sh and S33. Using the hurricane conditions in part (a), at what height dose the formula perdict 156-mph winds?
Y = log y 3.5 3 2.5 2 1.5
X = log x 0.2
0.4
0.8
0.6
Sound Loss (Db)
1 5 10 20 40 60 100 400
22.7 32.0 37.0 41.0 45.0 48.0 51.0 60.0
21. According to Adrian Bejan’s Constructal Theory, the optimal cruising speed v (in meters per second) of flying bodies (whether insects, birds, or airplanes) is directly proportional to the one-sixth power of body mass m (in kilograms). Use the accompanying graph below to confirm or refute this.
1 0.5 0
Wall Density (lb/sq. ft)
a. Graph these data with wall density on the horizontal axis. Try linear, log-linear (with the log scale on the horizontal and then on the vertical axis), and log-log plots. In which format do the data look most linear? b. Use the format of the data from part (a) to construct a formula for sound loss as a function of wall density. c. What does your result suggest about data that appear linear when graphed on a semi-log plot, where the log scale is on the horizontal axis?
19. a. Given the following graph, what kind of function is Y with respect to X?
0 1
b. Sketch in a best-fit line and then generate its equation using Y in terms of X. [Hint: Use the points (0, 0.3) and (1, 3.3).] c. Transform your equation in part (b) to one using y and x. What sort of function do you have now (for y in terms of x)?
1 10–5
10–4
10–2
Partridge
Pheasant Black vulture White stork Gannet Golden eagle Canada goose Wandering albatross Mute swan
Peregrine falcon Blue heron
Blue jay
10–1
Sparrow hawk
Skylark
House wren Burn swallow House sparrow Magnolia warbler Sand martin
Ruby-throated hummingbird
10–3
Goldcrest
Dung beetle
Bumblebee Hornet
Meat fly Honeybee House fly
v (m/sec)
10
Piper Warrior Beech Bonanza Beech Baron Beech King Air Leerjet-31 F-16 Mig-23 Fokker F-28 F-14 Boeing 737 Airbus A310 Douglas DC-10 Boeing 747
300
100
447
1
10
102
103
104
105
106
m (kg)
Source: Image from Wikipedia, derived from the work of Adrian Bejan, Professor of Mechanical Engineering at Duke University.
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E X P L O R AT I O N 7 . 1 Scaling Objects Objectives • find and use general formulas for scaling different types of objects Procedure 1. Scaling factors. When a two- or three-dimensional object is enlarged or shrunk, each linear dimension is multiplied by a constant called the scaling factor. For the two squares below, the scaling factor, F, is 3. That means that any linear measurement (for example, the length of the side or of the diagonal) is three times larger in the bigger square.
Scaling up a square by a factor of 3.
a. What is the relationship between the areas of the two squares above? Show that for all squares area scaled 5 soriginal aread ? F 2
where F is the scaling factor
b. Given a scaling factor, F, find an equation to represent the surface area S1 of a scaled-up cube, in terms of the surface area S0 of the original cube. c. Given a scaling factor, F, find an equation to represent the volume V1 of a scaled-up cube, in terms of the volume V0 of the original object. 2. Representing 3-D objects. We can describe the volume, V, of any three-dimensional object as V 5 kL3, where V depends on any length, L, that describes the size of the object. The coefficient, k, depends upon the shape of the object and the particular length L and the measurement units we choose. For a statue of a deer, for example, L could represent the deer’s width or the length of an antler or the tail. If we let L 0 represent the overall height of the deer in Figure 1, then L30 is the volume of a cube with edge length L 0 that contains the entire deer. The actual volume, V 0, of the deer is some fraction k of L30; that is, V0 5 kL 30 for some constant k.
Figure 1
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Figure 2
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3. Finding the scaling factor. If we scale up the deer to a height L1 (see Figure 2), then the volume of the scaled-up deer equals kL 31. The scaled-up object inside the cube will still occupy the same fraction, k, of the volume of the scaled-up cube. a. If an original object has volume V0 and some measure of its length is L 0, then we know V0 5 kL 30. Rewrite the equation, solving for the coefficient, k, in terms of L 0 and V 0. b. Write an equation for the volume, V1, of a replica scaled to have length L1. Substitute for k the expression you found in part (a). Simplify the expression such that V1 is expressed as some term times V 0. c. The ratio L1 /L0 is the scaling factor. A scaling factor is the number by which each linear dimension of an original object is multiplied when the size is changed. Write an equation for the volume of a scaled object, V1, in terms of V 0, the volume of the original object, and F, the scaling factor. 4. Using the scaling factor a. Draw a circle with a 1-inch radius. If you want to draw a circle with an area four times as large, what is the change in the radius? b. A photographer wants to blow up a 3-cm by 5-cm photograph. If he wants the final print to be double the area of the original photograph, what scaling factor should he use? c. A sculptor is commissioned to make a bronze statue of George Washington sitting on a horse. To fit into its intended location, the final statue must be 15 feet long from the tip of the horse’s nose to the end of its tail. In her studio, the sculptor experiments with smaller statues that are only 1 foot long. Suppose the final version of the small statue requires 0.15 cubic feet of molten bronze. When the sculptor is ready to plan construction of the larger statue, how can she figure out how much metal she needs for the full-size one? d. One of the famous problems of Greek antiquity was the duplication of the cube. Our knowledge of the history of the problem comes down to us from Eratosthenes (circa 284 to 192 B.C.), who is famous for his estimate of the circumference of Earth. According to him, the citizens of Delos were suffering from a plague. They consulted the oracle, who told them that to rid themselves of the plague, they must construct an altar to a particular god. That altar must be the same shape as the existing altar but double the volume. What should the scaling factor be for the new altar? [Adapted from COMAP, For All Practical Purposes (New York: W. H. Freeman, 1988), p. 370.] e. Pyramids have been built by cultures all over the world, but the largest and most famous ones were built in Egypt and Mexico. i. The Great Pyramid of Khufu at Giza in Egypt has a square base 755 feet on a side. It originally rose about 481 feet high (the top 31 feet have been destroyed over time). Find the original volume of the Great Pyramid. [The volume V of a pyramid is given by V 5 sB ? Hd/3, where B is the area of the base and H is its height.] ii. The third largest pyramid at Giza, the Pyramid of Menkaure, occupies approximately one-quarter of the land area covered by the Great Pyramid. Menkaure’s pyramid is the same shape as the original shape of the Great Pyramid, but it is scaled down. Find the volume of the Pyramid of Menkaure.
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E X P L O R AT I O N 7 . 2 Predicting Properties of Power Functions Objectives • construct power functions with positive integer exponents • find patterns in the graphs of power functions Materials • graphing calculator or function graphing program or “P1: k & p Sliders” in Power Functions • graph paper Procedure Working in Pairs 1. a. Construct the equation of a power function whose graph is symmetric across the y-axis. Construct a second power function whose graph is symmetric about the origin. b. Construct the equations for two power functions whose graphs are mirror images across the x-axis. Construct two more whose graphs are mirror images across the y-axis. 2. Using the same power for each function, construct two different power functions such that: a. Both functions have even powers and the graph of one function “hugs” the y-axis more closely than that of the other function. b. Both functions have odd powers and the graph of one function “hugs” the y-axis more closely than that of the other function. 3. Using the same value for the coefficient, k, construct two different power functions such that: a. Both functions have even powers and the graph of one function “hugs” the y-axis more closely than that of the other function. b. Both functions have odd powers and the graph of one function “hugs” the y-axis more closely than that of the other function. 4. a. Choose a value for k that is greater than 1 and construct the following functions: y 5 kx 2 y 5 kx4 y 5 kx 3 y 5 kx 5 i. y 5 kx0 5 k ii. y 5 kx1 b. Graph the functions in part (i) on the same grid. Choose a scale for your axes so you can examine what happens when 0 , x , 1. Now regraph, choosing a scale for your axes so you can examine what happens when x . 1. Then repeat for part (a. ii). c. Describe your findings. Class Discussion Compare your findings. As a class, develop a 60-second summary describing the results. Exploration-Linked Homework: A Challenge Problem In Chapter 7 we primarily studied power functions with integer exponents. Try extending your analyses to power functions with positive fractional exponents. Start with fractions that in reduced form have 1 in the numerator. What would the graphs of y 5 x1/2 and y 5 x1/4 look like? What are the domains? The ranges? What would the graphs of y 5 x1/3 or y 5 x1/5 look like? What about their domains and ranges? Generalize your results for fractions of the form 1n , where n is a positive integer. For an even harder challenge, consider fractional exponents of the form m/n.
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Objectives • examine the effect of k on negative integer power functions Materials • graphing calculator or function graphing program • graph paper Related Software • “P1: k & p Sliders” in Power Functions Procedure Class Demonstration: Constructing Negative Integer Power Functions 1. A power function has the form y 5 kx p, where k and p are constants. Consider the following power functions with negative integer exponents where k is 4 and 6, respectively: y 5 4x22
and
y 5 6x22
and
y5
We can also write these as y5
4 x2
6 x2
What are the constraints on the domain and the range for each of these functions? 2. Construct a table of values for both functions using positive and negative values for x. How do different values for k lead to different values for y for these two functions? Sketch a graph for each function. Check your graphs with a function graphing program or graphing calculator. a. Describe the overall behavior of these graphs. In each case, when is y increasing? When is y decreasing? How do the graphs behave for values of x near 0? b. Describe how these graphs are similar and how they are different. Working in Small Groups In the following exploration you will predict the effect of the coefficient k on the graphs of power functions with negative integer exponents. In each part, compare your findings, and then write down your observations. 1. a. Choose a value for p in the function y 5 xkp, where p is a positive integer. Construct several functions where p has the same value but k assumes different positive values (as in the example above, where k 5 4 and k 5 6). What effect do you think k has on the graphs of these equations? Graph the functions. As you choose larger and larger values for k, what happens to the graphs? Try choosing values for k between 0 and 1. What happens to the graphs? b. Using the same value of p as in part (a), construct several functions with negative values for the constant k. What effect do you think k has on the graphs of these equations? Graph the functions. Describe the effect of k on the graphs of your equations. Do you think your observations about k will hold for any value of p?
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c. Choose a new value for p and repeat your experiment. Are your observations still valid? Compare your observations with those of your partners. Have you examined both odd and even negative integer powers? In your own words, describe the effect of k on the graphs of functions of the form y 5 xkp , where p is a positive integer. What is the effect of the sign of the coefficient k? 2. a. Choose a value for k where k . 1 and rewrite each function in the form y 5 xkp. i. y 5 kx0 5 k y 5 kx22 y 5 kx24 ii. y 5 kx21 y 5 kx23 y 5 kx25 b. Graph the functions in part (i) on the same grid. Choose a scale for your graphs such that you can examine what happens when 0 , x , 1. Now choose scales for your axes so you can examine what happens when x . 1. c. Graph the functions in part (ii) on the same grid. Choose scales for your graphs such that you can examine what happens when 0 , x , 1. Now choose scales for your axes so you can examine what happens when x . 1. d. Describe your findings. Class Discussion Compare the findings of each of the small groups. As a class, develop a 60-second summary on the effect of k on power functions in the form y 5 xkp, where p is a positive integer.
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QUADRATICS, POLYNOMIALS, AND BEYOND OVERVIEW In this chapter we learn how to transform familiar functions into new ones. We start by adding power functions to create quadratics, polynomials of degree 2, which can be used to describe motion, model traffic flow, or predict the spread of a wildfire. We continue adding power functions to create polynomial functions, and then take the ratio of two polynomial functions to create a rational function. Finally, we look at many ways of transforming or combining functions from all of the families of functions we have studied. After reading this chapter you should be able to • understand the behavior and construct graphs of quadratic and other polynomial functions • determine the vertex and intercepts of a quadratic function • convert quadratic functions from one form to another • transform any function using stretches, compressions, shifts, and reflections • combine any two functions using basic algebraic operations or composition • identify when a function has an inverse
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8.1 An Introduction to Quadratic Functions The Simplest Quadratic A power function of degree 2 (in the form y 5 ax2) is also the simplest member of a family of functions called quadratics. We’ve already encountered quadratics such as S 6x2 (the surface area of a cube) and S 5 r 2 (the area of a circle). Since y 5 ax2 has an even integer power, its graph has the classic ∪ -shaped curve called a parabola. (See Figure 8.1.) Certain properties are suggested by the graph: • The parabola is concave up if a . 0 (and concave down if a , 0). • It has a minimum (or maximum) point called the vertex. • It is symmetric across a vertical line called an axis of symmetry that runs through the vertex. A parabola has another unique and useful property. There is an associated point off any 1 parabola called the focus or focal point. It is located u 4a u units above (or below) the vertex and lies within the arms of the parabola. y Axis of symmetry
y = ax 2 Focus
x Vertex
Figure 8.1 The parabolic graph
of y 5 ax2, where a . 0. 1
Determine whether the graph of y 5 2x 2 is concave up or down, find its vertex and focal point, and then graph the function.
SOLUTION
Since 2x 2 $ 0, the minimum value for y 5 2x 2 is 0, which will occur when x 5 0. So the function has a minimum and its vertex is at the origin (0, 0). Since a . 0, the graph is concave up. The coefficient, a, of x 2 is 2, so the focal point is 1/s4ad 5 1/s4 ? 2d 5 1/8 or 0.125 units above the vertex. The coordinates of the focal point (or focus) are (0, 0.125). Figure 8.2 shows the graph of the function.
EXAMPLE
y
Focus (0, 0.125)
y = 2x 2
x Vertex
Figure 8.2 Graph of y 5 2x 2
with vertex at (0, 0) and focal point at (0, 0.125).
The focal point merits its name. A three-dimensional parabolic bowl has vertical cross sections that are all the same parabola. If the bowl is built of a reflective material, the parabolic shape will concentrate parallel rays from the sun or a satellite TV channel at the focal point (see Figure 8.3). This property is used to focus the sun’s rays to construct primitive cooking devices or to concentrate electromagnetic waves from satellites or distant stars.
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Focus at dot
Figure 8.3 A parabola with parallel incoming rays concentrating at its focus.
The course software “Q10: Parabolic Reflector” gives an interactive demonstration of the reflective property of parabolas.
This process also works in reverse. For example, if you have a light bulb centered at the focus of a parabola, the light rays will bounce off a reflective parabolic surface as a set of parallel rays. This property is used to construct lamps, car headlights, or radio transmitters that send messages to robot rovers on Mars or broadcast our existence to the universe. Designing parabolic devices The easiest way to design a parabolic device is to place the vertex of the parabola at the origin (0, 0)—in other words, to construct a function of the form y 5 ax 2.
EXAMPLE
2
Designing a cooking device In developing countries, such as Burkina Faso,1 solar parabolic cooking devices are cheap ways to cook food and pasteurize water. The cooking devices can be made any size and can even be constructed of hardboard and aluminum foil. If a large parabolic cooking device is built that is 4r wide and 1r deep, where should the cooking pot be centered for maximum efficiency? y Focus
(2, 1)
1'
x 4'
Figure 8.4 Parabolic cooking device. SOLUTION
The cooking pot should be placed at the focal point of the parabola, where the sun’s rays will be concentrated. If we think of the bottom of the cooker as placed at (0, 0), we can construct a function of the form y 5 ax 2 and use it to find the focal point (see Figure 8.4). To find the value of a, we need the coordinates of a point on the parabolic cross section. Since the cooker is 4r wide, if we move 2r to the right of the cooker’s base, and 1r up, we will be at the point (2, 1), which lies on the rim of the cross section. Given the equation let x 5 2 and y 5 1 evaluate solve for a
1
y 5 ax 2 1 5 a ? 22 15a?4 1 a5 4
In 2004 the La Trame documentary film studio released Bon Appétit, Monsieur Soleil, a film about solar cooking in Burkina Faso. The studio has produced several versions in different languages, including English. The film covers the essential points of solar cooking, the acceptance of these devices by the population, the advantages achieved by using them (savings in money and time), and their contribution to fighting deforestation. The film is a good tool to inform people living in countries with energy problems about other ways to cook. It also serves to introduce this cooking method to decision makers in government agencies and NGOs.
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So the formula for the parabolic oven cross section is y 5 A14B x 2. The focal point is at
) 4a1 ) 5 ) 4 ?1A B ) 5 11 or 1 foot above the vertex, the bottom of the cooker. Since the height 1 4
of the cooker is also 1 foot, then the pot should be suspended 1 foot above the center, at the level of the rim, for maximum efficiency.
The General Quadratic A quadratic function may include a linear term added to the ax2 term. The general quadratic in standard or a-b-c form is y 5 ax 2 1 bx 1 c
sa 2 0d
where a, b, and c are constants. A quadratic is also called a polynomial of degree 2. For example, the quadratic function y 5 3x 2 1 7 is a polynomial of degree 2 where a 5 3, b 5 0, and c 5 7. EXAMPLE
3
A quadratic model for tuition revenue Because of state financial problems, many publicly funded colleges are faced with large budget cuts. To raise more revenue, one state college plans to raise both tuition and student body size over the next 10 years. Its goal is to increase the current tuition of $6000 by $600 a year and increase the student population of 1200 by 40 students a year. a. Generate two equations, one to describe the projected increases in tuition and the other to describe increases in student body size over time. b. Generate an equation to describe the projected total tuition revenue over time. c. Compare the current tuition revenue to the projected revenue in 10 years.
SOLUTION
a. If we let S 5 student body size and N 5 number of years after the present, then S 5 1200 1 40N. If we let T 5 cost of tuition in thousands of dollars, then T 5 6 1 0.6N. b. The total tuition revenue, R, is the product of the number of students, S, times the cost of tuition, T. So R5 5 5 5
Substitute multiply out2 combine terms
S?T (1200 1 40N)(6 1 0.6N) 7200 1 720N 1 240N 1 24N 2 7200 1 960N 1 24N 2
So the total revenue R (in thousands of dollars) from tuition is given by the quadratic function R 5 7200 1 960N 1 24N 2 c. At the present time N 5 0, so the total tuition revenue R 5 $7200 thousand or $7.2 million. In 10 years, N 5 10, so R 5 7200 1 (960 ? 10) 1 (24 ? 102 ) 5 7200 1 9600 1 2400 5 19,200 So in 10 years the projected total tuition revenue is $19,200 thousand or $19.2 million. 2 Recall that the product sa 1 bd sc 1 dd is the sum of four terms: the product of the first (F) two terms, the outside (O) terms, the inside (I) terms, and the last (L) two terms of the factors. F L
sa 1 bdsc 1 dd 5 ac 1 ad 1 bc 1 bd I O
F
O
I
L
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Properties of Quadratic Functions The vertex and the axis of symmetry The graph of the general quadratic, y 5 ax 2 1 bx 1 c, is a parabola that is symmetric about its axis of symmetry. For any quadratic other than the simplest (of the form y 5 ax 2), the axis of symmetry is not the vertical axis. The vertex lies on the axis of symmetry and is a maximum (if the parabola is concave down) or a minimum (if the parabola is concave up) (see Figure 8.5). In Section 8.2 we’ll learn how to find the coordinates of the vertex. y Axis of symmetry
x
Vertex
Figure 8.5 Each parabola has a vertex that lies on an axis of symmetry.
Vertical and horizontal intercepts The general properties of vertical and horizontal intercepts are the same for all functions. Intercepts of a Function For any function ƒ(x): the vertical intercept is at ƒ(0) the horizontal intercepts occur at values of x, called zeros, where ƒ(x) 5 0 For a quadratic function ƒ(x) 5 ax 2 1 bx 1 c, we have ƒ(0) 5 c. So ƒ(x) has a vertical intercept at (0, c). We usually shorten this to say that the vertical intercept is at c, since we know the other coordinate is 0. Because of the ∪ shape of its graph, a quadratic function may have no, one, or two horizontal intercepts (see Figure 8.6). If (r, 0) are the coordinates of a horizontal intercept, we abbreviate this to say that the horizontal intercept is at r. These values of r are also called zeros of the function since f(r) 5 0. In Section 8.3 we’ll learn how to find the horizontal intercepts. y
y
x
(a) No x-intercepts
y
x
(b) One x-intercept
x
(c) Two x-intercepts
Figure 8.6 Graphs of quadratic functions showing the three
possible cases for the number of horizontal or x-intercepts.
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Since a parabola is symmetric about its axis of symmetry, each point on one arm of the parabola has a mirror image on the other arm. In particular, if there are two horizontal intercepts at r1 and r2, then (r1, 0) and (r2, 0) are mirror images across the parabola’s axis of symmetry. The horizontal intercepts will lie at an equal distance, d, to the left and right of the axis of symmetry (see Figure 8.7). Equivalently, the axis of symmetry crosses the horizontal axis halfway between r1 and r2, at the one-dimensional point s r1 1 r2 d . So the equation of the axis of symmetry is the two-dimensional line 2 s r1 1 r2 d . 2
where x is x5
y
x-intercepts y-intercept d r1
d r2
x
Axis of symmetry
Figure 8.7 Each parabola has a
vertical intercept and zero, one, or two horizontal intercepts.
?
SOMETHING TO THINK ABOUT
Does increasing ua u bring the focus closer to or farther away from the vertex?
The focal point 1 The focal point for the general parabola y 5 ax 2 1 bx 1 c is still u 4a u units above (or 1 below) the vertex. The distance u 4a u from the vertex to the focal point is called the focal length.
The Quadratic Function and Its Graph A quadratic function can be written in standard form as ƒsxd 5 ax2 1 bx 1 c
(where a 2 0)
Its graph: is called a parabola is symmetric about its axis of symmetry has a minimum (if the parabola is concave up) or a maximum (if the parabola is concave down) point called its vertex has one vertical intercept, but may have zero, one, or two horizontal intercepts 1 u units above the minimum (or below the maximum); has a focal point u 4a 1 u is called the focal length. the value u 4a
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Estimating the Vertex and Horizontal Intercepts Quadratics arise naturally in area and motion problems. The earliest problems we know of that led to quadratic equations are on Babylonian tablets dating from 1700 B.C. The writings suggest a problem similar to the following example.
EXAMPLE
4
SOLUTION
Maximizing area What is the maximum rectangular area you can enclose within a fixed perimeter of 24 meters? What are the dimensions of the rectangle with the maximum area? If the rectangular region has length L and width W, then 2L 1 2W 2L 1 2W L1W W
Substitute divide by 2 subtract L from both sides
5 5 5 5
perimeter 24 12 12 2 L
A5L?W 5 L ? (12 2 L) 5 12L 2 L2
Since area, A, is given by substitute for W multiply through
So area, A, is a quadratic function of L. Table 8.1 shows values of L from 0 to 12 meters and corresponding values for W and A. Figure 8.8 shows a graph of area, A, versus length, L. In the equation A 5 12L 2 L2 the coefficient of L2 is negative (21), so the parabola is concave down. From the table and the graph, it appears that the vertex of the parabola is at (6, 36); that is, at a length of 6 meters the area reaches a maximum of 36 square meters. Since W 5 12 2 L, when L 5 6, W 5 6. Hence the rectangle has maximum area when the length equals the width, or in other words when the rectangle is a square.
0 1 2 3 4 5 6 7 8 9 10 11 12 Table 8.1
EXAMPLE
5
Width, W (m)
Area, A (m2)
12 11 10 9 8 7 6 5 4 3 2 1 0
0 11 20 27 32 35 36 35 32 27 20 11 0
40 35 30 Area (m2)
Length, L (m)
25 20 15 10 5 0 0
5
10
15
Length (m)
Figure 8.8 Graph of area vs. length for a rectangle.
The trajectory of a projectile Figure 8.9 shows a plot of the height above the ground (in feet) of a projectile for the first 5 seconds of its trajectory. How could we: a. Estimate the height from which the projectile was launched? b. Estimate when the projectile will hit the ground?
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Height (ft)
180 160 140 120 100 80 60 40 20
t 0.0
1.0
2.0
3.0 4.0 Time (sec)
5.0
6.0
7.0
Figure 8.9 The height of a projectile. SOLUTION
a. The initial height of the projectile corresponds to the vertical intercept of the parabola (where time t 5 0). A quick look at Figure 8.9 gives us an estimate somewhere between 10 and 20 feet. If we use technology to generate a best-fit quadratic, we get H(t) 5 216t 2 1 96t 1 15, where t 5 time (in seconds) and H(t) 5 height (in feet). Since H(0) 5 15, then 15 is the vertical intercept. In other words, when t 5 0 seconds, the initial height is H(0) 5 15 feet. b. Figure 8.10 overlays on Figure 8.9 the graph of H(t) 5 216t 2 1 96t 1 15, our function model for the projectile’s path. The projectile will hit the ground when the height above the ground H(t) 5 0. This occurs at a horizontal intercept of the parabola. One horizontal intercept (not shown) would occur at a negative value of time, t, which would be meaningless here. From the graph we can estimate that the other intercept occurs at t < 6.2 seconds.
Height (ft)
“The Mathematics of Motion” following this chapter offers an extended exploration into constructing mathematical models to describe the motion of freely falling bodies.
180 160 140 120 100 80 60 40
H(t) = –16t 2 + 96t + 15
20
t 0
1
2
3
4
5
6
7
Time (sec)
Figure 8.10 A graph of the function model for the projectile’s path.
Algebra Aerobics 8.1 1. Find the coordinates of the vertex and the focal point for each quadratic function. Then specify whether each vertex is a maximum or minimum. 1 2 a. y 5 3x 2 c. y 5 24 x b. y 5 26x 2
y
10'
1 2 d. y 5 212 x
2. A designer is planning an outdoor concert place for solo performers. The stage area is to have a parabolic wall 30r wide by 10r deep. If the performer stands at the focal point, the sound will be reflected out in parallel sound waves directly to the audience in front. See Figure 8.11.
x 30'
Figure 8.11 A parabolic design for an outdoor concert stage back wall.
a. Where would the focal point be? b. What is the equation for the reflecting wall?
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c. Do you see any problem with this idea for concert acoustics? (Hint: What if the audience is noisy?) 3. Evaluate each of the following quadratic functions at 2, 22, 0, and z. a. g(x) 5 x 2 d. m(s) 5 5 1 2s 2 3s 2 b. h(w) 5 2w 2 e. D(r) 5 2(r 2 3)2 1 4 2 c. Q(t) 5 2t 23t 1 1 f. k(x) 5 5 2 x 2 4. For each of the graphs of the quadratic functions f(x), g(x), h(x), and j(x) given below, determine: i. If the parabola is concave up or concave down ii. If the parabola has a maximum or a minimum iii. The equation of the axis of symmetry iv. The coordinates of the vertex v. Estimated coordinates of the horizontal and vertical intercepts
5. A manufacturer wants to make a camp stove using a parabolic reflector to concentrate sun rays at the focal point of the parabola, where the food to be cooked would be placed. The reflector must be no wider than 24s , and the depth is to be the same as the focal length, f, the distance from the focal point to the bottom. See Figure 8.12. y (12, d ) Focus
f
Depth
x 24"
Figure 8.12 A parabolic design for
a camp stove.
y
y
461
5
5
x
x –5
–5
5
7
–5
–5
a. Graph of f ( x)
c. Graph of h( x)
a. What focal length would the camp stove need to 1 u ). have? (Hint: The focal length is u 4a b. What formula is needed for manufacturing the stove reflector? 6. Put the following quadratic functions into standard a-b-c form:
y
y 6
5
ƒ(x) 5 (x 2 3)(x 1 5)
g(x) 5 (2x 1 5)(x 1 1)
h(x) 5 10(x 2 3)(x 2 5)
j(x) 5 2(x 2 3)(x 1 3)
x –10
10
x –5
7
–5 –16
b . Graph of g( x)
d . Graph of k( x)
Exercises for Section 8.1 The “Extended Exploration: The Mathematics of Motion,” which follows this chapter, contains many additional exercises using quadratics to describe freely falling and thrown bodies.
y
a.
–5
A graphing program is recommended or required for several exercises. 1. From the graph of each quadratic function, identify whether the parabola is concave up or down and hence whether the function has a maximum or minimum. Then estimate the vertex, the axis of symmetry, and any horizontal and vertical intercepts.
b.
5
5
10
c.
y
10
y
x
x –8
12
x –12
–10
–5
6 –10
2. Using the quadratic function y 5 18 x 2: a. Fill in the values in the table. x y
212
28
24
0
4
8
12
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b. Plot the points by hand and sketch the graph of the function. c. Determine the coordinates of the focal point and place the focal point on the graph. d. What is the equation of the axis of symmetry? e. What are the coordinates of the vertex? Is the vertex a maximum or minimum? 3. (Graphing program recommended.) On the same graph, plot the three functions. y1 5 x 2
y2 5 41 x 2
1 2 y3 5 12 x
a. Calculate the focal point for each parabola. b. Compare y2 and y3 with y1. c. Complete this sentence: As the value of k a k gets smaller, the focal point gets ________ the vertex and the graph gets ________.
a. What is the equation for the cross section of the parabolic dish? b. What is the focal length (the distance between the vertex and the focal point)? c. What is the diameter of the dish at the focal point? 11. An electric heater is designed as a parabolic reflector that is 5s deep. To prevent accidental burns, the centerline of the heating element, placed at the focus, must be set in 1.5s from the base of the reflector. a. What equation could you use to design the reflector? (Hint: Tip the parabolic reflector so it opens upward.) b. How wide would the reflector be at its rim?
4. (Graphing program recommended.) On the same graph, plot the three functions. y1 5 x 2
y2 5 4x 2
y3 5 8x 2
5"
a. Calculate the focal point for each parabola. b. Compare y3 and y2 with y1. c. Complete this sentence: As the value of k a k gets larger, the focal point gets ________ to the vertex and the graph of the parabola gets ________ . 5. (Graphing program optional.) Given a point P on a parabola of the form y 5 ax2, find the equation of the parabola. If available, use technology to verify your equations by plotting each parabola. a. P 5 (12, 6) c. P 5 (4, 12) b. P 5 (212, 26) d. P 5 (3, 144) 6. Find the coordinates of the vertex for each quadratic function listed. Then specify whether each vertex is a maximum or minimum. 1 2 c. Psnd 5 A12 Bn a. y 5 4x2 1 2 d. Qstd 5 2 A24 Bt b. ƒ(x) 5 28x2
1.5"
12. A parabolic reflector 3s in diameter and 2s deep is proposed for a spotlight.
0.25"
2"
3"
7. Given the following focal points, write the equation of a parabola in the form y 5 ax2 by finding a. 1 B a. (0, 4) c. A0, 16 1 B b. (0, 28) d. A0, 224
a. What formula is needed for manufacturing the reflector? b. Where would the focus be? c. Will a 14 s -diameter light source fit if it is centered at the focus?
8. Find the focal length (the distance from the focal point to the vertex) for each of the following. b. ƒ(x) 5 3x2 c. ƒ(x) 5 13 x 2 a. ƒ(x) 5 x2
13. A slimline fluorescent bulb 12 s in diameter needs 1s clearance top and bottom in a parabolic reflecting shade. a. What are the coordinates of the focus for this parabola? b. What is the equation for the parabolic curve of the reflector? c. What is the diameter of the opening of the shade?
9. For each of the following functions, evaluate ƒ(2) and ƒ(22). a. ƒ(x) 5 x2 2 5x 2 2 b. ƒ(x) 5 3x2 2x c. ƒ(x) 5 2x2 1 4x 2 2 10. A designer proposes a parabolic satellite dish 5 feet in diameter and 15 inches deep.
1" 1/2"
15" 1"
5' (or 60")
Diameter
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14. If we know the radius and depth of a parabolic reflector, we also know where the focus is. a. Find a generic formula for the focal length ƒ of a parabolic reflector expressed in terms of its radius R and depth D. 1 The focal length @ 4a @ is the distance between the vertex and the focal point. Assume a . 0. b. Under what conditions does ƒ 5 D?
463
21. Which of the following are true statements for quadratic functions? a. The vertex and focal point always lie on the axis of symmetry. b. The graph of a parabola could have three horizontal intercepts. c. The graph of a parabola does not necessarily have a vertical intercept. d. If ƒ(2) 5 0, then ƒ has a horizontal intercept at 2.
Focus
e. The focal point always lies above the vertex. f
D
R
15. Construct several of your own equations of the form y 5 ax 2 and then describe in words how the focal length varies depending on how open or closed the parabolic curve is. 16. For each of the following quadratics with their respective vertices, calculate the distance from the vertex to the focal point. Then determine the coordinates of the focal point. a. ƒ(x) 5 x 2 2 2x 2 3 with vertex at (1, 24) b. g(t) 5 2t 2 2 16t 1 24 with vertex at (4, 28) 17. Put each of the following quadratics into standard form. a. ƒ(x) 5 (x 1 3)(x 2 1) b. P(t) 5 (t 2 5)(t 1 2) c. H(z) 5 (2 1 z)(1 2 z) 18. Put each of the following quadratics into standard form. a. g(x) 5 (2x 2 1)(x 1 3) b. h(r) 5 (5r 1 2)(2r 1 5) c. R(t) 5 (5 2 2t)(3 2 4t) 19. Determine the dimensions for enclosing the maximum area of a rectangle if: a. The perimeter is held constant at 200 meters. b. The perimeter is held constant at P meters. 20. A gardener wants to grow carrots along the side of her house. To protect the carrots from wild rabbits, the plot must be enclosed by a wire fence. The gardener wants to use 16 feet of fence material left over from a previous project. Assuming that she constructs a rectangular plot, using the side of her house as one edge, estimate the area of the largest plot she can construct.
22. The management of a company is negotiating with a union over salary increases for the company’s employees for the next 5 years. One plan under consideration gives each worker a bonus of $1500 per year. The company currently employs 1025 workers and pays them an average salary of $30,000 a year. It also plans to increase its workforce by 20 workers a year. a. Construct a function C(t) that models the projected cost of this plan (in dollars) as a function of time t (in years). b. What will the annual cost be in 5 years? 23. (Graphing program required for part (c).) A landlady currently rents each of her 50 apartments for $1250 per month. She estimates that for each $100 increase in rent, two additional apartments will remain vacant. a. Construct a function that represents the revenue R(n) as a function of the number of rent increases, n. (Hint: Find the rent per unit after n increases and the number of units rented after n increases.) b. After how many rent increases will all the apartments be empty? What is a reasonable domain for this function? c. Using technology, plot the function. From the graph, estimate the maximum revenue. Then estimate the number of rent increases that would give you the maximum revenue. 24. (Graphing program required for part (c).) a. In economics, revenue R is defined as the amount of money derived from the sale of a product and is equal to the number x of units sold times the selling price p of each unit. What is the equation for revenue? b. If the selling price is given by the equation 1 p 5 210 x 1 20, express revenue R as a function of the number x of units sold. c. Using technology, plot the function and estimate the number of units that need to be sold to achieve maximum revenue. Then estimate the maximum revenue.
8.2 Finding the Vertex: Transformations of y = x 2 In this section we show that every quadratic can be generated through transformations of the basic quadratic y 5 x2. We start by investigating what happens to the equation of a function when we change its graph.
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Stretching and Compressing Vertically When we stretch or compress a graph we change its shape, creating a graph of a new function. To find the equation of this new function, we need to ask, “What is the relationship between the graphs of the new function and the old function?” We can build on what we know from Chapter 7, where we learned that multiplying a power function by a constant stretches or compresses its graph. Specifically, for power functions in the form y 5 kx p, the value of the coefficient k compresses or stretches the graph of y 5 x p. A quadratic in the form y 5 ax 2 is a power function, so the graph of y 5 ax 2 is the graph of y 5 x 2 vertically stretched by a factor of a, if a . 1 vertically compressed by a factor of a, if 0 , a , 1 The magnitude (or absolute value) of a tells us how much the graph of y 5 x2 is stretched or compressed. As k a k increases, a acts as a vertical stretch factor, which pulls harder and harder on the arms of the parabola anchored at the vertex, narrowing the graph. As k a k decreases, a acts as a compression factor, which weighs down the graph, flattening the parabola’s arms. So the value of a determines the shape of the parabola. See Figure 8.13. 5x 2 2x 2
y 10
x2 0.5x 2 0.25x 2
x –4
4 –4
Figure 8.13 The graph of y 5 x 2 vertically
compressed and stretched.
Reflections across the Horizontal Axis If a , 0, the graph of y 5 ax 2 is the graph of y 5 x 2 stretched or compressed by a factor of k a k and then reflected across the x-axis. In general, the graphs of y 5 ax 2 and y 5 2ax 2 are reflections of each other across the x-axis. See Figure 8.14. The sign of a tells us whether the parabola opens up or down and if the vertex represents a maximum or a minimum value of the function. If a is positive, the parabola is concave up (opens upward) and the vertex is a minimum. If a is negative, the parabola is concave down (opens downward) and the vertex is a maximum. y
y = 5x 2
10
y = 0.5x 2
x –4
4
–10
y = –5x 2
y = –0.5x 2
Figure 8.14 Graphs that are reflected across
the x-axis.
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1
Describe how the following pairs of functions are related to each other and to y 5 x 2. Sketch each pair and y 5 x 2 on the same grid. a. y 5 3x 2 and y 5 23x 2 b. y 5 0.3x 2 and y 5 20.3x 2
SOLUTION
a. The functions y 5 3x 2 and y 5 23x 2 are reflections of each other across the x-axis. For both functions, the absolute value of the coefficient a of x 2 is 3, so in each case k a k . 1. Hence both graphs are the graph of y 5 x 2 stretched by a factor of 3. See Graph A in Figure 8.15. b. The functions y 5 0.3x2 and y 5 20.3x2 are also reflections of each other across the x-axis. For both functions k a k 5 0.3, so in each case 0 , k a k , 1. Hence both of their graphs are the graph of y 5 x 2 compressed by a factor of 0.3. See Graph B in Figure 8.15.
EXAMPLE
y
y = 3x 2
y
y = x2
y = x2
10
10
y = 0.3x 2
x
x –4
4
–4
4
y = – 0.3x 2 –10
y = –3x 2
Graph A
–10 Graph B
Figure 8.15 Stretching and compressing y 5 x 2.
Stretching, Compressing, and Reflecting the Graph of Any Function If ƒ is a function and a is a constant, then the graph of a ? ƒ(x) is the graph of ƒ(x) vertically stretched by a factor of a, if a . 1 vertically compressed by a factor of a, if 0 , a , 1 vertically stretched or compressed by a factor of k a k and reflected across x-axis, if a , 0
Shifting Vertically and Horizontally What happens when we keep the shape of the graph of y 5 ax 2 but change its position on the grid, shifting the graph vertically or horizontally? Clearly we will get a graph of a new function where the coordinates of the vertex are no longer (0, 0). Let’s see how to construct the equation of the new function from the original function. Shifting a graph vertically How does a function change when its graph is shifted up or down? Examine the graphs in Figure 8.16, Graphs A and B.
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y = ax2 + 2 y
y
5
y = ax2
5
y = ax2 – 2
y = f (x) + 2 y = f (x) x 5 y = f (x) – 2
–5
x 5
–5
–5
–5
Graph A
Graph B
Figure 8.16 Graphs that are shifted vertically.
What happens to the output value, y, when the graph of y 5 ƒ(x) is shifted up two units? Every y value increases by two units. Translating this shift into equation form, our new function is y 5 ƒ(x) 1 2. (See Figure 8.16, Graph A.) What happens when the graph of y 5 ƒ(x) is shifted down two units? Our new function is of the form y 5 ƒ(x) 2 2. In general, if the vertical shift is k units, then y 5 ƒ(x) 1 k is the graph of y 5 ƒ(x) shifted up by k units if k is positive and shifted down by k units if k is negative. Figure 8.16, Graph B shows the graph of ƒ(x) 5 ax 2 raised or lowered by two units. If the vertical shift is k units, then the graph of y 5 ax 2 1 k is the graph of y 5 ax 2 shifted up by k units if k is positive and shifted down by k units if k is negative. The vertex is at (0, k).
Algebra Aerobics 8.2a In Problems 125, without drawing the graphs, compare the graph of part (b) to the graph of part (a). 1. a. r(x) 5 x 2 1 2 b. s(x) 5 2x 2 1 2 2 2. a. h(t) 5 t 1 5 b. k(t) 5 2t 2 1 5 3. a. ƒ(z) 5 25z 2 b. g(z) 5 20.5z 2 2 4. a. ƒ(x) 5 x 1 3x 1 2 b. g(x) 5 x 2 1 3x 1 8 5. a. ƒ(t) 5 23t 2 1 t 2 5 b. g(t) 5 23t 2 1 t 2 2 6. Create a quadratic equation of the form y 5 ax 2 1 k with the given values for a and the vertex. Sketch by hand the graph of each equation. a. a 5 3 and the vertex is at (0, 5). b. a 5 13 and the vertex is at (0, 22). c. a 5 22 and the vertex is at (0, 4)
7. Create new functions by performing the following transformations on ƒ(x) 5 x 2. a. g(x) is ƒ(x) stretched by a factor of 3. b. h(x) is ƒ(x) stretched by a factor of 5 and reflected across the x-axis. c. j(x) is ƒ(x) compressed by a factor of 1/2. d. k(x) is ƒ(x) reflected across the x-axis. 8. Create a function in the form y 5 ax 2 1 k for each of the following transformations of ƒ(x) 5 x 2: a. Stretched by a factor of 5 and shifted down 2 units b. Concave up and shifted 3 units up c. Multiplied by a factor of 0.5, concave down, and shifted 4.7 units down d. Opens up and is ƒ(x) shifted 71 units down
Shifting a graph horizontally How does a function change when its graph is shifted to the right or left? Examine the graphs in Figure 8.17, Graphs A and B.
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y y (3, y2)
(5, y2)
Original f (x) (1, y1)
(3, y2)
Original New parabola f (x) parabola
(2, y1)
New
(3, y1)
(5, y2)
(4, y1)
x 0
1
2
3
4
5
6
7
8
x –5
(0, 0)
Graph A
(2, 0)
5
Graph B
Figure 8.17 Graphs that are shifted horizontally to the right.
When we shift the graph of y 5 ƒ(x) two units to the right, how is the new function related to our original function? Recall that adding a constant to the output value of a function shifts its graph vertically. When we shift the graph of ƒ(x) horizontally, the input value, x, is shifted. The dotted lines in Figure 8.17 show that for any particular output value, y, the corresponding input value for the new function is (x 2 2). So the graph of ƒ(x) has been shifted to the right two units. Translating this shift into equation form, we get y 5 ƒ(x 2 2) At first this may seem counterintuitive. But remember, we are expressing the new function in terms of the original function. If you evaluate the new function at x, the same y-value of the original function ƒ(x) is now at ƒ(x 2 2). For example, in Figure 8.17 if you evaluate each of the new functions in Graph A and Graph B at x 5 5, you get an output of y 2. The same output for the original function occurs at 5 2 2 5 3. What if we shift y 5 ƒ(x) to the left two units? We need to think about the effect of the shift and the direction of the shift. If you evaluate the new function at x, the comparable point on the graph of ƒ(x) is ƒ(x 1 2). We can think of this shift as replacing x with x 2 (22) or x 1 2, so our new function is y 5 ƒ(x 1 2) Figure 8.18 shows the graph of y 5 3x 2 after being shifted to the right and then to the left 2 units. Note that the value of the x-coordinate of the vertex changed by 62 units. If the horizontal shift is h units, then y 5 a(x 2 h) 2 is the graph of y 5 ax 2 shifted right by h if h is positive, and shifted left by h if h is negative. The vertex is now at (h, 0). Note that if h 5 2, the expression (x 2 h) becomes x 2 2, and if h 5 22, the expression (x 2 h) becomes x 2 (22) 5 x 1 2.
y = 3(x + 2)2
5
y y = 3x2
y = 3(x – 2)2
x –5
0
5
Figure 8.18 Graph of y 5 3x 2 shifted horizontally
to the left and to the right two units.
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EXAMPLE
2
a. Identify the vertex for each of the following functions and indicate whether it represents a maximum or minimum value: g(x) 5 5(x 2 3)2 h(x) 5 25(x 1 3) 2
j(x) 5 5x 2 1 3 k(x) 5 25x 2 2 3
b. Describe how the graphs in part (a) are related to the graph of ƒ(x) 5 5x 2. Specify the order of any transformations. SOLUTION
a. Vertex for: g is at (3, 0); j is at (0, 3); h is at (23, 0); k is at (0, 23). Vertices for g and j represent minimum values since in both cases the coefficient for x2 is positive (a 5 5). Vertices for h and k represent maximum values since in both cases the coefficient for x2 is negative (a 5 25). b. The graph of ƒ(x) 5 5x 2 shifted horizontally to the right three units is the graph of g(x) shifted vertically up three units is the graph of j(x) shifted horizontally to the left three units and then reflected across the x-axis is the graph of h(x) reflected across the x-axis and then shifted vertically down three units is the graph of k(x) We can generalize to any function ƒ(x). Horizontal and Vertical Shifts The graph of ƒ(x) 1 k is the graph of ƒ(x) shifted vertically k k k units. If k is positive, the shift is up; if k is negative, the shift is down. The graph of ƒ(x 2 h) is the graph of ƒ(x) shifted horizontally k h k units. If h is positive, the shift is to the right; if h is negative, the shift is to the left.
Using Transformations to Get the Vertex Form We can use the previous transformations to generate a quadratic function in what is called the vertex form. We start by stretching or compressing the graph of y 5 x2 by a factor of a. So our new function will be y 5 ax 2 “Q3: a, h, k Sliders,” “Q8: y 5 ax 2 vs. y 5 a(x – h)2 + k,” and “Q9: Finding 3 Points: a-h-k Form” in Quadratic Functions can help you visualize quadratic functions in the a-h-k form.
Next we shift the graph of the function y 5 ax2 horizontally h units and vertically k units, to get y 5 a(x 2 h)2 1 k The vertex is now at (h, k). This quadratic is in the vertex or a-h-k form. Its axis of symmetry is the vertical line at x 5 h. See Figure 8.19. x = –2
x=2 y 5
y1
(2, 5)
y2 x 5
–5
(–2, –4)
–5
Figure 8.19 Graphs of y1 5 3(x 1 2)2 2 4 and
y2 5 23(x 2 2)2 1 5.
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The Vertex Form of a Quadratic Function The vertex or a-h-k form of the quadratic function is ƒ(x) 5 a(x 2 h)2 1 k where the vertex is at (h, k).
EXAMPLE
3
Transforming graphs Given ƒ(x) 5 x 2 and g(x) 5 22(x 1 3)2 1 5, show how to transform the graph of ƒ(x) into the graph of g(x).
SOLUTION
One way to transform the graph of ƒ(x) into the graph of g(x) is shown in Figure 8.20.
y
y
10
y
10
10
f (x) = x 2 f (x + 3) = (x + 3)2
–2f (x + 3) + 5 = –2(x + 3)2 + 5 or g(x)
x –6
x –6
6
–2f (x + 3) = –2(x + 3)2
–10
6
x –6
6
–10
Graph A
–10
Graph B
Graph C
Figure 8.20 The graph of ƒ(x) 5 x in A is shifted horizontally to the left three units, then in B stretched vertically by a factor of 2 and reflected across the x-axis, and finally in C shifted up vertically five units to generate g(x) 5 22(x 1 3)2 1 5. 2
EXAMPLE
4
Finding the function from its graph Figure 8.21 shows the graph of ƒ(x) 5 2x 2 transformed into three new parabolas. Assume that each of the three new graphs retains the shape of ƒ(x). a. Estimate the coordinates of the vertex for each new parabola. b. Use your estimates from part (a) to write equations for each parabola in Figure 8.21. y 10
g(x)
f (x)
x –5
5
h(x) k(x) –10
Figure 8.21 Three transformations of
ƒ(x) 5 2x 2 SOLUTION
a. Vertex for: g(x) is at (3, 4); h(x) is at (3, 21); k(x) is at (21, 24) b. g(x) 5 2(x 2 3) 2 1 4; h(x) 5 22(x 2 3) 2 2 1; k(x) 5 22(x 1 1) 2 2 4
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EXAMPLE
5
SOLUTION
Identifying the vertex For the following functions, identify the coordinates of the vertex and specify whether the vertex represents a maximum or minimum. a. y 5 25x 2 b. y 5 2(x 2 3) 2 2 15 c. y 5 23(x 1 5) 2 1 10 a. The vertex is at (0, 0) and represents a maximum. b. The vertex is at (3, 215) and represents a minimum. c. The vertex is at (25, 10) and represents a maximum.
Algebra Aerobics 8.2b For Problems 1 to 3, without graphing, compare the positions of the vertices of parts (b) and (c) to that of part (a). c. y 5 (x 2 2)2 1. a. y 5 x 2 b. y 5 (x 1 3)2 2. a. ƒ(x) 5 0.5x 2 c. ƒ(x) 5 0.5(x 1 4)2 b. ƒ(x) 5 0.5(x 2 1)2 3. a. r 5 22t 2 c. r 5 22(t 2 0.9)2 b. r 5 22(t 1 1.2)2 For Problems 4 and 5, without graphing, compare the graphs of parts (b), (c), and (d) to that of part (a). 4. a. y 5 x 2 c. y 5 (x 2 2)2 1 4 b. y 5 (x 2 2)2 d. y 5 (x 2 2)2 2 3 2 5. a. y 5 2x c. y 5 2(x 1 3)2 2 1 b. y 5 2(x 1 3)2 d. y 5 2(x 1 3)2 1 4
6. Create new functions by performing the following transformations on ƒ(x) 5 x 2. Give the coordinates of the vertex for each new parabola. a. g(x) is ƒ(x) shifted right 2 units, stretched by a factor of 3, and then shifted down by 1 unit. b. h(x) is ƒ(x) shifted left 3 units, stretched by a factor of 2, then reflected across the x-axis, and finally shifted up by 5 units. c. j(x) is ƒ(x) shifted left 4 units, compressed by a factor of 5, and then shifted down by 3.5 units. d. k(x) is ƒ(x) shifted right 1 unit, reflected across the x-axis, and then shifted up by 4 units. 7. Give the coordinates of the vertex for each of the following functions and indicate whether the vertex represents a maximum or minimum. a. y 5 2(x 2 3)2 2 4 c. y 5 20.5(x 2 4)2 2 b. y 5 23(x 1 1) 1 5 d. y 5 23x 2 2 7
Finding the Vertex from the Standard Form What if our quadratic function is in the standard a-b-c form and we want to find the vertex of its graph? We can use a formula to find the coordinates of the vertex or, as we show in the next section, we can convert the quadratic from the standard to vertex form, in which the vertex is easy to identify. Using a formula to find the vertex The following formula can be used to find the coordinates of the vertex of a parabola when the function is in standard form.
See the reading “Why the Formula for the Vertex and the Quadratic Formula Work” for a derivation of this vertex formula.
Formula for Finding the Vertex from the Standard Form The vertex of a quadratic function in the form ƒ(x) 5 ax 2 1 bx 1 c b b has coordinates A22a , ƒ A22a BB
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EXAMPLE
6
SOLUTION
471
Finding the vertex Find the vertex and sketch the graph of ƒ(x) 5 x 2 2 10x 1 100. The function ƒ is in standard form, ƒ(x) 5 ax 2 1 bx 1 c. So a 5 1, b 5 210, and c 5 100. Since a is positive, the graph is concave up, so the vertex represents a minimum. Using the formula for the horizontal coordinate of the vertex, we have (210) 10 b 52 5 55 2a 2(1) 2 To find the vertical coordinate of the vertex, we need to find the value of ƒ(x) when x 5 5. x52
ƒ(x) 5 x 2 2 10x 1 100 ƒ(5) 5 52 2 10(5) 1 100 5 75
Given let x 5 5 simplify
The coordinates of the vertex are (5, 75) and the y-intercept is at ƒ(0) 5 100. Figure 8.22 shows a sketch of the graph. y 200
100 (5, 75) Vertex
x –20
–10
0
10
20
Figure 8.22 Graph of
ƒ(x) 5 x 2 2 10x 1 100. EXAMPLE
7
Measuring traffic flow Urban planners and highway designers are interested in maximizing the number of cars that pass along a section of roadway in a certain amount of time. Observations indicate that the primary variable controlling traffic flow (and hence a good choice for the independent variable) is the density of cars on the roadway: the closer each driver is to the car ahead, the more slowly he or she drives. The following quadratic relationship between traffic flow rate and density of cars was derived from observing traffic patterns in the Lincoln Tunnel, which connects New York and New Jersey: t 5 20.21d 2 1 34.66d where t 5 traffic flow rate (cars/hour) and d 5 density of cars (cars/mile). a. Find the coordinates of the vertex of the parabola. b. What does the vertex represent in terms of the traffic flow rate? c. Graph the function and estimate the value of the horizontal intercepts. d. What do the horizontal intercepts mean in terms of the traffic flow and density of cars?
SOLUTION
a. Our equation t 5 20.21d 2 1 34.66d is in standard form (t 5 ad 2 1 bd 1 c), so we can use the formula for finding the coordinates of the vertex. Since a 5 20.21, b 5 34.66, and c 5 0, the horizontal coordinate of the vertex is 2 simplify round to the nearest integer
b 234.66 5 2a 2(20.21) < 82.52 < 83
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We can find the corresponding value of t by substituting d 5 83 into our equation: t 5 20.21d 2 1 34.66d t < 20.21(83)2 1 34.66(83) 5 1430.09 < 1430
Given substitute for d simplify round to the nearest integer
So the coordinates of the vertex are approximately (83, 1430). b. A vertex at (83, 1430) means that when the density is 83 cars/mile, the traffic flow rate reaches a maximum of 1430 cars/hour. When the density is above or below 83 cars/mile, the traffic flow rate is less than 1430 cars/hour. c. Figure 8.23 shows the graph of t 5 20.21d 2 1 34.66d. The horizontal intercepts are the values of d when t 5 0. From the graph we estimate that when t 5 0, then d is either 0 or about 165. t 1600 Traffic flow rate (cars per hour)
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(83, 1430)
1400 1200 1000 800 600 400 200 0
20
60
100
140
d
180
Density (cars per mile)
Figure 8.23 The quadratic relationship between traffic
flow rate and density. Source: Adapted from G. B. Whitman, Linear and Nonlinear Waves (New York: John Wiley, 1974), p. 68.
d. When the traffic flow rate is equal to zero, traffic is at a standstill. This happens when the density is either zero (when there are no cars) or approximately 165 cars per mile.
Converting between Standard and Vertex Forms Every quadratic can be written in either standard or vertex form. In this section we examine how to convert from one form to the other. Converting from a-h-k to a-b-c form Every function written in the vertex or a-h-k form can be rewritten as a function in the standard or a-b-c form if we multiply out and group terms with the same power of x. EXAMPLE
8
SOLUTION
Converting from vertex to standard form Rewrite the quadratic ƒ(x) 5 3(x 1 7) 2 29 in the a-b-c form. The function ƒ(x) is in the a-h-k form where a 5 3, h 5 27, and k 5 29. Given write out the factors multiply the factors distribute the 3 group the constant terms
ƒ(x) 5 3(x 1 7)2 2 9 5 3(x 1 7)(x 1 7) 2 9 5 3(x 2 1 14x 1 49) 2 9 5 3x 2 1 42x 1 147 2 9 5 3x 2 1 42x 1 138
This function is now in the a-b-c format with a 5 3, b 5 42, and c 5 138.
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For a graphic illustration of the shift from a-b-c to a-h-k form, see “Q7: From a-b-c to a-h-k Form” in Quadratic Functions.
473
Converting from a-b-c to a-h-k form Here we examine two strategies that can be used to convert from the standard or a-b-c form to the vertex or a-h-k form. Strategy 1: “Completing the Square.” We can convert the function ƒ(x) 5 x 2 1 14x 1 9 into a-h-k form using a method called completing the square. When a function is in a-h-k form, the term (x 2 h)2 is a perfect square; that is, (x 2 h)2 is the product of the expression x 2 h times itself. We can examine separately the expression x 2 1 14x and ask what constant term we would need to add to it in order to make it a perfect square. A perfect square is in the form (x 1 m)2 5 x 2 1 2mx 1 m 2 for some number m. Notice that the coefficient of x is two times the number m in our expression for a perfect square. So to turn x 2 1 14x into a perfect square (x 1 m)2, we need to find m. Since the coefficient of x corresponds to 2m, then 2m 5 14 1 m 5 7 1 m2 5 49. So if we add 49 to x 2 1 14x we have a perfect square. (x 1 7)2 5 x 2 1 14x 1 49 To translate ƒ(x) 5 x 2 1 14x 1 9 into the a-k-h form, we add 49 to make a perfect square, and then subtract 49 to preserve equality. So we have
add and subtract 49 regroup terms factor and simplify
ƒ(x) 5 x 2 1 14x 1 9 5 x 2 1 14x 1 (49 2 49) 1 9 5 (x 2 1 14x 1 49) 2 49 1 9 5 (x 1 7)2 2 40
Now ƒ(x) is in a-h-k form. The vertex is at (27, 240). EXAMPLE
9
SOLUTION
Converting from standard to vertex form Convert the function g(t) 5 22t 2 1 12t 2 23 to a-h-k form. This function is more difficult to convert by completing the square, since a is not 1. We first need to factor out 22 from the t terms only, getting g(t) 5 22(t 2 2 6t)223
A12
(1)
It is the expression t 2 2 6t for which we must complete the square. Since ? 6B 2 5 9, then adding 9 to t 2 2 6t gives us t 2 2 6t 1 9 5 (t 2 3)2
We must add the constant term 9 inside the parentheses in Equation (1) in order to make t 2 2 6t a perfect square. Since everything inside the parentheses is multiplied by 22, we have essentially subtracted 18 from our original function, so we need to add 18 outside the parentheses to preserve equality. We have Given factor out 22 from t terms add 9 inside parentheses and 18 outside parentheses factor and simplify
g(t) 5 22t 2 1 12t 2 23 5 22(t 2 2 6t) 2 23 5 22(t 2 2 6t 1 9) 1 18 2 23 5 22(t 2 3)2 2 5
We now have g(t) in a-h-k form. The vertex is at (3, 25).
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Strategy 2: Using the Formula for the Vertex. We could also convert g(x) 5 3x2 2 12x 1 5 to the a-h-k form by using the formula for the coordinates of the vertex. Since the coefficient a is the same in both the a-b-c and the a-h-k forms, we have a 5 3. The coordinates of the vertex of a quadratic ƒ(x) in the b a-b-c form are given by A22a , ƒ A2b 2a BB . For g(x) we have a 5 3, b 5 212, and c 5 5. So 2
b s212d 12 52 5 52 2a 2?3 6
and g(2) 5 (3 ? 22) 2 (12 ? 2) 1 5 5 12 2 24 1 5 5 27 The vertex is at (2, 27). In the a-h-k form the vertex is at (h, k), so h 5 2 and k 5 27. Substituting for a, h, and k in g(x) 5 a(x 2 h)2 1 k, we get g(x) 5 3(x 2 2)2 2 7 and the transformation is complete. EXAMPLE
10
Mystery parabola Find the equation of the parabola in Figure 8.24. Write it in a-h-k and in a-b-c form. y 12
–10
10
x
–4
Figure 8.24 A mystery
parabola. SOLUTION
The vertex of the graph appears to be at (2, 1) and the graph is concave up. We can substitute the coordinates of the vertex in the a-h-k form of the quadratic equation to get y 5 a(x 2 2)2 1 1
(1)
How can we find the value for a? If we can identify values for any other point (x, y) that lies on the parabola, we can substitute these values into Equation (1) and solve it for a. The y-intercept, estimated at (0, 3), is a convenient point to pick. Setting x 5 0 and y 5 3, we get
so
3 5 a(0 2 2)2 1 1 3 5 4a 1 1 2 5 4a a 5 0.5
The equation in the a-h-k form is y 5 0.5(x 2 2)2 1 1 If we wanted it in the equivalent a-b-c form, we could square, multiply, and collect like terms to get y 5 0.5x 2 2 2x 1 3
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The Vertex of a Quadratic Function A quadratic function in vertex form ƒ(x) 5 a(x 2 h)2 1 k
has a vertex at (h, k)
A quadratic function in standard form ƒ(x) 5 ax2 1 bx 1 c
?
SOMETHING TO THINK ABOUT
11
b b , ƒa2 b b 2a 2a
Whether in standard or vertex form, the value of the coefficient a determines if the parabola is relatively narrow or flat.
Using “Q1: a, b, c Sliders” in Quadratic Functions in the course software, can you describe the effect on the parabola of changing the value for b while you hold a and c fixed?
EXAMPLE
has a vertex at a2
If a . 0, the parabola is concave up and has a minimum at the vertex. If a , 0, the parabola is concave down and has a maximum at the vertex.
Match the following functions to the graphs in Figure 8.25. What allows you to match them easily? a. ƒ(x) 5 x 2 2 8x 1 18 c. h(x) 5 20.5x 2 1 4x 2 6 b. g(x) 5 2x 2 2 16x 1 34 d. j(x) 5 2x 2 1 8x 2 14 y A
B
x
C
D
Figure 8.25 Graphs of four quadratic functions. SOLUTION
All the graphs share the same vertex. Graphs A and B are concave up, so they could match with ƒ(x) or g(x), which both have a positive value of a (the coefficient of x 2). Graphs C and D are concave down, so they could match with h(x) or j(x), which both have a negative value of a. Graph A is narrower than B, so g(x) matches with graph A and ƒ(x) matches with graph B. Graph C is narrower than D, so j(x) matches with graph C and h(x) matches with graph D. Reasoning: The larger the absolute value of a, the narrower the graph.
Algebra Aerobics 8.2c Use of a graphing program is optional for Problems 3 and 13. 1. Find the vertex of the graph of each of the following quadratic functions: a. ƒ(x) 5 2x2 2 4 c. w 5 4t 2 1 1 2 b. g(z) 5 2z 1 6 2. Find the vertex of the graph of each of the following functions and then sketch the graphs on the same grid: a. y 5 x 2 1 3 b. y 5 2x 2 1 3
3. In parts (a) to (d) determine the vertex and whether the graph is concave up or down. Then predict the number of x-intercepts. Graph the function to confirm your answer. Use technology if available. a. ƒ(x) 5 23x 2 c. ƒ(x) 5 x 2 1 4x 2 7 b. ƒ(x) 5 22x 2 2 5 d. ƒ(x) 5 4 2 x 2 2x2
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4. Without drawing the graph, describe whether the graph of each of the following functions has a maximum or minimum at the vertex and is narrower or broader than y 5 x 2. a. y 5 2x 2 2 5 b. y 5 0.5x 2 1 2x 2 10 c. y 5 3 1 x 2 4x 2 d. y 5 20.2x 2 1 11x 1 8 5. For each of the following functions, use the formula b h 5 22a to find the horizontal coordinate of the vertex, and then find its vertical coordinate. Draw a rough sketch of each function. a. y 5 x 2 1 3x 1 2 b. ƒ(x) 5 2x 2 2 4x 1 5 c. g(t) 5 2t 2 2 4t 2 7 6. Find the coordinates of the vertex and the vertical intercept, and then sketch the function by hand. a. y 5 0.1(x 1 5)2 2 11 b. y 5 22(x 2 1)2 1 4 7. Rewrite each expression in the form (x 1 m)2 2 m2. a. x 2 1 6x c. x 2 2 30x 2 b. x 2 10x d. x 2 1 x 8. Convert the following functions to vertex form by completing the square. Identify the stretch factor and the vertex. a. ƒ(x) 5 x2 1 2x 2 1 b. j(z) 5 4z2 2 8z 2 6 c. h(x) 5 23x2 2 12x
d. h(t) 5 216t 2 1 96t e. h(t) 5 24.9t 2 2 98t 1 200 9. Express each of the following functions in standard form y 5 ax 2 1 bx 1 c. Identify the stretch or compression factor and the vertex. 2 a. y 5 2 Ax 2 12B 1 5 b. y 5 213 sx 1 2d 2 1 4 c. y 5 10(x 2 5)2 1 12 10. 11.
12.
13.
d. y 5 0.1(x 1 0.2)2 1 3.8 Express each of the following functions in vertex form. a. y 5 x 2 1 6x 1 7 b. y 5 2x 2 1 4x 2 11 Convert the following quadratic functions to vertex form. Identify the coordinates of the vertex. a. y 5 x 2 1 8x 1 11 b. y 5 3x 2 1 4x 2 2 The daily profit, f (in dollars), of a hot pretzel stand is a function of the price per pretzel, p (in dollars), given by ƒ(p) 5 21875p2 1 4500p 2 2400. a. Find the coordinates of the vertex of the parabola. b. Give the maximum profit and the price per pretzel that gives that profit. Find the equations in vertex form of the parabolas that satisfy the following conditions. Then check your solutions by using a graphing program, if available. a. The vertex is at (21, 4) and the parabola passes through the point (0, 2). b. The vertex is at (1, 23) and one of the parabola’s two horizontal intercepts is at (22, 0).
Exercises for Section 8.2 A graphing program is required or recommended for several exercises. Technology is needed in Exercises 30, 36, 38, and 40 to generate a best-fit quadratic. 1. Which of the following quadratics have parabolic graphs that are concave up? Concave down? Explain your reasoning. a. y 5 20 1 x 2 x 2 c. y 5 23x 1 2 1 x 2 2 b. y 5 0.5t 2 2 d. y 5 3(5 2 x)(x 1 2) 2. Match each function with its graph. Explain your reasoning for each choice. ƒ(x) 5 x 2 1 3 8
g(x) 5 x 2 2 4
y
8
h(x) 5 0.25x 2 1 3
y
8
x –8
8
y
x –8
8
x –8
8
–8
–8
–8
Graph A
Graph B
Graph C
3. On the same graph, sketch by hand the plots of the following functions and label each with its equation. y 5 22x 2 1 3 y 5 2x 2 2 y 5 22x 2 2 3 y 5 22x 4. In each case sketch by hand a quadratic function y 5 ƒ(x) with the indicated characteristics. a. Does not cross the x-axis and has a negative vertical intercept b. Has a vertex at (h, k) where h , 0 and k . 0 and a positive y-intercept c. Crosses the x-axis at x 5 2 and x 5 24 and is concave up d. Has the same shape as y 5 x2 2 3x but is raised up four units e. Has the same y-intercept as y 5 x2 2 3x 2 2 but is concave down and narrower 2 f. Has the same shape as y 5 x2 , but is shifted to the left by three units
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5. Identify the stretch/compression factor and the vertex for each of the following. a. y1 5 0.3(x 2 1)2 1 8 c. y3 5 0.01(x 1 20)2 2 b. y2 5 30x 2 11 d. y4 5 26x2 1 12x 6. For each of the following functions, identify the vertex and specify whether it represents a maximum or minimum, and then sketch its graph by hand. a. y 5 (x 2 2)2 c. y 5 22(x 1 1)2 1 5 1 2 b. y 5 2 sx 2 2d 1 3 d. y 5 20.4(x 23)2 2 1 7. For each of the following quadratic functions, find the vertex (h, k) and determine if it represents the maximum or minimum of the function. c. ƒ(x) 5 25(x 1 4)2 2 7 a. ƒ(x) 5 22(x 2 3)2 1 5 2 d. ƒ(x) 5 8(x 2 2)2 2 6 b. ƒ(x) 5 1.6(x 1 1) 1 8 8. For each quadratic function identify the vertex and specify whether it represents a maximum or minimum. Evaluate the function at 0 and then sketch a graph of the function by hand. a. ƒ(t) 5 0.25(t 2 2)2 1 1 c. h(x) 5 2(x 1 3)2 1 4 b. g(x) 5 3 2 (x 2 5)2
d. ksxd 5
sx 1 5d 2 3
2 13
9. (Graphing program optional.) Create a quadratic function in the vertex form y 5 a(x 2 h)2 1 k, given the specified values for a and the vertex (h, k). Then rewrite the function in the standard form y 5 ax 2 1 bx 1 c. If available, use technology to check that the graphs of the two forms are the same. a. a 5 1, (h, k) 5 (2, 24) b. a 5 21, (h, k) 5 (4, 3)
c. a 5 22, (h, k) 5 (23, 1) d. a 5 12 , (h, k) 5 (24, 6)
10. Transform the function ƒ(x) 5 x 2 into a new function g(x) by compressing ƒ(x) by a factor of 14, then shifting the result horizontally left three units, and finally shifting it down by six units. Find the equation of g(x) and sketch it by hand. 11. Transform the function ƒ(x) 5 3x 2 into a new function h(x) by shifting ƒ(x) horizontally to the right four units, reflecting the result across the x-axis, and then shifting it up by five units. a. What is the equation for h(x)? b. What is the vertex of h(x)? c. What is the vertical intercept of h(x)? 12. For the following quadratic functions in vertex form, ƒ(x) 5 a(x 2 h)2 1 k, determine the values for a, h, and k. Then compare each to ƒ(x) 5 x 2, and identify which constants represent a stretch/compression factor, or a shift in a particular direction. a. p(x) 5 5(x 2 4)2 2 2 b. gsxd 5 13 sx 1 5d 2 1 4 2 c. hsxd 5 20.25 Ax 2 12B 1 6 d. ksxd 5 23sx 1 4d 2 2 3 13. Using the strategy of “completing the square,” fill in the missing numbers that would make the statement true. a. x2 1 6x 1 _______ 5 (x 1 _____)2 b. x2 1 8x 1 _______ 5 (x 1 _____)2 c. x2 2 4x 1 _______ 5 (x 2 _____)2 d. x2 2 3x 1 _______ 5 (x 2 _____)2
477
e. 2(x2 1 2x 1 _______) 5 2(x 1 _____)2 f. 23(x2 1 x 1 _______) 5 23(x 1 _____)2 14. (Graphing program optional.) For each quadratic function use the method of “completing the square” to convert to the a-h-k form, and then identify the vertex. If available, use technology to confirm that the two forms are the same. a. y 5 x 2 1 8x 1 15 d. r(s) 5 25s2 1 20s 2 10 2 e. z 5 2m2 1 6m 2 5 b. ƒ(x) 5 x 2 4x 2 5 c. p(t) 5 t 2 2 3t 1 2 15. (Graphing program optional.) For each quadratic function b convert to a-h-k form by using h 5 22a and then find k. If available, use technology to graph the two forms of the function to confirm that they are the same. a. y 5 x2 1 6x 1 13 d. p(r) 5 23r 2 1 18r 2 9 2 e. m(z) 5 2z 2 1 8z 2 5 b. ƒ(x) 5 x 2 5x 2 5 2 c. g(x) 5 x 2 3x 1 6 16. Match each of the following graphs with one of following equations. Explain your reasoning. (Hint: Find the vertex.) Note that one function does not have a match. h(x) 5 0.5x2 2 2x 1 3 ƒ(x) 5 2x2 2 8x 2 2 2 i(x) 5 0.5x2 2 2x 1 8 g(x) 5 2x 2 8x 1 3 y
y
12
10 –12 Graph A
12
x
x
x –10
y
12
–10
10 –12 Graph B
–10
10 –12 Graph C
17. Without drawing the graph, list the following parabolas in order, from the narrowest to the broadest. Verify your results with technology. a. y 5 x 2 1 20 d. y 5 4x 2 2 b. y 5 0.5x 2 1 e. y 5 0.1x 2 1 2 c. y 5 13x 2 1 x 1 1 f. y 5 22x 2 2 5x 1 4 18. Convert the following functions from the a-b-c or standard form to the a-h-k or vertex form. d. y 5 2x 2 1 3x 2 5 a. ƒ(x) 5 x 2 1 6x 1 5 e. h(x) 5 3x 2 1 6x 1 5 b. g(x) 5 x 2 2 3x 1 7 2 f. y 5 2x 2 15x 2 2 c. y 5 3x 2 12x 1 12 19. (Graphing program required.) Given ƒ(x) 5 2x2 1 8x 2 15: a. Estimate by graphing: the x-intercepts, the y-intercept, and the vertex. b. Calculate the coordinates of the vertex. 20. Write each of the following quadratic equations in function form (i.e., solve for y in terms of x). Find the vertex and the y-intercept using any method. Finally, using these points, draw a rough sketch of the quadratic function. a. y 1 12 5 x(x 1 1) d. y 2 8x 5 x2 1 15 2 b. 2x 1 6x 1 14.4 2 2y 5 0 e. y 1 15 (x 2 2)(x 1 5) f. y 1 2x(x 2 6) 5 20 c. y 1 x2 2 5x 5 26.25
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21. a. Find the equation of the parabola with a vertex of (2, 4) that passes through the point (1, 7). b. Construct two different quadratic functions both with a vertex at (2, 23) such that the graph of one function is concave up and the graph of the other function is concave down. c. Find two different equations of a parabola that passes through the points (22, 5) and (4, 5) and that opens downward. (Hint: Find the axis of symmetry.) 22. For each part construct a function that satisfies the given conditions. a. Has a constant rate of increase of $15,000/year b. Is a quadratic that opens upward and has a vertex at (1, 24) c. Is a quadratic that opens downward and the vertex is on the x-axis d. Is a quadratic with a minimum at the point (10, 50) and a stretch factor of 3 e. Is a quadratic with a vertical intercept of (0, 3) that is also the vertex 23. If a parabola is the graph of the equation y 5 a(x 2 4)2 2 5: a. What are the coordinates of the vertex? Will the vertex change if a changes? b. What is the value of stretch factor a if the y-intercept is (0, 3)? c. What is the value of stretch factor a if the graph goes through the point (1, 223)? 24. Construct an equation for each of the accompanying parabolas. 5
y 16
30. (Technology required to generate a best-fit quadratic.) The accompanying graph of the data file INCLINE D A T A shows the motion of a cart, initially at the bottom of an inclined plane, after it was given a push toward a motion detector positioned at the top of the plane. The distance (in feet) of the cart from the top of the plane is plotted vs. time (in seconds). The motion can be modeled with a quadratic function. Motion of a Cart Moving Up and Down an Inclined Plane 9 8 7 6 5 4 3 2 1 0.0
1.0
2.0
3.0
4.0
5.0
6.0
Time (sec)
31. (Graphing program required.) A baseball hit straight up in the air is at a height
2 x –10
–2
2
–3 Graph A
29. A manager has determined that the revenue R(x) (in millions of dollars) made on the sale of supercomputers is given by R(x) 5 48x 2 3x 2, where x represents the number of supercomputers sold. How many supercomputers must be sold to maximize revenue? According to this model, what is the maximum revenue (in millions of dollars)?
a. Estimate the coordinates of the vertex. Describe what is happening to the cart at the vertex. b. Use technology to generate a best-fit quadratic to the data. c. Calculate the coordinates of the vertex.
y
x –4
28. Tom has a taste for adventure. He decides that he wants to bungee-jump off the Missouri River bridge. At any time t (in seconds from the moment he jumps) his height h(t) (in feet above the water level) is given by the function h(t) 5 20.5t 2 2 123t 1 190.5. How close to the water will Tom get?
Distance from top of plane (ft)
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Graph B
25. Determine the equation of the parabola whose vertex is at (2, 3) and that passes through the point (4, 21). Show your work, including a sketch of the parabola. 26. Students noticed that the path of water from a water fountain seemed to form a parabolic arc. They set a flat surface at the level of the water spout and measured the maximum height of the water from the flat surface as 8 inches and the distance from the spout to where the water hit the flat surface as 10 inches. Construct a function model for the stream of water. 27. Marketing research by a company has shown that the profit, P(x) (in thousands of dollars), made by the company is related to the amount spent on advertising, x (in thousands of dollars), by the equation P(x) 5 230 1 20x 2 0.5x 2. What expenditure (in thousands of dollars) for advertising gives the maximum profit? What is the maximum profit?
h 5 4 1 50t 216t 2 feet above ground level at time t seconds after being hit. (This formula is valid for t $ 0 until the ball hits the ground.) a. What is the value of h when t 5 0? What does this value represent in this context? b. Construct a table of values for t 5 0, 1, 2, 3, 4. Roughly when does the ball hit the ground? How can you tell? c. Graph the function. Does the graph confirm your estimate in part (b)? d. Explain why negative values for h make no sense in this situation. e. Estimate the maximum height that the baseball reaches. When does it reach that height? 32. (Graphing program optional.) The following function represents the relationship between time t (in seconds) and height h (in feet) for objects thrown upward on Pluto. For an initial velocity of 20 ft/sec and an initial height above the ground of 25 feet, we get h 5 2t 2 1 20t 1 25
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a. Find the coordinates of the point where the graph intersects the h-axis. b. Find the coordinates of the vertex of the parabola. c. Sketch the graph. Label the axes. d. Interpret the vertex in terms of time and height. e. For what values of t does the mathematical model make sense?
34. A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle (see the accompanying figure). If the perimeter of the window is 20 feet (including the semicircle), what dimensions will admit the most light (maximize the area)? (Hint: Express L in terms of r. Recall that the circumference of a circle 5 2pr, and the area of a circle 5 pr 2, where r is the radius of the circle.)
r
L
35. A pilot has crashed in the Sahara Desert. She still has her maps and knows her position, but her radio is destroyed. Her only hope for rescue is to hike out to a highway that passes near her position. She needs to determine the closest point on the highway and how far away it is. a. The highway is a straight line passing through a point 15 miles due north of her and another point 20 miles due east. Draw a sketch of the situation on graph paper, placing the pilot at the origin and labeling the two points on the highway. b. Construct an equation that represents the highway (using x for miles east and y for miles north). c. Now use the Pythagorean Theorem to describe the square of the distance, d, of the pilot to any point (x, y) on the highway.
d. Substitute the expression for y from part (b) into the equation from part (c) in order to write d 2 as a quadratic in x. e. If we minimize d 2, we minimize the distance d. So let D 5 d 2 and write D as a quadratic function in x. Now find the minimum value for D. f. What are the coordinates of the closest point on the highway, and what is the distance, d, to that point? 36. (Requires technology to find a best-fit quadratic.) The DATA accompanying figure is a plot of the data in the file BOUNCE, which shows the height of a bouncing racquetball (in feet) over time (in seconds). The path of the ball between each pair of bounces can be modeled using a quadratic function. 2.5
2.0
Height (ft)
33. In ancient times, after a bloody defeat that made her flee her city, the queen of Carthage, Dido, found refuge on the shores of Northern Africa. Sympathetic to her plight, the local inhabitants offered to build her a new Carthage along the shores of the Mediterranean Sea. However, her city had to be rectangular in shape, and its perimeter (excluding the coastal side) could be no larger than the length of a ball of string that she could make using fine strips from only one cow hide. Queen Dido made the thinnest string possible, whose length was one mile. Dido used the string to create three non-coastal sides enclosing a rectangular piece of land (assuming the coastline was straight). She made the width exactly half the length. This way, she claimed, she would have the maximum possible area the ball of string would allow her to enclose. Was Dido right?
479
1.5
1.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Time (sec)
a. Select from the file BOUNCE the subset of the data that represents the motion of the ball between the first and second bounces (that is, between the first and second times the ball hits the floor). Use technology to generate a best-fit quadratic function for this subset. b. From the graph, estimate the maximum height the ball reaches between the first and second bounces. c. Use the best-fit function to calculate the maximum height the ball reaches between the first and second bounces. 37. (Graphing program required.) At low speeds an automobile engine is not at its peak efficiency; efficiency initially rises with speed and then declines at higher speeds. When efficiency is at its maximum, the consumption rate of gas (measured in gallons per hour) is at a minimum. The gas consumption rate of a particular car can be modeled by the following equation, where G is the gas consumption rate in gallons per hour and M is speed in miles per hour: G 5 0.0002M 2 2 0.013M 1 1.07 a. Construct a graph of gas consumption rate versus speed. Estimate the minimum gas consumption rate from your graph and the speed at which it occurs. b. Using the equation for G, calculate the speed at which the gas consumption rate is at its minimum. What is the minimum gas consumption rate? c. If you travel for 2 hours at peak efficiency, how much gas will you use and how far will you go? d. If you travel at 60 mph, what is your gas consumption rate? How long does it take to go the same distance that you calculated in part (c)? (Recall that travel distance 5 speed 3 time traveled.) How much gas is required for the trip?
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e. Compare the answers for parts (c) and (d), which tell you how much gas is used for the same-length trip at two different speeds. Is gas actually saved for the trip by traveling at the speed that gives the minimum gas consumption rate? f. Using the function G, generate data for gas consumption rate measured in gallons per mile by completing the following table. Plot gallons per mile (on the vertical axis) vs. miles per hour (on the horizontal axis). At what speed is gallons per mile at a minimum?
Speed of Car (mph)
Measures of the Rate of Gas Consumption (gal/hr)
a. Plot weight (on vertical axis) vs. age (on horizontal axis). Then use technology to find the best-fit quadratic model to approximate the data. b. According to your model, what would an average 32-week-old embryo weigh? c. Comment on the domain for which your formula is reliable. d. Plot length versus age; then use technology to construct an appropriate mathematical model for the length as a function of embryo age from 20 to 40 weeks. e. Using your model, compute the age at which an embryo would be 42.5 centimeters long. 39. A shot-put athlete releases the shot at a speed of 14 meters per second, at an angle of 45 degrees to the horizontal (ground level). The height y (in meters above the ground) of the shot is given by the function
(gal/hr)/mph 5 gal/mile
0 10 20 30 40 50 60 70 80
1 2 x y 5 2 1 x 2 20
g. Add a fourth column to the data table. This time compute miles/gal 5 mph/(gal/hr). Plot miles per gallon vs. miles per hour. At what speed is miles per gallon at a maximum? This is the inverse of the preceding question; we are normally used to maximizing miles per gallon instead of minimizing gallons per mile. Does your answer make sense in terms of what you found for parts (b) and (f)? 38. (Requires technology to create best-fit functions.) The following data show the average growth of the human embryo prior to birth. Embryo Age (weeks)
Weight (g)
Length (cm)
8 12 20 28 36 40
3 36 330 1000 2400 3200
2.5 9 25 35 45 50
Source: Reprinted with permission from Kimber et al., Textbook oƒ Anatomy and Physiology (Upper Saddle River, NJ: Prentice Hall, 1955), “Embryo Age, Weight and Height,” p. 785.
where x is the horizontal distance the shot has traveled (in meters). a. What was the height of the shot at the moment of release? b. How high is the shot after it has traveled 4 meters horizontally from the release point? 16 meters? c. Find the highest point reached by the shot in its flight. d. Draw a sketch of the height of the shot and indicate how far the shot is from the athlete when it lands. 40. (Technology required to create a best-fit quadratic.) When people meet for the first time, it is customary for all people in the group to shake hands. Below is a table that shows the number of handshakes that occur depending on the size of the group. Group size Total number of handshakes
2 1
3 3
4 6
5 10
6 15
a. Draw a scatterplot of the data in the table (with group size on the horizontal axis) and find a quadratic model of best fit. b. Use the model to find the total number of handshakes for a group of seven people, and for a group of ten people. c. Factor the model of best fit. Describe in words how you could find the number of handshakes knowing the group size.
8.3 Finding the Horizontal Intercepts We often want to find the exact values for horizontal intercepts when we use a function to model a real-world situation. We know that a quadratic function may have two, one, or no horizontal intercepts. To find the horizontal intercepts (if any) of a function ƒ(x), we need to find the zeros of the function, the value(s) of x such that ƒ(x) 5 0. If ƒ(x) 5 ax2 1 bx 1 c, then setting ƒ(x) 5 0, we have 0 5 ax2 1 bx 1 c. If, as in the following example, we can factor ax2 1 bx 1 c as a product of two linear factors (each with an x term), solving for the intercepts is easy. Later on in this chapter we’ll use the quadratic formula to find the horizontal intercepts when the function is not easily factored.
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Using Factoring to Find the Horizontal Intercepts EXAMPLE
1
A quadratic model for a battleship gun range Iowa class battleships have large naval guns that have a muzzle velocity of about 2000 feet per second. If the gun is set to maximize range, then the relationship between the height of the projectile fired and the distance it travels is h(d) 5 d 2 8·1026d 2, where d 5 horizontal distance in feet from the battleship, and h(d) 5 height in feet of the projectile. What is the maximum range of the battleship gunfire? Height
Projectile path
Distance from battleship
Maximum range
Figure 8.26 Graph of height versus distance for a
projectile fired from a battleship. SOLUTION
Assuming that the battleship and the target are at the same level, to find the maximum range we need to find where the projectile will hit the ground. At that point h(d) 5 0, so the point represents a horizontal intercept on the graph (see Figure 8.26). To find the horizontal intercepts we set h(d) 5 0 and solve for d. Given the function if we set h(d) 5 0 and factor out d, we get
h(d) 5 d 2 8·1026d 2 0 5 d 2 8·1026d 2 0 5 d · (1 2 8·1026d)
So when h(d) 5 0, then either d 5 0 (at the gun)
or
1 2 8·1026d 5 0 (when the projectile hits the ground)
To solve the second expression for d subtract 1 from each side
1 2 8·1026d 5 0 28·1026d 5 21
divide each side by 28·1026 and simplify to get
d5
21 s28?1026 d
5 0.125·10 6 5 125,000 feet
So a projectile fired from this gun is able to hit a target that is 125,000 feet, or almost 24 miles, from the battleship. Finding the Horizontal Intercepts of a Function Given a function ƒ(x): To find the x-intercepts, set ƒ(x) 5 0 and solve for x. Every x-intercept is a zero of the function. One of the properties of real numbers we used in the previous example was the “zero product rule”—the notion that if the product of two terms is 0, then at least one term must be 0. Zero Product Rule For any two numbers r and s, if the product rs 5 0, then r or s or both must equal 0. We’ll use this rule repeatedly in our search for the horizontal intercepts of a function.
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Factoring Quadratics In the battleship example, we wrote h(d) 5 d 2 8·1026d 2, which is in standard form, as h(d) 5 d(1 2 8·1026d), which is in factored form. If a quadratic is in factored form, it’s easy to find the horizontal intercepts. We set the product equal to 0 and use the zero product rule. Factoring review To convert ax 2 1 bx 1 c to factored form requires thinking, practice, and a few hints. It is often a trial-and-error process. We usually restrict ourselves to finding factors with integer coefficients. First, look for common factors in all of the terms. For example, 10x 2 1 2x can be factored as 2x(5x 1 1). Second, look for two linear factors. This is easiest to do when the coefficient of x 2 is 1. For example, to factor x 2 1 7x 1 12, we want to rewrite it as (x 1 m)(x 1 n) for some m and n. Note that the coefficients of both x’s in the factors equal 1, since x times x is equal to the x 2 in the original expression. If we multiply out the two factors, we get x 2 1 (m 1 n)x 1 m ? n. So we need m 1 n 5 7 and m ? n 5 12. Since 12 is positive, m and n must have the same sign, either both positive or both negative. But since m 1 n (57) is positive, m and n must both be positive. So we consider pairs of positive integers whose product is 12, namely, 1 and 12, 2 and 6, or 3 and 4. We can then narrow our list of factors of 12 to those whose sum equals 7, the coefficient of the x term. Only the factors 3 and 4 fit this criterion. We can factor our polynomial as: x 2 1 7x 1 12 5 (x 1 3)(x 1 4) We can check that these factors work by multiplying them out. Third, look for the special case of the difference of two squares. In this case the middle terms cancel out when multiplying: x 2 2 25 5 (x 2 5)(x 1 5) 5 x 2 2 5x 1 5x 2 25 5 x 2 2 25 In general,
x 2 2 n 2 5 (x 2 n)(x 1 n)
Guidelines for Factoring ax 2 1 bx 1 c First factor out any common terms. If a 5 1, Find the factors of c that add to give b. x 2 1 bx 1 c 5 (x 1 m)(x 1 n),
where c 5 mn and b 5 m 1 n
If the quadratic is the difference of two squares (so b 5 0 and c 5 n 2 for some number n), factor it into the product of a sum and a difference. x 2 2 n 2 5 (x 2 n)(x 1 n)
It’s a good idea to double-check your answer by multiplying out the factored terms.
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EXAMPLE
483
2
Finding factors Put each of the following functions into factored form and then identify any horizontal intercepts. a. ƒ(x) 5 300x2 1 195x c. g(z) 5 23z 2 1 12z 2 12 e. Q(w) 5 2w2 2 3w 2 5 2 2 d. H(v) 5 22v 1 18 b. h(t) 5 t 2 5t 1 6
SOLUTION
a. In factored form ƒ(x) 5 15x(20x 1 13). (You can double-check this by multiplying the factors to return to 300x2 1 195x.) To find the horizontal intercepts, we need to find values for x such that ƒ(x) 5 0. If we set 15x(20x 1 13) 5 0, then according to the zero product rule, either 15x 5 0 x50
or
20x 1 13 5 0 20x 5 213 213 x5 20
or
20.65
So there are two horizontal intercepts, at x 5 0 and x 5 20.65 b. In factored form h(t) 5 (t 2 3)(t 2 2). (Again we can double-check our answer by multiplying the two factors to get t 2 2 3t 2 2t 1 6 5 t 2 2 5t 1 6.) To find the horizontal intercepts we need to find values for t such that h(t) 5 0. If we set (t 2 3)(t 2 2) 5 0, then either t2350 t53
or
t2250 t52
So there are two horizontal intercepts, at t 5 3 and t 5 2. c. In factored form g(z) 5 23(z 2 2 4z 1 4) 5 23(z 2 2)(z 2 2) 5 23(z 2 2)2 To find the horizontal intercepts, we need to find values for z such that g(z) 5 0. When 23(z 2 2)2 5 0, then z 2 2 5 0, so z 5 2. Hence g(z) has a single horizontal intercept at z 5 2. d. In factored form H(v) 5 22(v 2 2 9) 5 22(v 1 3)(v 2 3) When (v 1 3) (v 2 3) 5 0, then either v1350 v 5 23
or
v2350 v53
So H(v) has two horizontal intercepts, at 23 and 3. e. In factored form Q(w) 5 (2w 2 5)(w 1 1). When (2w 2 5)(w 1 1) 5 0, then either 2w 2 5 5 0 2w 5 5 w5
or
w1150 w 5 21
5 2
So Q(w) has two horizontal intercepts, at 52 (or 2.5) and at 21.
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Algebra Aerobics 8.3a 1. Put the function into factored form with integer coefficients and then identify any horizontal intercepts. e. y 5 15 2 8x 1 x 2 a. y 5 216t 2 1 50 t b. y 5 t 2 2 25 f. v(x) 5 x 2 1 2x 1 1 2 c. h(z) 5 z 2 3z 2 4 g. p(q) 5 q 2 2 6q 1 9 d. g(x) 5 4x 2 2 9 2. When possible, put the function into factored form with integer coefficients and then identify any horizontal intercepts. a. ƒ(x) 5 5 2 x 2 4x 2 d. z 5 4w 2 2 20w 1 25 2 e. y 5 2x 2 2 3x 2 5 b. h(t) 5 64 2 9t c. y 5 10 2 13t 2 3t 2 f. Q(t) 5 6t 2 1 11t 2 10 3. Identify which of the following quadratic functions can be factored into the product of a sum and a difference, y 5 (a 1 b)(a 2 b), which can be factored into the square of the sum or difference, y 5 (a 6 b) 2, and which can be factored into neither. a. y 5 x 2 2 9 e. y 5 x 2 2 8x 1 16 2 b. y 5 x 1 4x 1 4 f. y 5 16 2 25x 2 c. y 5 x 2 1 5x 1 25 g. y 5 4 1 16x 2 2 d. y 5 9x 2 25 4. Find any horizontal and vertical intercepts for the following functions and explain the meaning of each within the context of the problem.
a. An object is thrown vertically into the air at time t 5 0 seconds. Its height, h(t), in feet, is given by h(t) 5 216t 2 1 64t. b. The monthly profit P(q) (in dollars) is a function of q, the number of items sold. The relationship is described by P(q) 5 2q 2 1 60q 2 800. 5. Match the factored form of the quadratic function with its graph. y1 5 2x(x 2 2) y2 5 (x 2 2)(x 1 1) y 3 5 (x 1 4)(x 1 1) y
y
y
5
5
5
x –5
5
x
x –5
5
–5
5
–5
–5
–5
Graph A
Graph B
Graph C
6. Given ƒ(x) 5 x 2 1 x 2 30: a. Factor ƒ(x). b. Find the horizontal intercepts. c. Sketch the graph of ƒ(x). d. Describe the relationship between the factored form and the horizontal intercepts in this problem.
Using the Quadratic Formula to Find the Horizontal Intercepts Most quadratic functions are not easily factored. When this is the case, we still set ƒ(x) 5 0 and solve for x. But now we use the quadratic formula to find the solutions or roots to the equation 0 5 ax 2 1 bx 1 c.
The Quadratic Formula For any quadratic equation of the form 0 5 ax 2 1 bx 1 c (where a 2 0), the solutions, or roots, of the equation are given by x5
2b 6 "b 2 2 4ac 2a
The term under the radical sign, b2 2 4ac, is called the discriminant.
The symbol 6 lets us use one formula to write the two roots as See the reading “Why the Formula for the Vertex and the Quadratic Formula Work” for a derivation of this famous formula.
x5
2b 1 "b 2 2 4ac 2a
and
x5
2b 2 "b 2 2 4ac 2a
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Note on Terminology. The language of roots and zeros can be confusing. The numbers 3 and 23 are called the roots or solutions of the equation x 2 2 9 5 0 and the zeros of the function ƒ(x) 5 x 2 2 9. The zeros of the function ƒ(x) are the roots of the equation ƒ(x) 5 0. The discriminant One shortcut for predicting the number of horizontal intercepts is to use the discriminant, the term b 2 2 4ac.
Using the Discriminant A quadratic function ƒ(x) 5 ax 2 1 bx 1 c has a discriminant of b 2 2 4ac. If the discriminant . 0, there are two distinct real roots and hence two x-intercepts, at x1 5
2b 1 "discriminant 2a
x2 5
and
2b 2 "discriminant 2a
If the discriminant 5 0, there is only one distinct real root and hence only one x-intercept, at 2b x5 2a If the discriminant , 0, then "discriminant is not a real number; so there are no x-intercepts and the zeros of the function are not real numbers.
EXAMPLE
3
Identify whether each of the graphs in Figure 8.27 has a discriminant . 0, , 0, or equal to 0.
y
y
x
Graph A
y
x
Graph B
y
x
Graph C
x
Graph D
Figure 8.27 Graphs of four parabolas.
SOLUTION
EXAMPLE
4
Graph B has two horizontal intercepts, so the discriminant is . 0. Graph C has only one horizontal intercept, so the discriminant 5 0. Graphs A and D have no horizontal intercepts, so the discriminant is , 0.
For each of the following functions, use the discriminant to predict the number of horizontal intercepts. If there are any, use the quadratic formula to find them. Then using technology, graph the function to confirm your predictions. a. ƒ(z) 5 z 2 1 3z 1 2.25 c. g(x) 5 2x2 2 6x 2 10 2 b. h(t) 5 34 1 32t 2 16t
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SOLUTION
a. Setting ƒ(z) 5 0, we have 0 5 z2 1 3z 1 2.25. Here a 5 1, b 5 3, and c 5 2.25, so the discriminant is b2 2 4ac 5 32 2 (4·1·2.25) 5 9 2 9 5 0. Since "0 5 0, the quadratic formula says that there is only one root, at 2b /(2a) 5 23 /(2·1) 5 21.5. So ƒ(z) has one real zero, and hence one z-intercept at (21.5, 0), which must also be the vertex of the parabola. (See Figure 8.28.) 40
20
z –10
–5
5
10
–10
Figure 8.28 Graph of
ƒ(z) 5 z2 1 3z 1 2.25, with one horizontal intercept, at the vertex.
b. Setting h(t) 5 0, we have 0 5 34 1 32t 2 16t 2. If we rearrange the terms as 0 5 216t2 1 32t 1 34, it’s easier to see that we should set a 5 216, b 5 32, and c 5 34. The discriminant is b2 2 4ac 5 (32) 2 2 (4·(216)·34) 5 1024 1 2176 5 3200. So there are two real roots: one at
232 1 "3200 232 1 56.6 24.6 < 5 < 20.77 2(216) 232 232
the other at
232 2 "3200 232 2 56.6 288.6 < 5 < 2.77 2(216) 232 232
Therefore, the parabola for h(t) has two real zeros, and hence two t-intercepts, at approximately (20.77, 0) and (2.77, 0). (See Figure 8.29.) 60
t –2
4 –10
Figure 8.29 Graph of
h(t) 5 34 1 32t 216t 2, with two horizontal intercepts.
c. Setting g(x) 5 0, we have 0 5 2x 2 2 6x 2 10. Here a 5 21, b 5 26, and c 5 210, so the discriminant is b2 2 4ac 5 (26)2 2 4(21)(210) 5 36 2 40 5 24. The discriminant is negative, so taking its square root presents a problem. There is "24 no real number r such that r 2 5 24. Therefore, the roots at 6 622 are not real. Since there are no real zeros for the function, there are no horizontal intercepts. (See Figure 8.30.)
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1
487
y x
–10
5
–10
Figure 8.30 Graph of
g(x) 5 2x2 2 6x 2 10, with no horizontal intercepts.
Imaginary and complex numbers Mathematicians were uncomfortable with the notion that certain quadratic equations did not have solutions, so they literally invented a number system in which such equations would be solvable. In the process, they created new numbers, called imaginary numbers. The imaginary number i is defined as a number such that i 2 5 21 or equivalently
i 5 "21
A number such as "24 is also an imaginary number. We can write "24 as "s4ds21d 5 "4 "21 5 2 "21 5 2i When a number is called imaginary, it sounds as if it does not exist. But imaginary numbers are just as legitimate as real numbers. Imaginary numbers are used to extend the real number system to a larger system called the complex numbers.
Complex Numbers A complex number is defined as any number that can be written in the form z 5 a 1 bi where a and b are real numbers and i 5 "21. The real part of z is the number a, and the imaginary part is the number b.
Note that the real numbers are a subset of the complex numbers, since any real number a can be written as a 1 0 ? i. EXAMPLE
5
SOLUTION
Write each expression as a complex number of the form a 1 bi. a. 22 1 7i b. 4 1 "29 c. 13 2 "36 d. "225 a. b. c. d. e.
e. 5 1 3i 2
22 1 7i is complex and already in a 1 bi form. 4 1 "29 5 4 1 "9 "21 5 4 1 3i 13 2 "36 5 13 2 6 5 7 5 7 1 0 ? i (a real number) "225 5 "25 "21 5 5i 5 0 1 5i (an imaginary number) Since i 2 5 21, then 5 1 3i 2 5 5 1 3(21) 5 5 2 3 5 2 5 2 1 0 ? i (a real number)
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The Factored Form We started this section by finding the horizontal intercepts from the factored form. We now show that any quadratic function can be put into factored form, whether the zeros are real or complex. The Factor Theorem relates the zeros of a function to the factors of a function.
The Factor Theorem Given a function ƒ(x), if ƒ(r) 5 0, then r is a zero of the function and (x 2 r) is a factor of ƒ(x).
Using the Factor Theorem, if r1 and r2 are zeros of the function ƒ(x), then both (x 2 r1) and (x 2 r2) are factors of ƒ(x). So we can write a quadratic function ƒ(x) 5 ax 2 1 bx 1 c in factored form as ƒ(x) 5 a(x 2 r1)(x 2 r2). Note that if you multiply out the factored form, the coefficient of x 2 is a, as it is in the standard form. The factored form is useful when we want to emphasize the zeros of a function.
The Factored Form The quadratic function ƒ(x) 5 ax 2 1 bx 1 c can be written in factored form as ƒ(x) 5 a(x 2 r1)(x 2 r2) where r1 and r2 are the zeros of ƒ(x). If r1 and r2 are real numbers, then ƒ(x) has x-intercepts at r1 and r2.
EXAMPLE
6
Constructing a quadratic function with complex zeros a. Construct a quadratic function that is concave up and has zeros at 5 1 i and 5 2 i. Put it into standard form. b. What do we know about the graph of this function? Are there any other functions with the same characteristics?
SOLUTION
a. The function ƒ(x) 5 (x 2 (5 1 i)) (x 2 (5 2 i)) has zeros at 5 1 i and 5 2 i. If we multiply it out (using the FOIL technique), we get (x 2 (5 1 i))(x 2 (5 2 i)) 5 x 2 2 (5 2 i)x 2 (5 1 i)x 1 (5 1 i)(5 2 i) use the distributive law and FOIL again simplify and substitute 21 for i 2
5 x 2 2 5x 1 ix 2 5x 2 ix 1 (25 2 5i 1 5i 2 i 2) 5 x 2 2 10x 1 (25 2 i 2) 5 x 2 2 10x 1 26
So we can rewrite ƒ(x) in standard form as ƒ(x) 5 x 2 2 10x 1 26. (You can doublecheck this by using the quadratic formula.) Since the coefficient of x 2 is 1, and hence positive, the graph is concave up. b. Since the zeros are not real, the graph of the function has no x-intercepts. Any function of the form aƒ(x) 5 a(x 2 2 10x 1 26), where a . 0, will be concave up and have zeros at 5 1 i and 5 2 i. Since there are an infinite number of values of a, there are an infinite number of functions with these characteristics.
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7
A parabola has horizontal intercepts at d 5 21 and d 5 2, and passes through the point (d, h) 5 (1.5, 1.25). a. Find the equation for the parabola. b. Graph the parabola using a dotted line. Now restrict the domain from d 5 0 to the horizontal intercept that is positive and then color in the corresponding section of the parabola with a solid line. This section of the parabola, drawn with a solid line, describes the path of a water jet located at the center of a circular fountain, where d 5 distance (in feet) from the base of the fountain, and h 5 height (in feet) of the water. c. At what height is the nozzle of the water jet? d. What is the greatest height the stream of water reaches?
SOLUTION
a. If the parabola has horizontal intercepts at 21 and 2, then the equation for the parabola is in the form h 5 a(d 2 (21))(d 2 2) or a(d 1 1)(d 2 2). The point (1.5, 1.25) lies on the parabola, so it must satisfy the equation. Hence
EXAMPLE
given that if we substitute 1.5 for d and 1.25 for h simplify multiply and divide by 21.25, we have
h 5 a(d 1 1)(d 2 2) 1.25 5 a (1.5 1 1)(1.5 2 2) 1.25 5 a(2.5)(20.5) 1.25 5 a(21.25) 21 5 a h 5 2(d 1 1)(d 2 2).
So the equation of the parabola is b. See Figure 8.31.
h 3
d –2
3 –1
Figure 8.31 Graph of
h 5 2(d 1 1)(d 2 2).
c. When d 5 0, h 5 2(0 1 1)(0 2 2) 5 2. So the water nozzle is at 2 feet. d. There are several strategies for finding the vertex. You could calculate it using the formula in Section 8.2. Or recall that the vertex lies on the axis of symmetry, s21 1 2d halfway between the two horizontal intercepts (21 and 2) at d 5 5 12 or 2 0.5. Substituting 0.5 for d in the equation in part (a), we get h 5 2(0.5 1 1) (0.5 2 2) 5 2(1.5)(21.5) 5 2.25 The coordinates of the vertex are (0.5, 2.25). So the maximum height of the water (at the vertex of the parabola) is 2.25 feet (or 27s ), which occurs at 0.5 feet (or 6s ) from the base of the fountain.
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The Horizontal Intercepts of a Quadratic Function The graph of a quadratic function may have zero, one, or two horizontal intercepts, which can be found by factoring or using the quadratic formula. The discriminant in the quadratic formula can be used to predict the number of horizontal intercepts. Any quadratic function in standard form ƒ(x) 5 ax 2 1 bx 1 c can be written in factored form as ƒ(x) 5 a(x 2 r1)(x 2 r2) where r1 and r2 are the zeros of ƒ(x) If the zeros r1 and r2 are real numbers, they are the horizontal intercept(s) of ƒ(x).
Algebra Aerobics 8.3b 1. Find any real numbers that satisfy the following equations. a. 4x 1 7 5 0 e. (2x 2 11)2 5 0 2 b. 4x 2 7 5 0 f. (x 1 1)2 5 81 c. 4x 2 2 7x 5 0 g. x 5 x 2 2 5 2 d. 2(x 1 3) 5 x 2. Evaluate the discriminant b2 2 4ac for each of the following quadratic functions of the form ƒ(x) 5 ax 2 1 bx 1 c. Use the discriminant to determine the nature of the zeros of the function and the number (if any) of horizontal intercepts. a. ƒ(x) 5 2x 2 1 3x 2 1 c. ƒ(x) 5 4x 2 1 4x 1 1 b. ƒ(x) 5 x 2 1 7x 1 2 d. ƒ(x) 5 2x 2 1 x 1 5 3. Find and interpret the horizontal and vertical intercepts for the following height equations. a. h 5 24.9t 2 1 50t 1 80 (h is in meters and t is in seconds) b. h 5 150 2 80t 2 490t 2 (h is in centimeters and t is in seconds) 2 c. h 5 216t 1 64t 1 3 (h is in feet and t is in seconds) d. h 5 64t 2 16t 2 (h is in feet and t is in seconds) 4. Evaluate the discriminant and then predict the number of x-intercepts for each function. Use the quadratic formula to find all the zeros of each function and then identify the coordinates of any x-intercept(s).
a. y 5 4 2 x 2 5x 2 d. y 5 2x 2 2 3x 2 1 2 b. y 5 4x 2 28x 1 49 e. y 5 2 2 3x 2 c. y 5 2x 2 1 5x 1 4 From the descriptions given in parts (a) and (b), determine the coordinates of the vertex, find the equation of the parabola, and then sketch the parabola. a. A parabola with horizontal intercepts at x 5 22 and x 5 4 that passes through the point (3, 2). b. A parabola with horizontal intercepts at x 5 2 and x 5 8, and a vertical intercept at y 5 10. Identify the number of x-intercepts for the following functions without converting into standard form. a. y 5 3(x 2 1) 2 1 5 c. y 5 25(x 1 3)2 b. y 5 22(x 1 4) 2 2 1 d. y 5 3(x 2 1) 2 2 2 Write an equation for a quadratic function, in factored form, with the specified zeros. a. 2 and 23 b. 0 and 25 c. 8 For each quadratic function in the accompanying graphs, specify the number of real zeros and whether the corresponding discriminant would be positive, negative, or zero.
5.
6.
7.
8.
y
y
y
1
x
5
2
–5
1
x –2
3
x
–3 Graph A
0
Graph B
5
–5 Graph C
Exercises for Section 8.3 Several exercises either require or recommend the use of a graphing program. 1. Solve the following quadratic equations by factoring. e. 4x 2 1 9 5 12x a. x 2 2 9 5 0 f. 3x 2 5 13x 1 10 b. x 2 2 4x 5 0 2 g. (x 1 1)(x 1 3) 5 21 c. 3x 5 25x h. x(x 1 2) 5 3x(x 2 1) 2 3 d. x 2 1 x 5 20
2. Find the x-intercepts for each of the following functions. Will the vertex lie above, below, or on the x-axis? Find the vertex and sketch the graph, labeling the x-intercepts. a. y 5 (x 1 2)(x 1 1) b. y 5 3(1 2 2x)(x 1 3) c. y 5 24(x 1 3)2
d. y 5 12 (x)(x 2 5) e. q(x) 5 2(x 2 3)(x 1 2) f. ƒ(x) 5 22(5 2 x)(3 2 2x)
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3. Factor the quadratic expression and then sketch the graph of the function, labeling the axes and horizontal intercepts. a. y 5 x 2 1 6x 1 8 d. w 5 t 2 2 25 e. r 5 4s 2 2 100 b. z 5 3x 2 2 6x 2 9 2 f. g(x) 5 3x 2 2 x 2 4 c. ƒ(x) 5 x 2 3x 2 10 4. a. Construct a quadratic function with zeros at x 5 1 and x 5 2. b. Is there more than one possible quadratic function for part (a)? Why or why not? 5. (Graphing program required.) Using a graphing program, estimate the real solutions to the following equations. (Hint: Think of the equations as resulting from setting ƒ(x) 5 0.) Verify by factoring, if possible. a. x 2 2 5x 1 6 5 0 d. 23x 2 2 12x 1 15 5 0 2 e. 0.05x 2 1 1.1x 5 0 b. 3x 2 2x 1 5 5 0 2 f. 22x 2 2 x 1 3 5 0 c. 3x 2 12x 1 12 5 0 6. Write each function in factored form, if possible, using integer coefficients. d. k(p) 5 p 2 1 5p 1 7 a. ƒ(x) 5 x 2 1 2x 2 15 e. l(s) 5 5s 2 2 37s 2 24 b. g(x) 5 x 2 2 6x 1 9 2 f. m(t) 5 5t 2 1 t 1 1 c. h(a) 5 a 1 6a 2 16 7. Solve the following equations using the quadratic formula. (Hint: Rewrite each equation so that one side of the equation is zero.) a. 6t 2 2 7t 5 5 e. 6s 2 2 10 5 217s b. 3x(3x 2 4) 5 24 f. 2t 2 5 3t 1 9 c. (z 1 1)(3z 22) 5 2z 1 7 g. 5 5 (4x 1 1)(x 2 3) d. (x 1 2)(x 1 4) 5 1 h. (2x 2 3)2 5 7
491
11. In each part (a) to (e), graph a parabola with the given characteristics. Then write an equation of the form y 5 ax 2 1 bx 1 c for that parabola. a. a . 0, b 2 2 4ac . 0, c . 0 b. a . 0, b 2 2 4ac . 0, c , 0 c. a . 0, b 2 2 4ac , 0, b 2 0 d. a , 0, b2 2 4ac 5 0, c 2 0 e. b 2 0, c , 0, b2 2 4ac . 0 12. For each part, draw a rough sketch of a graph of a function of the type ƒ(x) 5 ax 2 1 bx 1 c a. Where a . 0, c . 0, and the function has no real zeros. b. Where a , 0, c . 0, and the function has two real zeros. c. Where a . 0 and the function has one real zero. 13. a. Construct a quadratic function Q(t) with exactly one zero at t 5 21 and a vertical intercept at 24. b. Is there more than one possible quadratic function for part (a)? Why or why not? c. Determine the axis of symmetry. Describe the vertex. 14. a. Construct a quadratic function P(s) that goes through the point (5, 222) and has two real zeros, one at s 5 26 and the other at s 5 4. b. What is the axis of symmetry? c. What are the coordinates of the vertex? d. What is the vertical intercept? 15. Construct a quadratic function for each of the given graphs. Write the function in both factored form and standard form.
10
y
10
y
8. Solve using the quadratic formula. a. x 2 2 3x 5 12 b. 3x 5 4x 1 2 2
c. 3(x 2 1 1) 5 x 1 2
e.
x11 1 5 x22 x21
x2 x 1 1 2 50 f. 3 2 6 1 3 1 g. 2 2 5 x x 6
d. (3x 2 1)(x 1 2) 5 4 9. Calculate the coordinates of the x- and y-intercepts for the following quadratics. c. y 5 (5 2 2x)(3 1 5x) a. y 5 3x 2 1 2x 2 1 2 d. ƒ(x) 5 x 2 2 5 b. y 5 3(x 2 2) 2 1 10. Use the discriminant to predict the number of horizontal intercepts for each function. Then use the quadratic formula to find all the zeros. Identify the coordinates of any horizontal or vertical intercepts. a. y 5 2x 2 1 3x 2 5 b. ƒ(x) 5 216 1 8x 2 x2 c. ƒ(x) 5 x 2 1 2x 1 2 d. y 5 2(x 2 1) 2 1 1 e. g(z) 5 5 2 3z 2 z 2 f. ƒ(t) 5 (t 1 2)(t 2 4) 1 9
x –10
10
x –10
10
–14
–14
Graph A
Graph B
16. Complete the following table, and then summarize your findings. i 1 5 "21 5 i i 2 5 i ? i 5 21 i 3 5 i ? i2 5 ? i 4 5 i2 ? i2 5 ?
i5 5 ? i6 5 ? i7 5 ? i8 5 ?
17. Complex number expressions can be simplified by combining the real parts and then the imaginary parts. Add (or subtract) the following complex numbers and then simplify. a. (4 1 3i) 1 (25 1 7i) c. (7i 2 3) 1 (2 2 4i) b. (22 1 3i) 2 (23 1 3i) d. (7i 2 3) 2 (2 2 4i)
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18. Complex number expressions can be multiplied using the distributive property or the FOIL technique. Multiply and simplify the following. (Note: i2 5 21.) a. (3 1 2i)(22 1 3i) d. (5 2 3i)(5 1 3i) b. (4 2 2i)(3 1 i) e. (3 2 i)2 c. (2 1 i)(2 2 i) f. (4 1 5i)2
25. Find the equation of the graph of a parabola that has the following properties: • The x-intercepts of the graph are at (2, 0) and (3, 0), and • The parabola is the graph of y 5 x 2 vertically stretched by a factor of 4. Explain your reasoning. Sketch the parabola.
19. A quadratic function has two complex roots, r1 5 1 1 i and r2 5 1 2 i. Use the Factor Theorem to find the equation of this quadratic, assuming a 5 1, and then put it into standard form.
26. (Graphing program required for part (b)). We dealt previously with systems of lines and ways to determine the coordinates of points where lines intersect. Once you know the quadratic formula, it’s possible to determine where a line and a parabola, or two parabolas, intersect. As with two straight lines, at the point where the graphs of two functions intersect (if they intersect), the functions share the same x value and the same y value. a. Find the intersection of the parabola y 5 2x 2 2 3x 1 5.1 and the line y 5 24.3x 1 10. b. Plot both functions, labeling any intersection point(s).
20. The factored form of a quadratic function is y 5 22(x 2 (3 1 i))(x 2 (3 2 i)). Answer the following. a. Will the graph open up or down? Explain. b. What are the zeros of the quadratic function? c. Does the graph cross the x-axis? Explain. d. Write the quadratic in standard form. (Hint: Multiply out; see Exercise 18.) e. Verify your answer in part (b) by using the quadratic formula and your answer for part (d). 21. Use the quadratic formula to find the zeros of the function ƒ(x) 5 x2 2 4x 1 13 and then write the function in factored form. Without graphing this function, how can you tell if it intersects the x-axis? 22. Let (h, k) be the coordinates of the vertex of a parabola. Then h is equal to the average of the two real zeros of the function (if they exist). For parts (a) and (b) use this to find h, and then construct an equation in vertex form, y 5 a(x 2 h)2 1 k. a. A parabola with x-intercepts of 4 and 8, and a y-intercept of 32 b. A parabola with x-intercepts of 23 and 1, and a y-intercept of 21 c. Can you find the equation of a parabola knowing only its x-intercepts? Explain. 23. Let (h, k) be the coordinates of the vertex of a parabola. Then h equals the average of the two real zeros of the function (if they exist). For each of the following use this to find h, and then put the equations into the vertex form, y 5 a(x 2 h)2 1 k. a. A parabola with equation y 5 x 2 1 2x 2 8 b. A parabola with equation y 5 2x 2 2 3x 1 4 24. (Graphing program optional) a. Write each of the following functions in both the a-b-c and the a-h-k forms. Is one form easier than the other for finding the vertex? The x- and y-intercepts? y1 5 2x 2 2 3x 2 20 y3 5 3x 2 1 6x 1 3 y4 5 2(2x 1 4)(x 2 3) y2 5 22(x 2 1)2 2 3 b. Find the vertex and x- and y-intercepts and construct a graph by hand for each function in part (a). If you have access to a graphing program, check your work.
27. (Graphing program recommended for part (b)). a. Find the intersection of the two parabolas y 5 7x2 2 5x 2 9 and y 5 22x2 1 4x 1 9 . b. Plot both functions, labeling any intersection points. 28. Market research suggests that if a particular item is priced at x dollars, then the weekly profit P(x), in thousands of dollars, is given by the function 1 2 P(x) 5 29 1 11 2 x 2 2x
a. What price range would yield a profit for this item? b. Describe what happens to the profit as the price increases. Why is a quadratic function an appropriate model for profit as a function of price? c. What price would yield a maximum profit? 29. A dairy farmer has 1500 feet of fencing. He wants to use all 1500 feet to construct a rectangle and two interior separators that together form three rectangular pens. See the accompanying figure.
W
L
a. If W is the width of the larger rectangle, express the length, L, of the larger rectangle in terms of W. b. Express the total area, A(W), of the three pens as a polynomial in terms of W. c. What is the domain of the function A(W)? d. What are the dimensions of the larger rectangle that give a maximum area? What is the maximum area?
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8.4 The Average Rate of Change of a Quadratic Function In previous chapters we argued that the average rate of change of a linear function is constant, and that the average rate of change of an exponential function is exponential. What about the average rate of change of a quadratic function?
EXAMPLE
1
Finding the average rate of change of the simplest quadratic function Given y 5 x 2, calculate the average rate of change of y with respect to x at unit intervals from 23 to 3. Then calculate the average rate of change of the average rate of change. What do these data points suggest?
SOLUTION
Column 3 in Table 8.2 shows the values for the average rate of change of y with respect to x. For these values the average rate of change is linear, since adding 1 to x increases the average rate of change by 2 over each interval. Column 4 computes the average rate of change of column 3 with respect to x. Since these values all are constant at 2, this confirms that column 3 (the average rate of change) is linear with respect to x. This suggests that when y 5 x 2 the average rate of change of y with respect to x is a linear function.
Average Rate of Change
Average Rate of Change of the Average Rate of Change
x
y 5 x2
23
9
22
4
4 2 9 22 2 (23)
5 25
21
1
(1 2 4) 1
5 23
23 2 (25) 21 2 (22)
52
0
0
(0 2 1) 1
5 21
(21 2 (23) ) 1
52
1
1
(1 2 0) 1
51
(1 2 (21) ) 1
52
2
4
(4 2 1) 1
53
(3 2 1) 1
52
3
9
(9 2 4) 1
55
(5 2 3) 1
52
n.a.
n.a. n.a.
Table 8.2
EXAMPLE
2
Finding the average rate of change of a quadratic function Given y 5 3x 2 2 8x 2 23, calculate the average rate of change of y with respect to x at unit intervals from 23 to 3. Then calculate the average rate of change of the average rate of change. What do the data suggest about the original function?
SOLUTION
Column 3 in Table 8.3 shows the values for the average rate of change. Again the values suggest that the average rate of change is linear, since adding 1 to x increases the average rate of change by 6 over each interval. This is confirmed by column 4, which shows that the average rate of change of the third column with respect to x is constant at 6. This suggests that when y 5 3x 2 2 8x 2 23, the average rate of change of y with respect to x is a linear function.
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x
Average Rate of Change
y 5 3x 2 2 8x 2 23
23 22 21 0 1 2 3
28 5 212 223 228 227 220
Average Rate of Change of the Average Rate of Change
n.a. 223 217 211 25 1 7
n.a. n.a. 6 6 6 6 6
Table 8.3
We have just seen two numerical examples that suggest that the average rate of change of a quadratic function is a linear function. We now show algebraically that this is true for every quadratic function. Suppose we have y 5 ƒ(x) 5 ax 2 1 bx 1 c. In the previous examples we fixed an interval size of 1 over which to calculate the average rate of change, since it is easy to make comparisons. Now we pick a constant interval size r and for each position x compute the average rate of change of ƒ over the interval from x to x 1 r. First we must compute ƒ(x 1 r): ƒ(x 1 r) 5 a(x 1 r)2 1 b(x 1 r) 1 c apply exponent 5 a(x 2 1 2rx 1 r 2) 1 b(x 1 r) 1 c multiply through 5 ax 2 1 2arx 1 ar 2 1 bx 1 br 1 c regroup terms 5 ax 2 1 bx 1 c 1 (2ax 1 b)r 1 ar 2 Then the average rate of change of ƒ(x) between x and x 1 r is change in ƒ(x) ƒ(x 1 r) 2 ƒ(x) 5 change in x (x 1 r) 2 x [ax2 1 bx 1 c 1 (2ax 1 b)r 1 ar2] 2 (ax2 1 bx 1 c) 5 r (2ax 1 b)r 1 ar2 5 r 5 2ax 1 (b 1 ar) The average rate of change between (x, ƒ(x)) and (x 1 r, ƒ(x 1 r)) is a linear function of x in its own right, where slope 5 2a
y-intercept 5 b 1 ar
Note that the slope depends only on the value of a in the original equation. The y-intercept depends not only on a and b in the original equation, but also on r, the interval size over which we calculate the average rate of change. (See Figure 8.32.) y f (x + r)
(x + r, f (x + r))
f (x)
(x, f (x))
x
r
x x+r
Figure 8.32 The slope of the line
segment connecting two points on the parabola separated by a horizontal distance of r is 2ax 1 (b 1 ar).
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If we took smaller and smaller values for r, then the term ar would get closer and closer to zero, and hence the y-intercept b 1 ar would get closer and closer to b. For very small r’s the average rate of change would get closer and closer to the linear expression 2ax 1 b. So given a quadratic function ƒ(x) 5 ax2 1 bx 1 c, the average rate of change over small intervals around any x value can be approximated by the linear expression 2ax 1 b. So we talk about the function g(x) 5 2ax 1 b as representing the average rate of change of ƒ(x) with respect to x. This is a central idea in calculus.
The Average Rate of Change of a Quadratic Function Given a quadratic function ƒ(x) 5 ax2 1 bx 1 c, we can think of the linear function g(x) 5 2ax 1 b as representing the average rate of change of ƒ(x) with respect to x.
EXAMPLE
3
Graph each of the following functions. Construct and graph the equation for the average rate of change in each case. a. ƒ(x) 5 x 2 b. ƒ(x) 5 3x 2 2 8x 2 23
SOLUTION
a. The average rate of change of the quadratic function ƒ(x) 5 x2 is the linear function g(x) 5 2x (see Figure 8.33).
4
y
4
y
f (x) = x2
g (x) = 2x x
–4
x
4
–4
–4
4
–4
Figure 8.33 Graph of ƒ(x) 5 x and graph of its average rate of 2
change.
b. The average rate of change of the quadratic function ƒ(x) 5 3x 2 2 8x 2 23 is the linear function g(x) 5 6x 2 8 (see Figure 8.34). y
y 10
50
x –10
10
x –5
5
y = 6x – 8
y = 3 x 2 – 8 x – 23 –50
–10
Figure 8.34 Graph of y 5 3x 2 2 8x 2 23 and graph of its average
rate of change.
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Algebra Aerobics 8.4 1. a. Complete the table below for the function y 5 5 2 x 2.
x
y
21 0 1 2 3 4
4 5 4 1
Average Rate of Change
Average Rate of Change of Average Rate of Change
n.a. 1 21
n.a. n.a. 22
b. x
y
Average Rate of Change
21 0 1 2 3 4
7 4 1 22 25 28
n.a. 23 23 23 23 23
n.a. n.a. 0 0 0 0
x
y
Average Rate of Change
Average Rate of Change of Average Rate of Change
21 0 1 2 3 4
0.5 1 2 4 8 16
n.a. 0.5 1 2 4 8
n.a. n.a. 0.5 1 2 4
c.
b. What do the third and fourth columns tell you? 2. Determine from each of the three following tables whether you would expect the original function y 5 ƒ(x) to be linear, exponential, or quadratic. a. Average Average Rate of x
y
Rate of Change
Change of Average Rate of Change
21 0 1 2 3 4
2 1 6 17 34 57
n.a. 21 5 11 17 23
n.a. n.a. 6 6 6 6
Average Rate of Change of Average Rate of Change
3. Construct a function that represents the average rate of change for the following three quadratic functions. c. ƒ(x) 5 5x 2 1 2x 1 7 a. ƒ(t) 5 t 2 1 t b. ƒ(x) 5 3x 2 1 5x
Exercises for Section 8.4 Exercise 6 requires a graphing program. 1. Complete these sentences. a. For a quadratic function, the graph of its average rate of change represents a ___________________ function. b. When a quadratic function is increasing, its average rate of change is _____________, and when a quadratic function is decreasing, its average rate of change is _________________. 2. Complete the table for the function y 5 3 2 x 2 x 2.
x 23
y 23
Average Rate of Change
Average Rate of Change of Average Rate of Change
n.a.
n.a.
22
1
1 2 s23d 22 2 s23d
21
3
3 2 1 21 2 1s22d
5 21 5 2
0
3
3 2 3 0 2 s21d
5 01 5 0
1 2 3
1
5
4 1
54
n.a. 2 2 4 21 2 s22d
5 22 1 5 22
0 2 2 0 2 s21d
5
What type of function does the average rate of change represent? What is its slope (the rate of change of the average rate of change)? 3. a. What is the average rate of change of the linear function y 5 3x 2 1? Of y 5 22x 1 5? Of y 5 ax 1 b? What equation could represent the average rate of change of the general linear function y 5 ax 1 b? Describe its graph. b. We have seen that the average rate of change of a quadratic function y 5 ax2 1 bx 1 c can be represented by a linear function of the form y 5 2ax 1 b (where a 2 0). Describe the linear graph. Could such a graph ever represent the average rate of change of a linear function? c. What would you guess the equation representing the average rate of change of the cubic function y 5 ax 3 1 bx 2 1 cx 1 d to be?
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4. a. Complete the table for the function ƒ(x) 5 3x 2 2 2x 2 5.
x
y
23 22
28 11
21
0
0 1 2 3
25
Average Rate of Change of Average Rate of Change
Average Rate of Change n.a. 11 2 28 s22d 2 s23d 0 2 11 s21d 2 s22d
s211d 2 s217d s21d 2 s22d
5 211
8. Having found the matched pairs of graphs in Exercise 7, explain the relationship between the horizontal intercept of the linear function and the vertex of the quadratic function. 9. Construct a function that represents the average rate of change for each given function. a. ƒ(t) 5 3t 2 1 t b. g(x) 5 25x 2 1 0.4x 1 3 c. h(z) 5 2z 1 7
n.a. n.a.
5 217
497
56
10. Construct a function that represents the average rate of change for each given function. a. r(t) 5 23t2 2 7 c. w(x) 5 25x 1 0.25
b. What is the slope of the linear function that represents the average rate of change? How is this slope related to the average rate of change of the average rate of change? c. What type of function represents the average rate of change of ƒ(x)? 5. Complete the table for the function Q 5 2t2 1 t 1 1. a. Plot Q 5 2t2 1 t 1 1. What type of function is this? b. What does the third column tell you about the function that represents the average rate of change for Q? t
b. s(v) 5 14v2 2 32v 1 5
11. Determine from each of the tables whether you would expect the original function y 5 ƒ(x) to be linear, exponential, or quadratic. a.
Q Average Rate of Change
23 22 21 0 1 2 3
n.a.
7. Match the quadratic function with the graph of its average rate of change. y 6
6
y
y
6
x –6
6
x –6
6
–6
–6
–6
Graph A
Graph B
Graph C
y
y
6
y
6
6
x –6
6
x
y
21 0 1 2 3 4
22 0 8 22 42 68
n.a. 2 8 14 20 26
n.a. n.a. 6 6 6 6
x
y
Average Rate of Change
Average Rate of Change of Average Rate of Change
21 0 1 2 3 4
9 5 1 23 27 211
n.a. 24 24 24 24 24
n.a. n.a. 0 0 0 0
x
y
Average Rate of Change
Average Rate of Change of Average Rate of Change
21 0 1 2 3 4
0.25 1 4 16 64 256
n.a. 0.75 3 12 48 192
n.a. n.a. 2.25 9 36 144
x –6
6
–6
–6
Graph D
Graph E
c.
6
x
x –6
6 –6 Graph F
Average Rate of Change of Average Rate of Change
Average Rate of Change
b.
6. a. (Graphing program required.) Plot the function h(t) 5 4 1 50t 2 16t 2 for the restricted domain 0 # t # 3. b. For what interval is this function increasing? Decreasing? c. Estimate the maximum point. d. Construct and graph a function g(t) that represents the average rate of change of h(t). e. What does g(t) tell you about the function h(t)?
–6
d. m(p) 5 4p 2 12
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8.5 An Introduction to Polynomial Functions Defining a Polynomial Function Linear, power, and quadratic functions are all part of a larger family of functions called polynomials. We can think of a polynomial as a sum of power functions with positive integer powers. For example, if we sum the power functions y1 5 2x4, y2 5 x3, y3 5 22x, y 5 y1 1 y2 1 y3 1 y4 y 5 2x4 1 x3 2 2x 1 11
to get we obtain
and y4 5 11
This function is called a polynomial of degree 4, since the highest power of the independent variable (x) is 4. Each of the separate power functions is called a term of the polynomial. In standard form we arrange the terms from the highest power down to the lowest.
Definition of a Polynomial Function A polynomial function of degree n is of the form ƒ(x) 5 an x n 1 an21 x n21 1 . . . 1 a1 x 1 a0 where n is a nonnegative integer and an 2 0. The constants an, an21, . . . , a0 are called the coefficients. a n x n is called the leading term. a0 is called the constant term.
Polynomials of certain degrees have special names.
Polynomials of Degree
Are Called
0 1 2 3 4 5
Constant Linear Quadratic Cubic Quartic Quintic
Example y53 y 5 24x 2 8 y 5 3x 2 1 5x 2 10 y 5 5x 3 2 4x 2 14 y 5 22x 4 2 x 3 1 4x 2 1 4x 2 14 y 5 8x 5 2 3x 4 1 4x 2 2 4
Notice that the example for the cubic function, y 5 5x3 2 4x 2 14, does not include an x2 term. In this case the coefficient for the x2 term is zero, since this function could be rewritten as y 5 5x3 1 0x2 2 4x 2 14. Early algebraists believed that higher-degree polynomials were not relevant to the physical world and hence were useless. “Going beyond the cube just as if there were more than three dimensions . . . is against nature.”3 But as we shall see shortly, there are real applications for polynomials with degrees greater than 3.
3
Stifel, as cited by M. Kline, Mathematical Thought from Ancient to Modern Times (Oxford: Oxford University Press, 1972).
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EXAMPLE
499
1
Identifying function types Specify the type(s) of each function (linear, exponential, power, quadratic, and/or polynomial). If it’s polynomial, identify the degree and any special name. a. y 5 3x 1 5 c. Q(t) 5 5 ? 2 t e. g(z) 5 2(z2 2 1)(z2 1 3) 2 2 5 d. v 5 1 2 13w 2 5w 1 2w b. ƒ(x) 5 4x
SOLUTION
All of the functions are polynomials, except for (c). a. This function is a polynomial of degree 1, so it’s a linear function. b. This function is a polynomial of degree 2, so it’s quadratic. It’s also a power function. c. This is an exponential function. d. This function is a polynomial of degree 5, called a quintic. e. Multiplying out, we get g(z) 5 2z4 1 4z2 2 6. So this function is a polynomial of degree 4, called a quartic.
EXAMPLE
2
Curing poison ivy A patient with an acute poison ivy rash is given a 5-day “prednisone taper.” The daily dosage of the drug prednisone is respectively 20 mg, 15 mg, 10 mg, 5 mg, and 5 mg. The daily decay rate r (in decimal form) is the rate at which the drug is naturally absorbed and removed from the body. The rate varies from patient to patient. The decay factor, x 5 1 2 r, gives the percentage (in decimal form) of the drug left in the patient’s body after each day. a. If one patient has a prednisone decay rate of 0.35 (or 35%) per day, what is the patient’s decay factor? If another has a decay rate of 50%, what would this patient’s decay factor be? b. Construct functions to describe a patient’s prednisone level as a function of x, the decay factor, for each of the 5 days of treatment. c. Use technology to graph the prednisone level on day 5. What type of function is this, and what is its domain in this context? d. If a patient’s daily decay factor is 40%, use the graph to estimate the total dosage of prednisone remaining on day 5. Then use the equation to calculate a more exact answer. e. Construct a function to describe the patient’s prednisone level after 6 days, after 7 days, and after n days. How do the functions for which n . 5 differ from those with n # 5?
SOLUTION
a. A decay rate of 0.35 corresponds to a decay factor of 0.65. A decay rate of 0.50 (or 50%) corresponds to a decay factor of 0.50. b. On day 1, the patient would receive 20 mg of prednisone. Then on day 2 the patient would have 20x remaining of the original 20 mg of prednisone, plus the additional 15 mg he took that day. On day 3 he’d have 20x2 left of his 20-mg dosage from day 1, plus 15x left from his 15-mg dosage from day 2, plus the 10 mg he took that day. And so on. If we let Dayi (x) 5 total amount of prednisone in the body for day i, then Day1(x) 5 20 Day2(x) 5 20x 1 15 Day3(x) 5 20x2 1 15x 1 10
Day4(x) 5 20x3 1 15x2 1 10x 1 5 Day5(x) 5 20x4 1 15x3 1 10x2 1 5x 1 5
c. The graph of Day5(x) is given in Figure 8.35. The function is a polynomial of degree 4. Since its domain is restricted to decimal values of decay rates, 0 , x , 1. d. Estimating from the graph, a patient with a decay factor of 0.40 has about 10 mg of prednisone left in the body on day 5. Evaluating with a calculator, the same patient on day 5 has Day5(0.40) 5 20(0.40)4 1 15(0.40)3 1 10(0.40)2 1 5(0.40) 1 5 < 10.07 mg. e. When n . 5 there are no more added dosages. Thus Day6(x) 5 Day5(x) ? x, and in general Dayn(x) 5 Dayn21(x) ? x, if n . 5. So for n $ 5, Dayn(x) 5 20xn21 1 15x n22 1 10x n23 1 5x n24 1 5x n25.
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Amount of prednisone (mg)
50
Day5(x)
(0.40, 10.07)
x 0
1 Daily decay factor
Figure 8.35 Amount of prednisone in the
body on day 5 for various decay factors.
Algebra Aerobics 8.5a 1. For each of the following polynomials, specify the degree and evaluate each function at x 5 21. a. ƒ(x) 5 4x 3 1 11x 5 2 11 c. g(x) 5 22x 4 2 20 b. y 5 1 1 7x 4 2 5x 3 d. z 5 3x 2 4 22x 2 2. Find the degree of the polynomial function without multiplying out. a. ƒ(x) 5 x 5(3x 2 2)3 (5x 2 1 1) b. g(x) 5 (x 2 2 1)(x 3 1 x 2 5)4
3. Given the polynomial function ƒ(t) 5 0.5 2 2t 5 1 4t 3 2 6t 2 2 t: a. What is the degree of the polynomial? What is the name for polynomials of this degree? b. What is the leading term? c. What is the constant term? d. What is ƒ(0)? ƒ(0.5)? ƒ(21)?
Visualizing Polynomial Functions What can we predict about the graph of a polynomial function from its equation? Examine the graphs of polynomials of different degrees in Figure 8.36. What can we observe from each of these pairs? Turning points The first thing we might notice is the number of turning points on each graph—the number of times the graph bends and changes direction. The quadratics bend once, the cubics seem to bend twice, the quartics three times, one quintic seems to bend four times, and the other quintic appears to bend twice. In general, a polynomial function of degree n will have at most n 2 1 turning points. Horizontal intercepts Second, we might notice the number of times each graph crosses the horizontal axis. Each quadratic crosses at most two times, the cubics each cross at most three times, the quartics cross at most four times, and the quintics cross at most five times. In general, a polynomial function of degree n will cross the horizontal axis at most n times. Global behavior Finally, imagine zooming out on a graph to look at it on a global scale. All the polynomials in Figure 8.36 were graphed using small values for x (at most between 210 and 10) in order for us to see all the turning points and horizontal intercepts. What
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y
y 200
10
y = –2x3 – 14x2 – 6x – 42 y = –2x2 – 12x – 16
y = 0.5x2 – 2x + 6 x
x –5
–10
5
10
y = 7x3 – 147x2 + 1001x – 2205 –200
–10 (a) Two quadratics
(b) Two cubics
y = x4 + 3x3 – 10x2 – 24x y
y 200
200
y = 0.25(x – 5)(x – 3)x(x + 3)(x + 5)
x
x –10
10
–10
10
y = –x4 + 24x2 + 25 –200
–200 (c) Two quartics
y = –0.1(x +7)(x4 + 100)
(d) Two quintics
Figure 8.36 Graphs of pairs of polynomial functions (a) of degree 2 (quadratics),
(b) of degree 3 (cubics), (c) of degree 4 (quartics), and (d ) of degree 5 (quintics).
if we graphed the functions using a much wider range of x values, say between 21000 and 1000 or between 21,000,000 and 1,000,000? Figure 8.37 shows two graphs of the same quintic, y 5 0.25(x 2 5)(x 2 3)x(x 1 3)(x 1 5), that is graphed in Figure 8.36(d). The scale for x on the left-hand graph is 210 to 10. The scale for x on the right-hand graph is 21000 to 1000. On the larger-scale graph on the right, the global behavior is clear, but the five horizontal intersection points and four turning points are no longer visible. The dominant features now are the arms of polynomials that extend infinitely upward on the right and downward on the left. Their direction is dictated by the leading term, which eventually dominates all the other terms. So at this large scale, the graph of our quintic function y 5 0.25(x2 5)(x 2 3)x(x 1 3)(x 1 5) looks like the graph of y 5 0.25x 5.
y
1.0 • 1014
200
y
x –10
10
–200 Graph A
x –1000
1000
1.0 • 10–14 Graph B
Figure 8.37 Two graphs of the same quintic using different
scales.
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For polynomial functions of odd degree, the two arms extend in opposite directions, one up and one down. For polynomial functions of even degree, both arms extend in the same direction, either both up or both down (see Figure 8.36). This is because given a polynomial ƒ(x) 5 a n x n 1 an21x n21 1 c 1 a1x1 1 a0 of degree n, for large values of x the values of the leading term an x n will dominate the values of the other terms of smaller degree. In other words, for large values of x the function behaves like the power function g(x) 5 an x n. The leading term of a polynomial determines its global shape. The Graph of a Polynomial Function The graph of a polynomial function of degree n will have at most n 2 1 turning points cross the horizontal axis at most n times For large values of x, the graph of ƒ(x) 5 an x n 1 an21x n21 1 g1 a1x 1 a0 will resemble the graph of the power function g(x) 5 an x n.
EXAMPLE
3
What are the number of visible turning points and the number of horizontal intercepts for each polynomial graphed in Figure 8.38? What is the minimal degree for each polynomial function?
20
y
10
y
10
y
x –10
10
x –10
10
x –10
10 –10 Graph A
–20 Graph B
–20 Graph C
Figure 8.38 Graphs of three polynomial functions.
SOLUTION
a. Graph A has two turning points, three horizontal intercepts, and the global shape of an increasing odd power function. This is at least a third-degree polynomial with a positive leading term. b. Graph B has three turning points but only two horizontal intercepts and has the global shape of an even power function. This is at least a fourth-degree polynomial with a positive leading term. c. Graph C has four turning points but only one horizontal intercept. It has the global shape of a decreasing odd power function. This is at least a fifth-degree polynomial with a negative leading term.
Finding the Vertical Intercept The polynomial function ƒ(x) 5 an x n 1 an21x n21 1 c 1 a1x 1 a0 has a vertical intercept at a0, since ƒ(0) 5 a0. For example, the function ƒ(x) 5 x 4 2 16 has a vertical intercept at 216 (see Figure 8.39).
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y 60
x –3
3 –20
Figure 8.39 The graph of ƒ(x) 5 x 4 2 16
has a vertical intercept at 216 and two horizontal intercepts at 2 and 22.
Finding the Horizontal Intercepts As with any function, to find the zeros of a polynomial function ƒ(x), we set ƒ(x) 5 0 and solve for x. If a zero is a real number, it represents a horizontal intercept. For example, revisiting the relatively simple function ƒ(x) 5 x4 2 16, we can set ƒ(x) 5 0 and solve the corresponding equation for x to find the zeros. Setting ƒ(x) 5 0 we have adding 16 to both sides taking the fourth root
0 5 x4 2 16 16 5 x4 x 5 62
So 22 and 12 are both real zeros, and hence x-intercepts, for the function ƒ(x) 5 x4 2 16 (see Figure 8.39). But for a general polynomial function ƒ(x), solving the equation resulting from setting ƒ(x) 5 0 can be difficult. There is no simple analogue to the quadratic formula for polynomials of degree $3. The formulas for finding the zeros for third- and fourth-degree polynomials are extremely complicated. For polynomials of degree 5 or higher, there are no general algebraic formulas for finding the zeros. But there are algebraic approximation methods that allow us to compute the values for the real zeros (the horizontal intercepts) accurate to as many decimal places as we wish. Estimating the horizontal intercepts The complicated algebraic strategies for finding real zeros for a polynomial function of degree $3 are beyond the scope of this course. However, we can estimate the horizontal intercepts from a graph. EXAMPLE
4
Estimating Estimate the horizontal intercepts of the polynomial function in Figure 8.40. What is the minimum degree of the polynomial? 4
t –4
6
–3
Figure 8.40 Graph of a polynomial
function Q(t).
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SOLUTION
EXAMPLE
The t-intercepts are at approximately 23.5, 20.25, 2, and 4. Since the polynomial has four horizontal intercepts and three turning points, it must be at least of degree 4.
5
Zooming in on a graph Suppose you make four yearly deposits of $2000 in a risky high-yield account, starting today. What annual interest must you earn if you want to have $10,000 in the account after 4 years?4
SOLUTION
Let’s track the four separate $2000 deposits made in years 1, 2, 3, and 4. If r is the annual interest rate (in decimal form), then x 5 1 1 r is the annual multiplier or the growth factor. The first $2000 invested at the beginning of year 1 will be multiplied by x, the annual growth factor, four times, once for each year. So at the end of 4 years this $2000 investment will be worth 2000x4 dollars. The second $2000 invested at the beginning of year 2 will only have 3 years in which to earn interest, years 2, 3, and 4. At the end of year 4, the second $2000 investment will have been multiplied by the growth factor x only three times, making it worth 2000x 3 dollars. Similarly, the $2000 investments made at the beginning of years 3 and 4 will be worth 2000x 2 and 2000x dollars, respectively, by the end of year 4. So at the end of the fourth year, assuming you do not make a final deposit, the total dollars in your account are 2000x 4 1 2000x 3 1 2000x 2 1 2000x Since you want to have $10,000 in your account after 4 years, our equation is 10,000 5 2000x 4 1 2000x 3 1 2000x 2 1 2000x
(1)
We need to solve Equation (1) for x to find the annual interest rate r (where x 5 1 1 r). If we subtract 10,000 from each side and then divide by 2000, we have 0 5 x4 1 x3 1 x2 1 x 2 5
(2)
A solution to Equation (2) will be a value x for the function ƒ(x) 5 x 4 1 x 3 1 x 2 1 x 2 5 such that ƒ(x) 5 0. So x-intercepts for ƒ(x) correspond to real solutions to Equation (2). We can estimate the x-intercepts of ƒ(x) by graphing the function (Figure 8.41) and then zooming in (Figure 8.42).
60
y y 4
x
0
30
(1.09, 0)
2
x –3
3 –10
Figure 8.41 Graph of
ƒ(x) 5 x4 1 x3 1 x2 1 x 2 5.
4
–10
Figure 8.42 Zooming in on the x-intercept between 1 and 2.
Example modeled after a problem in E. Connally, D. Hughes-Hallett, A. M. Gleason, et al., Functions Modeling Change: A Preparation for Calculus, 2nd ed. (New York: John Wiley & Sons, 1998), pp. 384–385.
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Only positive solutions have meaning in this model, and there is only one positive x-intercept, at approximately (1.09, 0). We have x 5 1 1 r, where r is the annual interest rate. Since x 5 1.09, then r 5 x 2 1 5 0.09 in decimal form, or 9% written as a percentage. So you would need an annual interest rate of approximately 9% to end up with $10,000 when investing $2000 each year for 4 years. Using factoring to find the horizontal intercepts The Factor Theorem extends to polynomials in general.
The Factored Form of a Polynomial If r1, r2, . . . , rn are zeros of a polynomial ƒ(x) 5 an x n 1 an21x n21 1 g1 a1x 1 a0, then ƒ(x) can be written in factored form as ƒ(x) 5 an(x 2 r1)(x 2 r2) ? g? (x 2 rn) Any real zero is an x-intercept for ƒ(x).
So if you are given the zeros, it’s easy to construct a function in factored form with those zeros. Conversely, if you have a function in factored form, it is easy to find the zeros.
EXAMPLE
6
Finding horizontal intercepts a. What are the t-intercepts of the function Q(t) 5 1.5t(t 2 7)(t 2 5)(t 1 3)? b. For large values of t (both positive and negative), the graph of Q(t) resembles what power function? Would the left arm of the parabola eventually extend up or down? And the right arm?
SOLUTION
a. The factors of Q(t) tell us it has real zeros and hence t-intercepts at 0, 7, 5, and 23. b. For large values of t, the leading term will dominate. If we put Q(t) into standard form, the leading term will be 1.5t 4, so in the long run Q(t) behaves like the power function ƒ(t) 5 1.5t 4. So both arms of the graph of Q(t) will eventually extend upward.
EXAMPLE
7
SOLUTION
Constructing polynomial functions Create a polynomial function with x-intercepts at 0, 22, and 1. If x 5 0 is an x-intercept, then x is a factor of the polynomial function. If x 5 22 is an x-intercept, then (x 2 (22)), or (x 1 2), is also a factor. If x 5 1 is an x-intercept, then (x 2 1) is a factor as well. So the function ƒ(x) 5 x(x 1 2)(x 2 1) has x-intercepts at 0, 22, and 1. There are many other polynomial functions with the same horizontal intercepts, as we will see in the next example.
EXAMPLE
8
Multiple functions with the same horizontal intercepts a. Describe two other polynomial functions of degree 3 that have x-intercepts at 0, 22, and 1. b. Construct one that passes through the point (2, 216).
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SOLUTION
a. The functions g(x) 5 3x(x 1 2)(x 2 1) and h(x) 5 25x(x 1 2)(x 2 1) both have x-intercepts at 0, 22, and 1 (see Figure 8.43). Note that both are multiples of the function ƒ(x) 5 x(x 1 2)(x 2 1); that is, g(x) 5 3ƒ(x) and h(x) 5 25ƒ(x). In general, any function of the form aƒ(x) 5 ax(x 1 2)(x 2 1) where a 2 0 will have x-intercepts at 0, 22, and 21. So there are infinitely many possibilities. (See Figure 8.43.)
15
g(x)=3f (x)
f (x)
x –4
4
h(x) = –5f (x)
–15
Figure 8.43 The graphs of three
polynomial functions all with x-intercepts at 0, 22, and 1.
b. Given the general form of a polynomial with x-intercepts at 0, 22, and 1, ƒ(x) 5 ax(x 1 2)(x 2 1) If we substitute in x 5 2 and f(x) 5 216 216 5 a(2)(2 1 2)(2 2 1) simplify 216 5 a(8) and solve for a, we have a 5 22 So the polynomial function ƒ(x) 5 22x(x 1 2)(x 2 1) has x-intercepts at 0, 22, and 1 and passes through the point (2, 216) (see Figure 8.44).
y 5
x –10
10
(2, –16) –20
Figure 8.44 A polynomial with
x-intercepts at 0, 22, and 1 that passes through (2, 216).
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?
SOMETHING TO THINK ABOUT
Are the only functions with x-intercepts at 0, 22, and 1 of the form k ? f(x), where k is a constant and f(x) 5 x(x 1 2)(x 2 1)?
507
From Examples 7 and 8, we see that we can create a polynomial function with any finite set of real numbers as its zeros. Looking at it another way, we can create new polynomials by multiplying together existing polynomials.
A Summary of Polynomial Behavior Given a polynomial of the form ƒ(x) 5 an x n 1 an21x n21 1 g1 a1x 1 a0, The graph has at most n 2 1 turning points and n horizontal intercepts. For large values of x, the graph will resemble the graph of g(x) 5 an x n. The zeros of ƒ(x) are the values of x such that ƒ(x) 5 0. If r1, r2, r3, . . . , rn are all the zeros of a function ƒ(x), then ƒ(x) can be written in factored form as ƒ(x) 5 an(x 2 r1)(x 2 r2) ? g? (x 2 rn). Each real zero corresponds to an x-intercept. The vertical intercept is at ƒ(0) 5 a 0.
Algebra Aerobics 8.5b 3. For each of the following functions, identify the vertical intercept. Then use a function graphing program (and its zoom feature) to estimate the number of horizontal intercepts and their approximate values. a. y 5 3x 3 2 2x 2 2 3 b. ƒ(x) 5 x 2 1 x 1 3 4. Identify the degree and the x-intercepts of each of the following polynomial functions. If technology is available, graph each function to verify your work. a. y 5 3x 1 6 b. ƒ(x) 5 (x 1 4)(x 2 1) c. g(x) 5 (x 1 5)(x 2 3)(2x 1 5) 5. Construct three polynomial functions, all with x-intercepts at 23, 0, 5, and 7. Use technology, if available, to plot them.
A graphing program is required for Problems 1 and 2, and recommended for Problems 4 and 5. 1. Use technology to graph the polynomial function y 5 22(x 2 3)2(x 1 4). a. What is the degree of the function? b. How many turning points does it have? c. What happens to y as x S 1`? As x S 2`? d. The graph looks like what power function for very large positive or negative values of x? e. What are the horizontal intercepts? f. When x 5 0, what is y? What is this point called? 2. Determine the lowest possible degree and the sign of the leading coefficient for the polynomial functions in the three graphs below. Justify your answers. (Hint: Use your knowledge of power functions.)
y
y
12
y
12
10
x –10
10
–10 Graph A
x –12
10
x –12
10
–10
–10
Graph B
Graph C
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Exercises for Section 8.5 y
y
A calculator that can evaluate powers is required. Several exercises require a graphing program.
5
25
y 10
1. Identify which of the following are polynomial functions and, for those that are, specify the degree. a. y 5 3x 1 2 d. y 5 3x 2 2 e. y 5 3x5 2 4x 3 2 6x 2 2 12 b. y 5 2 2 x 3 c. y 5 5t 2 2 3 "t5
x4 b. y 5 2x 5 1 2 3x 3
d. y 5 2t 3 1 5t 4 1 "2
3. Evaluate the following expressions for x 5 2 and x 5 22. b. 4x23 c. 24x23 d. 24x3 a. x23 4. Evaluate the following polynomials for x 5 2 and x 5 22, and specify the degree of each polynomial. a. y 5 3x 2 2 4x 1 10 b. y 5 x 3 2 5x 2 1 x 2 6 c. y 5 22x 4 2 x 2 1 3 5. Match each of the following functions with its graph. c. y 5 (x2 1 1)(x2 2 4) a. y 5 2x 2 3 b. y 5 3(2x) 60
y
y 25
3
x –3
30
3
x
x
–9
–4
Graph A
–10
4
–3
Graph B
0
3
Graph C
6. Match each of the following functions with its graph. c. ƒ(x) 5 212 x 3 1 x 2 3 a. ƒ(x) 5 x 2 1 3x 1 1 b. ƒ(x) 5 13 x 3 1 x 2 3 40
y
y
y 40
60
x –5
x
5
–5
x –50 Graph A
–5
5
x –10
10
–10 Graph B
5
5 –60 Graph C
7. For each of the graphs of polynomial functions, at the top of the next column, determine (assuming the arms extend indefinitely in the indicated direction): i. The number of turning points ii. The number of x-intercepts iii. The sign of the leading term iv. The minimum degree of the polynomial
x –10
f. y 5 6(x)(x 2 5)(2x 1 7)
2. Identify the degree of any of the following functions that are polynomials, and for those that are not polynomials, explain why. 2 a. y 5 3x 2 2 3 1 1 c. y 5 x5/3 1 x 2 x
y
x –5
–10 Graph A
–15 Graph B
10
–15 Graph C
8. Describe how g(x) and h(x) relate to ƒ(x). ƒ(x) 5 x 5 2 3x 2 1 4 g(x) 5 2x 5 1 3x 2 2 4 h(x) 5 x 5 2 3x2 2 2 9. Divide the following functions into groups having the same global shape for large values of x. Explain your groupings. a. y 5 (x2 2 3)(x 1 9) b. y 5 2x4 1 3 c. y 5 x5 2 3x4 2 11x3 1 3x2 1 10x d. y 5 x(3 2 x)(x 1 1)2 e. y 5 7x3 2 3x2 2 20x 1 5 f. y 5 2(x2 1 1)(x2 2 4) 10. Estimate the maximum number of turning points for each of the polynomial functions. If available, use technology to graph the function to verify the actual number. a. y 5 x 4 2 2x 2 2 5 c. y 5 x 3 2 3x 2 1 4 6 2 d. y 5 5 1 x b. y 5 4t 1 t 11. Describe the behavior of each polynomial function for large values (positive or negative) of the independent variable and estimate the maximum number of turning points. If available, use technology to verify the actual number. a. y 5 22x 4 1 4x 1 3 b. y 5 (t 2 1 1)(t 2 2 1) c. y 5 x 3 1 x 1 1 d. y 5 x 5 2 3x 4 2 11x 3 1 3x 2 1 10x 12. (Graphing program required.) Estimate the maximum number of horizontal intercepts for each of the polynomial functions. Then graph the function using technology to find the actual number. (See Exercise 10.) a. y 5 x 4 2 2x 2 2 5 c. y 5 x 3 2 3x 2 1 4 6 2 d. y 5 5 1 x b. y 5 4t 1 t 13. (Graphing program required.) Estimate the maximum number of horizontal intercepts for each of the polynomial functions. Then, using technology, graph the functions to find their approximate values. a. y 5 22x 2 1 4x 1 3 b. y 5 (t 2 1 1)(t 2 2 1) c. y 5 x 3 1 x 1 1 d. y 5 x 5 2 3x 4 2 11x 3 1 3x 2 1 10x
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14. a. (Graphing program required.) Use a function graphing program to estimate the x-intercepts for each of the following. Make a table showing the degree of the polynomial and the number of x-intercepts. What can you conclude? y 5 2x 1 1 y 5 x 3 25x 2 1 3x 1 5 2 y 5 0.5x 4 1 x 3 2 6x 2 1 x 1 3 y 5 x 2 3x 2 4 b. Repeat part (a) for the following functions. How do your results compare with those for part (a)? Are there any modifications you need to make to your conclusions in part (a)? y 5 3x 1 5 y 5 x 3 22x 2 2 4x 1 8 2 y 5 (x 2 2) 2 (x 1 1) 2 y 5 x 1 2x 1 3 15. (Graphing program required.) Use a function graphing program (and its zoom feature) to estimate the number of x-intercepts and their approximate values for: a. y 5 3x 3 2 2x2 2 3 b. ƒ(x) 5 x 2 1 5x 1 3 16. (Graphing program required.) Identify the x-intercepts of the following functions; then graph the functions to check your work. a. y 5 3x 1 6 b. y 5 (x 1 4)(x 2 1) c. y 5 (x 1 5)(x 2 3)(2x 1 5) 17. a. If the degree of a polynomial is odd, then at least one of its zeros must be real. Explain why this is true. b. Sketch a polynomial function that has no real zeros and whose degree is: i. 2 ii. 4 c. Sketch a polynomial function of degree 3 that has exactly: i. One real zero ii. Three real zeros d. Sketch a polynomial function of degree 4 that has exactly two real zeros. 18. In each part, construct a polynomial function with the indicated characteristics. a. Crosses the x-axis at least three times b. Crosses the x-axis at 21, 3, and 10 c. Has a y-intercept of 4 and degree of 3 d. Has a y-intercept of 24 and degree of 5 19. Which of the following statements are true about the graph of the polynomial function ƒ(x) 5 x 3 1 bx 2 1 cx 1 d a. b. c. d. e.
It intersects the vertical axis at one and only one point. It intersects the x-axis in at most three points. It intersects the x-axis at least once. The vertical intercept is positive. For large (positive or negative) values of x, the graph looks like y 5 x 3. f. The origin is a point on the graph. 20. (Graphing program required.) A manufacturer sells children’s wooden blocks packed tightly in a cubic tin box with a hinged lid. The blocks cost 3 cents a cubic inch to make. The box and
509
lid material cost 1 cent per square inch. (Assume the sides of the box are so thin that their thickness can be ignored.) It costs 2 cents per linear inch to assemble the box seams. The hinges and clasp on the lid cost $2.50, and the label costs 50 cents. (See the accompanying figure.)
s s
s
A cubic box with edge length s filled with blocks. a. If the edge length of the box is s inches, develop a formula for estimating the cost C(s) of making a box that’s filled with blocks. b. Graph the function C(s) for a domain of 0 to 20. What section of the graph corresponds to what the manufacturer actually produces—boxes between 4 and 16 inches in edge length? c. What is the cost of this product if the cube’s edge length is 8 inches? d. Using the graph of C(s), estimate the edge length of the cube when the total cost is $100. 21. Polynomial expressions of the form an3 1 bn2 1 cn 1 d can be used to express positive integers, such as 4573, using different powers of 10: 4573 5 4 ? 1000 1 5 ? 100 1 7 ? 10 1 3 or 4573 5 4 ? 10 3 1 5 ? 10 2 1 7 ? 10 1 1 3 ? 100 or if n 5 10, 4573 5 4 ? n 3 1 5 ? n 2 1 7 ? n1 1 3 ? n0 5 4n3 1 5n 2 1 7n 1 3 Notice that in order to represent any positive number, the coefficient multiplying each power of 10 must be an integer between 0 and 9. a. Express 8701 as a polynomial in n assuming n 5 10. b. Express 239 as a polynomial in n assuming n 5 10. Computers use a similar polynomial system, called binary numbers, to represent numbers as sums of powers of 2. The number 2 is used because each minuscule switch in a computer can have one of two states, on or off; the symbol 0 signifies off, and the symbol 1 signifies on. Each binary number is built up from a row of switch positions each set at 0 or 1 as multipliers for different powers of 2. For instance, in the binary number system 13 is represented as 1 1 0 1, which stands for 1 ? 2 3 1 1 ? 2 2 1 0 ? 21 1 1 ? 20 5 1? 811? 410? 211? 15 13
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c. What number does the binary notation 1 1 0 0 1 represent? d. Find a way to write 35 as the sum of powers of 2; then give the binary notation. 22. A typical retirement scheme for state employees is based on three things: age at retirement, highest salary attained, and total years on the job. Annual retirement allowance 5 a
retirement % of highest total years b ? a b ? a b age factor salary worked
where the maximum percentage is 80%. The highest salary is typically at retirement. We define: total years worked 5 retirement age 2 starting age
a. For an employee who started at age 30 in 1973 with a salary of $12,000 and who worked steadily, receiving a $2000 raise every year, find a formula to express retirement allowance, R, as a function of employee retirement age, A. b. Graph R versus A. c. Construct a function S that shows 80% of the employee’s salary at age A and add its graph to the graph of R. From the graph estimate the age at which the employee annual retirement allowance reaches the limit of 80% of the highest salary. d. If the rule changes so that instead of highest salary, you use the average of the three highest years of salary, how would your formula for R as a function of A change?
retirement age factor 5 0.001 ? (retirement age 2 40) salary at retirement 5 starting salary 1 all annual raises
8.6 New Functions from Old We’ve seen how the simple quadratic y 5 x2 can be transformed (through stretching, compressing, and shifting) to generate all quadratic functions. We’ve also added power functions to generate polynomial functions. In the next three sections we’ll look at many ways of transforming any function or combining any two or more functions to create new functions. Our focus, however, will be on combining members from the families of functions we have studied: linear, exponential, logarithmic, power, quadratic, and polynomial.
Transforming a Function Stretching, compressing, and shifting Recall from our discussion in Section 8.2 that to stretch or compress the graph of a function, we multiply the function by a constant. To shift the horizontal or vertical position of its graph, we add (or subtract) a constant to either the input or the output of the function.
Stretching, Compressing, or Shifting the Graph of f(x) To stretch or compress: Multiply the output of ƒ(x) by a constant to get aƒ(x). If k a k . 1, the graph of ƒ(x) is vertically stretched by a factor of a. If 0 , k a k , 1, the graph of ƒ(x) is vertically compressed by a factor of a. To shift vertically:
Add a constant to the output of ƒ(x) to get ƒ(x) 1 k. If k is positive, the graph of ƒ(x) is shifted up. If k is negative, the graph of ƒ(x) is shifted down.
To shift horizontally:
Subtract a constant from the input of ƒ(x) to get ƒ(x 2 h). If h is positive, the graph of ƒ(x) is shifted to the right. If h is negative, the graph of ƒ(x) is shifted to the left.
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Reflections across the horizontal axis When we studied power functions, we learned that the graphs of y 5 kx p and y 5 2kx p are reflections of each other across the x-axis. What happens to the equation of any function when we reflect its graph across the horizontal axis? Figure 8.45 shows two functions, ƒ and g, that are reflections of each other across the x-axis. We can see from the graph that when a point is reflected vertically across the x-axis, the x value stays fixed while the y value changes sign. This means that for each input, the corresponding outputs for f and g are opposites; that is, they have the same magnitude but opposite signs, and are on opposite sides of the x-axis. So g(x) 5 2ƒ(x). y
f (x) x g(x)
Figure 8.45 The graph of g(x) 5 2ƒ(x) is a reflection of the graph of ƒ(x) across the horizontal axis.
Reflections across the vertical axis What if we reflected a function’s graph across the vertical axis? What happens to the equation of the original function? Figure 8.46 shows two functions, ƒ and g, that are reflections of each other across the y-axis. We can see from the graph that when a point is reflected horizontally across the y-axis, the y value stays fixed while the x value changes sign. This means that when ƒ and g have the same output, then the corresponding inputs, or x values, are opposites; that is, they have the same magnitude but opposite signs, and are on opposite sides of the y-axis. So g(x) 5 ƒ(2x). g(x)
y
f (x)
x
Figure 8.46 The graph of
g(x) 5 ƒ(2x) is a reflection of the graph of ƒ(x) across the vertical axis.
Reflections across both the horizontal and vertical axes The net result of reflecting the graph of ƒ(x) across both the x- and y-axes is equivalent to rotating the graph 1808 about the origin, as shown in Figure 8.47. This is a double reflection, so we need to multiply both the input and the output by 21 to get 2ƒ(2x).
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y f (x) f (–x) x –f (–x)
Figure 8.47 The graph of
2f (2x) is the graph of ƒ(x) reflected across the y-axis and then across the x-axis.
Reflecting the graph of f(x) To reflect across the x-axis:
?
SOMETHING TO THINK ABOUT
Can a function be symmetric across the x-axis? Why or why not?
Multiply the output of ƒ(x) by 21 to get 2ƒ(x).
To reflect across the y-axis:
Multiply the input of ƒ(x) by 21 to get ƒ(2x).
To reflect across both the x- and y-axes:
Multiply both the input and output of ƒ(x) by 21 to get 2ƒ(2x).
Symmetry If we reflect the graph of ƒ(x) across the y-axis and get the graph of ƒ(x) again, then ƒ(x) is symmetric across the y-axis. Symmetry across the y-axis means that the part of the graph to the left of the y-axis is the mirror image of the part to the right of the y-axis. More formally, it means ƒ(x) 5 ƒ(2x) for all values of x in the domain of ƒ. If after we rotate the graph of ƒ(x) 1808 about the origin we get the graph of ƒ(x), then ƒ(x) is symmetric about the origin. We can get the same net result by reflecting the graph of ƒ(x) across both the x- and y-axes. So ƒ(x) 5 2ƒ(2x) for all values of x in the domain of ƒ. In Chapter 7 we learned that power functions with even integer powers are symmetric across the y-axis and those with odd integer powers are symmetric about the origin. Figure 8.48 shows the graphs of other functions that have symmetry.
y
y
x
(a) Symmetry across the y-axis
x
(b) Symmetry about the origin
Figure 8.48 Examples of symmetric graphs.
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Symmetry across the y-Axis and about the Origin If for all values of x in the domain of ƒ: ƒ(x) 5 ƒ(2x) ƒ(x) 5 2ƒ(2x)
EXAMPLE
1
then the graph of ƒ is symmetric across the y-axis then the graph of ƒ is symmetric about the origin
Reflections and symmetry a. Describe the functions g(x) and h(x) shown in Figure 8.49 in terms of transformations of ƒ(x).
y
y
y g(x)
f (x)
x
x
x h(x)
Figure 8.49 Two transformations of ƒ(x).
b. When two objects A and B collide, the force F1 exerted on object A is equal in magnitude and opposite in direction to the force F2 exerted on object B, as shown in Figure 8.50(a). The graph in Figure 8.50(b) shows F1 and F2 as functions of time. Describe F1 in terms of F2. Force During collision
F1 F1
F2
A
Time
B
F2
(a)
(b)
Figure 8.50 Objects colliding.
c. Identify any symmetries in the graphs in Figure 8.51.
y
y
x
x
Graph A
y
Graph B
Figure 8.51 Different symmetries.
x
Graph C
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SOLUTION
EXAMPLE
2
a. The graph of g is a reflection of the graph of ƒ across the y-axis, so g(x) 5 ƒ(2x). The graph of h is a refection of the graph of ƒ across both the x- and y-axes, so h(x) 5 2ƒ(2x). b. F1(t) 5 2F2(t) since their graphs are reflections of each other across the time axis. c. A is the graph of a function that is symmetric across the y-axis. B is the graph of a function that is symmetric about the origin. C is a graph that is symmetric across the x-axis, but it is not a function since there are two values of y for each x . 0. Shifting and stretching a. The time series in Figure 8.52 show two related measures of income inequality in the United States. Zero percent represents perfect equality. Construct an equation that approximates ZD1R in terms of ZD.
Income Inequality (USA)
% 22 21 20
ZD+R
19 18
ZD
17 16
2000
1990
1988
1986
1984
1982
1980
1978
1976
1974
1972
1970
1968
15 1966
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Figure 8.52 Measures of income inequality in the United States. Source: G. Kluge, Wealth and People: Inequality Measures (2002), www.poorcity.richcity.org/entkiss.htm.
b. Use the properties of logs and function transformations to predict how the graphs of the following functions will differ from each other. Confirm your predictions by graphing all of the functions on the same grid. y 5 log x SOLUTION
y 5 log 10x
y 5 log 100x
y 5 10 log x
a. A rough approximation is ZD1R 5 ZD 1 1.3. b. The graph of y 5 log 10x is the graph of y 5 log x shifted vertically up one unit, since Given Rule 1 of logs Evaluate log 10
y 5 log 10x 5 log x 1 log 10 5 log x 1 1
Similarly, the graph of y 5 log 100x is the graph of y 5 log x shifted vertically up two units, since y 5 log 100x 5 log x 1 log 100 5 log x 1 log(10 2) 5 log x 1 2 The graph of y 5 10 log x is the graph of y 5 log x stretched by a factor of 10. Figure 8.53 shows the graphs of the four functions on the same grid.
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515
y 5
y = 10 log x y = log 100x y = log 10x y = log x x
–1
3
–5
Figure 8.53 Four log
functions.
EXAMPLE
3
Multiple transformations a. Without using technology, sketch graphs of each of the following functions: ƒ(x) 5 x 3
g(x) 5 2ƒ(x 1 1)
h(x) 5 ƒ(2x) 2 2
b. Construct a new function for each the following transformations of g(t) 5 100(1.03) t: 3g(t 2 2) SOLUTION
EXAMPLE
4
and
y 5
0.5g(2t) 1 3
a. See Figure 8.54. b. Given g(t) 5 100(1.03)t 3g(t 2 2) 5 (3)(100)(1.03)(t22) 5 300(1.03)(t22) and 0.5g(2t) 1 3 5 (0.5)(100)(1.03)2t 1 3 5 50(1.03)2t 1 3 1 t 5 50 A1.03 B 13 < 50(0.97) t 1 3
g
h
f
x –3
3
–5
Figure 8.54 Two
transformations of ƒ(x).
Financial considerations: Horizontal shift of an exponential function Twins each receive $10,000 from their grandmother on their shared 21st birthday. One invests her money right away in high-grade corporate bonds that return an annual 6% in interest, which the sister keeps reinvesting. The other twin can’t decide what to do, so she puts her money in a non–interest-bearing checking account. At her 33rd birthday, after getting married and having kids, she decides she needs to save for the future. She now invests $10,000 in the equivalent bonds, receiving the same interest rate as her sister. a. Construct a function S1(t) that represents how much the first sister’s investment is worth t years after her 21st birthday. b. Use S1(t) to construct a function S2(t) for the second sister, who waits until she is 33 to invest her money. c. Graph S1(t) and S2(t) on the same grid. What are the differences and similarities between the graphs in practical terms? d. Use your graphs to estimate how long it will take for each investment to be worth $40,000. e. Use your graphs to estimate the difference in the worth of the sisters’ investments when they are both 65 years old.
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SOLUTION $200,000 Value of the investment
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$160,000
S1( t )
$120,000 $80,000 $40,000
S2( t ) t 0
10 20 30 40 50 Years since the twins turned 21
Figure 8.55 Twin investments.
EXAMPLE
5
a. S1(t) 5 10,000(1.06) t for t $ 0 b. S2(t) 5 S1(t 2 12) 5 10,000(1.06)(t212) for t $ 12 c. The graphs are drawn in Figure 8.55. The graph of S2 is the graph of S1 shifted 12 units (33 2 21 5 12) to the right. So at age 33, the second sister has $10,000 in her account, the same amount she had when she was 21 years old. Each successive year the second sister is 12 years behind her twin sister in terms of how much money her investment is worth. d. It will take about 24 years for each investment to be worth about $40,000. So the first twin would be about 45 and the second twin about 57, or 12 years older than her sister. Remember that S2(t) 5 S1(t 2 12), so S2(57) 5 S1(57 2 12) 5 S1(45). e. The difference between the investments when the twins are 65 years old (44 years after their 21st birthday) is about $65,000, since when t 5 44 years, S1(44) < $130,000 and S2(44) < $65,000. The first twin has almost twice as much as the second. Transformations of a power function a. Construct an equation and sketch a graph of the following transformations of ƒ(x) 5 1/x 2: 2ƒ(x 2 1)
3ƒ(x 1 2).
and
b. In what ways will the graph of g(x) 5 0.5ƒ(x 2 2) differ from the graph of ƒ(x)? How will it be similar? Check your predictions by graphing ƒ(x) and g(x) on the same grid. SOLUTION
a. The graphs are drawn in Figure 8.56, with dotted vertical lines indicating vertical asymptotes.
y
y
5
y
5
5
x –5
0 Graph of f (x) =
5 1
x2
x –5
0
5
Graph of 2f (x – 1) =
x –5
2 (x – 1)2
0 Graph of 3f (x + 2) =
5 3 (x + 2)2
Figure 8.56 Graph of ƒ(x) 5 1/x2 and two transformations of ƒ(x).
b. As shown in Figure 8.57, the graph of g(x) has the same overall shape (containing two parts) as the graph of ƒ(x), but g(x) is ƒ(x) compressed by a factor of 0.5 and shifted horizontally two units to the right. The y-axis is the vertical asymptote for ƒ(x), but the line x 5 2 is the vertical asymptote (dotted line) for g(x).
y 5
f ( x)
g( x)
x –5
0
Figure 8.57 Graphs of ƒ(x) 5 x12 and
g(x) 5 0.5ƒ(x 2 2).
5
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6
Cooling coffee: Vertical shift of an exponential function A cup of hot coffee (at 2008 Fahrenheit) is left to stand in a 708 Fahrenheit room. The difference in temperature between the coffee and the room can be modeled by an exponential function where the temperature decay rate is 10% per minute. a. Construct D(t), the temperature difference, as a function of time t in minutes that the coffee has been left standing. b. Using D(t), construct a new function C(t) that describes the coffee’s temperature as it gradually approaches the temperature of the room. c. What are the differences and similarities between the graphs of D(t) and C(t)? Check your answer by graphing D(t) and C(t) on the same grid. d. Use the graph of C(t) to estimate how long it will take for the coffee to cool to 1508F, a comfortable temperature for drinking.
SOLUTION
a. D(t) 5 130(0.9)t since the starting temperature difference is 2008 2 708 5 1308, and the decay factor is 1 2 0.1 5 0.9. b. C(t) 5 70 1 130(0.9)t c. The graph of C(t) is the graph of D(t) shifted vertically up 70 units (see Figure 8.58). Temperature (in degrees Fahrenheit)
EXAMPLE
517
200
C( t )
100
D( t ) t 0
10
20
40
30
Time (in minutes)
Figure 8.58
d. It will take approximately 5 minutes for the coffee to cool to 1508F.
Algebra Aerobics 8.6 Graphing program optional in Problems 6 and 7. 1. Graphs A, B, and C show f (x) and a transformation of f (x). Identify the transformation as: 2ƒ(x), ƒ(2x), or 2ƒ(2x).
3
2. Apply the transformations specified in parts (a)2(e) to ƒ(x) 5 2x 2 3 and ƒ(x) 5 1.5x. a. ƒ(x 1 2) b. 12 ƒsxd
e. 2ƒ(2x)
y
y
y
3
5
f
f
h
x –5
c. 2ƒ(x) d. ƒ(2x)
5
f x
x –5
5
–5
5
g
j –3
–5
–5
Graph A
Graph B
Graph C
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3. Identify which of the following graphs are symmetric across the vertical axis. Which are symmetric across the horizontal axis? Which are symmetric about the origin? y
y
P
q
x
Graph A
z
Graph B
Graph C
4. Construct a new function from h(t) 5 e t, where the graph of h(t) is: a. First shifted two units right
b. Then reflected across the t-axis c. Then shifted down by one unit 5. Construct a new function from Q(t) 5 2 ? 1.06 t where the graph of Q(t) is: a. First shifted two units to the left b. Then shifted down by one unit c. Then reflected across the t-axis (Graphing program optional.) Describe in each part of Problems 6 and 7 how g(x) is related to ƒ(x). If you have a graphing tool available, check your answer by looking at the graphs. g(x) 5 x 2 2 5 6. a. ƒ(x) 5 5 2 x 2 and x b. ƒ(x) 5 3 ? 2 and g(x) 5 3 ? 22x 1 7. a. ƒ(x) 5 x 2 and g(x) 5 3x 12 6 2 b. ƒ(x) 5 ln x and g(x) 5 ln 3 1 ln x 5 ln 3x
Exercises for Section 8.6 Graphing program optional for Exercises 7, 16, and 18, and required for Exercises 21 and 22. 1. In each part of the problem the graph of ƒ(x) to the left has been transformed into the graph of g(x) to the right. First describe whether the graph of ƒ(x) was stretched/compressed, reflected, and/or shifted vertically/horizontally to form g(x). Then write the equation for g(x) in terms of ƒ(x). y y a. 5
2. Match each of the following functions with its graph. Identify the parent (original) function p(x) and the transformation(s) that took place. a. ƒ(x) 5 (x 2 2)3 1 4 c. h(x) 5 2ln(x 1 1) d. k(x) 5 2e2x b. g(x) 5 2x 3 2 2 5
y
5
5
f (x) = 冑x
x –5
x –5
5
–5
5
y
5
y
f (x) = ex
x –5
–5
–5 Graph C
y
5
y
y
y 5
g (x)
f (x) = ln x x 5
x –5
5
x –5
5
5
–5
5
–5
–5
g (x)
5
–5
5
Graph A
5
x
c.
x –5
5
–5
–5
5
x –5
g (x)
b.
y
x –5
5
–5
–5
–5
Graph B
Graph D
3. Let ƒ(x) 5 x 3. a. Write the equation for the new function g(x) that results from each of the following transformations of ƒ(x). Explain in words the effect of the transformations. i. ƒ(2x) iii. ƒ(x 1 2) ii. 22ƒ(x) 2 1 iv. 2ƒ(2x) b. Sketch by hand the graph of ƒ(x) and each function in part (a).
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519
4. Explain in words the effect of the following transformations on the graph of g(t). a. 25g(t) c. 2g(2t) 2 4 1 b. 5 g(t 2 3) 1 1 d. 4g(2t) 2 2
d. For each function that you have identified as even or odd, what do you predict about the symmetry of its graph? If possible, check your predictions with a function graphing program.
5. Decide if each graph (although not necessarily a function) is symmetric across the x-axis, across the y-axis, and/or about the origin.
8. Use the function ƒ(x) to create a new function g(x) where the graph of g(x) is: a. The graph of ƒ(x) shifted three units to the left, then multiplied by 5, and finally shifted down by four units. b. The graph of ƒ(x) shifted three units to the right, then shifted up by four units, and finally multiplied by five.
y
y
x
x
9. For each function, construct a new function whose graph is the graph of the original function shifted left by two units, then multiplied by 13 , and then shifted down by five units. x a. ƒ(x) 5 60 A12B b. g(x) 5 12x 3 c. y 5 log x 1 10. If ƒ(x) 5 , evaluate: x
Graph A
Graph C
c. ƒ(t 2 1 2) d. ƒ(x 1 a) 2 ƒ(a)
a. ƒ(x 2 3) 1 5 b. 23ƒ(x 1 2) 1 12ƒ(x 2 1)
y
11. If ƒ(x) 5 a. b.
x
1 , determine: x2
ƒ(2 2 x) 2 1 1 2 ƒ(2t)
c. ƒ( "s 2 3) d. ƒ(x 1 h) 2 ƒ(x)
12. If ƒ(x) 5 1x , a. Describe the transformations of ƒ(x) used to create the new functions g(x), h(x), and k(x).
Graph B
6. Complete the partial graph shown in three different ways to create a graph that is: a. Symmetric across the x-axis b. Symmetric across the y-axis c. Symmetric about the origin y
x
g(x) 5
1 3x
h(x) 5 2
2 x23
k(x) 5
1 14 22x
b. Determine the domain of each function in part (a). c. Determine the equation of the vertical asymptote for each function in part (a). 13. Let k(s) 5 1s . Construct a new function j(s) that is the end result of the transformations of the graph of k(s) described in the following steps. Show your work for each transformation. a. First shift k(s) to the right by two units. b. Then compress your result by a factor of 1/3. c. Reflect across the s-axis. d. Finally, shift it up four units. 1 , construct an expression for: t2 a. p(t 1 1) c. 22(p(t 2 3) 1 5) b. 3p(t 1 2) 2 1
14. If p(t) 5
7. (Graphing program optional.) A function is said to be even if ƒ(2x) 5 ƒ(x) and odd if ƒ(2x) 5 2ƒ(x) for all x in ƒ’s domain. Use these definitions to: a. Show that the even power functions are even. b. Show that the odd power functions are odd. c. Show whether each of the following functions is even, odd, or neither. iii. h(x) 5 x4 1 x3 i. ƒ(x) 5 x 4 1 x 2 5 3 iv. g(x) 5 10.3 x ii. u(x) 5 x 1 x
15. Apply the transformations specified in parts (a)–(e) to 1 f(x) 5 ln x and f(x) 5 3 . x a. f(x 1 2) c. 2f(x) e. 2f(2x) b. 12 f(x) d. f (2x)
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16. (Graphing program optional.) a. Starting with the function f(x) 5 ex, create a new function g(x) by performing the following transformations. At each step show the transformation in terms of f(x) and e x, i. First shift the graph of f(x) to the left by three units, ii. Then compress your result by a factor of 1/4, iii. Next reflect it across the x-axis, iv. And finally shift it up by five units to create g(x). b. Graph f(x) and g(x) on the same grid. 17. Given the function g(t), identify the simplest function f(t) (linear, power, exponential, or logarithmic) from which g(t) could have been constructed. Describe the transformations that changed f(t) to g(t). t21 27 22 a. g(t) 5 c. g(t) 5 2 t25 1 t 14 b. g(t) 5 3 a b 2 18. (Graphing program optional.) a. On separate grids sketch the graphs of f(x) 5 "2x 1 2 and g(x) 5 2 "x 1 2. b. Using interval notation, describe the domains of f(x) and g(x). c. Using interval notation, describe the ranges of f(x) and g(x). d. What is the simplest function h(x) from which both f(x) and g(x) could be created? e. Describe the transformations of h(x) to obtain f(x). Of h(x) to obtain g(x). f. Does the graph of f(x) , g(x) , or h(x) have any symmetries (across the x- or y-axis, or about the origin)? 19. a. Given f(x) 5 ln x, describe the transformations that created g(x) 5 3f(x 1 2) 2 4. Find g(x). b. Use your knowledge of properties of logarithms to find any vertical and horizontal intercepts for the function g(x). 20. The following two graphs show the hours of daylight during the year for two different locations. One is for a latitude of 40 degrees above the equator (in the Northern Hemisphere), and the other for a latitude of 40 degrees below the equator (in the Southern Hemisphere). a. Which graph is associated with which hemisphere? b. Using the language of function transformation, describe Graph B in terms of Graph A. 24 Hours of daylight
20 16 12 8
March equinox
Sep equinox
4 0 Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Graph A
24 20 Hours of daylight
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16 12 8
March equinox
Sep equinox
4 0 Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Graph B
Source: www.vcaa.vic.edu.au/prep10/csf/support/sampleunits/ SolarHouseDaylight.xls.
21. (Graphing program required.) If an object is put in an environment with a fixed temperature A (the “ambient temperature”), then the object’s temperature, T, at time t is modeled by Newton’s Law of Cooling: T 5 A 1 Ce 2 k t , where k is a positive constant. (Note that T is a function of t and as t S 1`, then e2kt S 0 , so the temperature T of the object gets closer and closer to the ambient temperature, A.) A corpse is discovered in a motel room at midnight. The corpse’s temperature is 808 and the room temperature is 608. Two hours later the temperature of the corpse had dropped to 758. (Problem adapted from one in the public domain site S.O.S. Math.) a. Using Newton’s Law of Cooling, construct an equation to model the temperature T of the corpse over time, t, in hours since the corpse was found. b. Then determine the time of death. (Assume the normal body temperature is 98.68.) c. Graph the function from t 5 25 to t 5 5, and identify when the person was alive, and the coordinates where the temperature of the corpse was 98.68, 808, and 758. 22. (Graphing program required.) Newton’s Law of Cooling (see Exercise 21) also works for objects being heated. Suppose you place a frozen pizza (at 328) into a preheated oven set at 3508. Thirty minutes later the pizza is at 3208— ready to eat. a. Determine the constants A, C, and k in Newton’s Law. b. Sketch a graph of your function. c. From your graph, estimate when the pizza will be at 2008 and then calculate the time. d. According to your model, if you kept the pizza in the oven indefinitely, would the pizza ever reach 3508? What would be a reasonable domain for the function as a model for cooking pizza?
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8.7 Combining Two Functions We can combine functions the way we combine numbers; that is, we can add, subtract, multiply, or divide two functions.
The Algebra of Functions If ƒ(x) and g(x) are two functions with the same domain, we can define new functions by: Adding to get ƒ(x) 1 g(x) 5 (ƒ 1 g)(x) Subtracting to get ƒ(x) 2 g(x) 5 (ƒ 2 g)(x) Multiplying to get ƒ(x) ? g(x) 5 (ƒ ? g)(x) ƒsxd
ƒ
Dividing to get gsxd 5 AgB sxd
(where g(x) 2 0)
For example, if ƒ(x) 5 x 3 and g(x) 5 5(2)x, then (ƒ 1 g)(x) 5 ƒ(x) 1 g(x) 5 x 3 1 5(2)x (ƒ 2 g)(x) 5 ƒ(x) 2 g(x) 5 x3 2 5(2)x (ƒ ? g)(x) 5 ƒ(x) ? g(x) 5 x 3 ? 5(2)x ƒ ƒsxd x3 a b sxd 5 5 g gsxd 5s2d x We’ve already seen several examples of combining functions. We initially described a polynomial function as a sum of power functions. Once a polynomial is put into factored form, we can think of it as a product of linear functions. EXAMPLE
1
SOLUTION
Adding two functions: Skid distances In Example 7 of Section 7.2, we estimated the skid distance (in feet) of a car after the brakes were applied as S 2/30, where S (in mph) is the initial speed of the car. But the total braking distance has an additional component, the distance due to the driver’s reaction time before hitting the brakes. a. It takes the average driver about 0.75 second to react before putting on the brakes. If the car is traveling at S mph, how many feet has the car traveled during the driver’s reaction time? Construct the equation for the reaction distance dr(S) in feet. b. Construct an equation for the total braking distance as a function of car speed S. c. If you suddenly see that there is a massive traffic accident 100 feet ahead, can you stop in time if you are traveling at 20 mph? At 40 mph? At 60 mph? a. If you are traveling at S mph, the distance traveled during 0.75 second is s0.75 secd ? aS
miles 5280 feet b 5 s0.75 secd ? aS b 5 1.1S feet hour 3600 sec
So dr(S) 5 1.1S. b. The total braking distance, dt(S), is the sum of two functions: the reaction distance dr(S) and the skid distance ds(S). dt(S) 5 dr (S) 1 ds(S)
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dt(S) 5 1.1S 1
substituting, we get
S2 30
where S is in mph and dt(S) is in feet. c. If the car is traveling at 20 mph, then the total braking distance is dt s20d 5 1.1s20d 1
s20d 2 400 5 22 1 < 22 1 13 5 35 feet 30 30
So at 20 mph, you could stop in 35 feet, well before the accident. At 40 mph, your total braking distance is dt s40d 5 1.1s40d 1
s40d 2 1600 5 44 1 < 44 1 53 5 97 feet 30 30
So at 40 mph, you would be cutting it very close, stopping only about 100 2 97 5 3 feet from the accident (too close really, since all the functions are estimates). At 60 mph, your total braking distance is dt s60d 5 1.1s60d 1
s60d 2 3600 5 66 1 5 66 1 120 5 186 feet 30 30
So you will not be able to stop in time.
EXAMPLE
2
SOLUTION
Subtracting one function from another: Making a profit You are the Chair of the Board of the “Friends” of a historical house in Philadelphia. You are trying to increase membership by sending out a mass mailing. The up-front fixed costs are $1000 for designing a logo and printing 5000 brochures (the minimum for a discount rate). The mailing costs are $0.51 per person, which includes stuffing the envelope, adhering address labels, and paying postage. Each person who joins would pay a $35 annual membership fee; however, the usual response rate for such mass mailings is about 3%. How large should your mailing be for you to recover the costs of the mailing through the membership fees? If n 5 the number of brochures mailed, then C(n), the cost of the mailing, is C(n) 5 fixed costs 1 (cost per mailing) ? (number of brochures) 5 $1000 1 $0.51n The total revenue R(n) is R(n) 5 (number of responses) ? (membership fee) 5 (3% of n) ? ($35) 5 (0.03n) ? ($35) 5 $1.05n Your profit P(n) is the difference between the revenue and cost functions. Profit 5 Revenue 2 Costs P(n) 5 R(n) 2 C(n) 5 $1.05n 2 (1000 1 $0.51n) 5 $0.54n 2 1000 P(n) is a linear function of n, the number of mailings, with a vertical intercept at 2$1000 and a positive slope of $0.54 per mailing (see Figure 8.59).
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$2000 $1500 $1000 $500
Profit
n
$0
5000 –$500
Loss
–$1000 –$1500
Figure 8.59 Graph of Profit 5 Revenue 2 Costs with breakeven point at n 5 1850.
What is the breakeven point, where revenue equals costs? This happens when the profit (or loss) is 0. Setting substitute for P(n) add 1000 to each side divide and round to the nearest ten
P(n) 5 0 $0.54n 2 1000 5 0 $0.54n 5 1000 n < 1850
The model suggests that you should mail out letters to at least 1850 people to cover your costs. If you mail to more than 1850 people, the model predicts you will have a profit. Mailing to fewer people, you would take a loss, so financially the mass mailing would not be worth doing. EXAMPLE
3
SOLUTION
The quotient of two functions: The federal debt per person Since 1940 both the gross federal debt and the U.S. population have been growing exponentially. Best-fit curves describe the debt (in millions of dollars) as D(t) 5 78,800(1.073) t and the U.S. population size (in millions) as P(t) 5 141(1.012) t, where for both functions t 5 years since 1940. a. Construct a function for the estimated gross federal debt per person over time. b. What was the estimated federal debt per person in 1990? In 2007? The projected federal debt per person in 2015? c. Is the model realistic? a. F(t) 5 federal debt per person over time, which equals federal debt each year D(t) 78,800(1.073) t 5 5 U.S. population each year P(t) 141(1.012) t b. In 1990 we have t 5 50. So F(50) 5
D(50) 78,800(1.073) 50 $2,670,000 million 5 < $10,430 per person 50 < P(50) 141(1.012) 256 million people
This means that if we had decided to pay off the gross federal debt in 1990, we would have had to spend the equivalent of over $10,000 per person. In 2007 we have t 5 67. So F(67) 5
D(67) 78,800(1.073) 67 $88,445,200 million 5 < < $28,210 per person P(67) 141(1.012) 67 314 million people
So only 17 years later, the federal debt had almost tripled to over $28,000 per person.
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If the same rates of growth continue, then in 2015 (when t 5 75), we will have F(75) 5
D(75) 78,800(1.073) 75 $15,542,000 million 5 < $45,050 per person 75 < P(75) 141(1.012) 345 million people
or over $45,000 of gross federal debt for every man, woman, and child in the United States. c. There are several U.S. “national debt clocks” (such www.brilligcom/debt_clock) that continuously track the gross federal debt and the debt per person. In May 2007 one clock estimated the gross federal debt per person at $29,790, which is slightly over our model’s value of $28,000. So the model is reasonably accurate. Check at least one such site and compare its figure for the current debt per person with what our model predicts for the current year.
Rational Functions: The Quotient of Two Polynomials The quotient of two polynomial functions is called a rational function (from the word “ratio”). We’ve seen simple examples of rational functions before in Chapter 7 in the form of power functions with negative integer powers. For example, R(x) 5 6x21 5 6x , the ratio of surface area to volume of a cube. A more complex example would be the function S(x) 5
x3 2 x 1 5 2x2 2 7
Rational Functions A rational function R(x) is of the form R(x) 5
p(x) q(x)
(q(x) 2 0)
where p(x) and q(x) are both polynomials.
EXAMPLE
4
Cost per unit The start-up costs for a small pizza company are $100,000, and it costs $3 to produce each additional pizza. (Economists call $3 the marginal cost.) a. Construct a function C(x) for the total cost of producing x pizzas. b. Then create a function P(x) for the total cost per pizza. c. As more pizzas are produced, what happens to the cost per pizza?
SOLUTION
a. C(x) 5 100,000 1 3x 100,000 1 3x b. P(x) 5 x c. The cost of producing one pizza is $100,003! But as more pizzas are produced, the total production cost per pizza goes down—allowing the producer to sell pizzas at a cheaper price.
EXAMPLE
5
A rational function in disguise 1 1 3 is a rational function. a. Show that T(x) 5 x22 b. Graph the function. What is the domain of T(x)?
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SOLUTION
1 13 x22 1 3(x 2 2) 5 1 x22 (x 2 2) 1 3x 2 6 5 1 (x 2 2) (x 2 2)
T(x) 5
a. Given multiply 3 by
525
x22 (51) x22
use the distributive law
5
combine terms with common denominators
3x 2 5 x22
T(x) is now rewritten as the quotient of two polynomials. b. The graph is shown in Figure 8.60. The domain is all real numbers except 2, since the denominator cannot equal zero.
10 y
T(x)
x –10
10
–10
Figure 8.60 The graph of
T(x) 5
3x 2 5 . x22
Visualizing Rational Functions Horizontal intercepts and vertical asymptotes The graphs of rational functions can be quite complex. However, if ƒ(x) and g(x) have no f(x) common factors, we do know two things about the graph R(x) 5 g(x) . • First, the horizontal intercepts of ƒ(x) are the horizontal intercepts of R(x), since when ƒ(x) 5 0, R(x) 5 0. f(x) • Second, when g(x) 5 0, then R(x) 5 g(x) is not defined since the denominator is 0. The function “blows up” at that value of x, which is called a singularity. The vertical line through that value of x is a vertical asymptote for the graph; that is, as the values of x approach the vertical asymptote, the values of the function approach 6` .
EXAMPLE
6
Finding horizontal intercepts and vertical asymptotes Given the rational function ƒ(x) 5 3x 2 2
2 13x 2 10 : x 2 6x 2 7
a. b. c. d.
Put the numerator and denominator into factored form. What are the horizontal intercepts of f (x)? What is the domain of the function? What are its vertical asymptotes? Use technology to graph the function and confirm your answers in parts (b) and (c).
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SOLUTION
a. Factoring, we get ƒsxd 5
3x 2 2 13x 2 10 x 2 2 6x 2 7
5
s3x 1 2d sx 2 5d sx 1 1d sx 2 7d .
b. The horizontal intercepts occur when ƒ(x) 5 0 or, equivalently, when the numerator (3x 1 2)(x 2 5) 5 0. This means either 3x 1 2 5 0 3x 5 22
or
x2550 x55
x 5 223 So ƒ(x) has two horizontal intercepts, at (223, 0) and (5, 0). c. ƒ(x) is not defined when the denominator (x 1 1)(x 27) is 0. This happens when either x1150 x 5 21
or
x2750 x57
So the domain is all real numbers except the singularities at x 5 21 and x 5 7. The graph of f (x) “blows up” at its singularities, creating two vertical asymptotes at the vertical lines x 5 21 and x 5 7. d. Figure 8.61 shows the graph of this rational function, which confirms our answers in parts (b) and (c).
y
20
x
10
–4
–10
Figure 8.61 Graph of the rational function
ƒsxd 5
s3x 1 2d sx 2 5d sx 1 1d sx 2 7d .
The end behavior of rational functions: Additional asymptotes p(x) What happens to the rational function R(x) 5 q(x) as x S 6` ? This is called its end behavior. As x S 6`, the polynomial expressions p(x) and q(x) become dominated by their leading term—the one with the highest degree. For example, if we look at the 2 rational function in Example 6, where ƒ(x) 5 3x 2 2 13x 2 10, then as x S 6`, 2
x 2 6x 2 7
ƒ(x) < 3x x2 5 3. So as x S 6`, f (x) S 3 (but never reaches 3). So f (x) is asymptotic to the horizontal line y 3. Examine Figure 8.61 to verify that this conclusion seems reasonable.
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EXAMPLE
7
527
End behavior Describe the behavior of the following rational functions as x S 6`. Then graph each function to confirm your descriptions. 1 6x 1 1 3x2 1 2x 2 4 a. f (x) 5 b. g(x) 5 c. h(x) 5 2x 2 3 3x 1 5 x21
SOLUTION
As x S 6`, a rational function’s behavior can be approximated by the ratio of the leading terms of the numerator and denominator. 1 1 < . So as x S 6`, f (x) S 0 (but never a. As x S 6`, f (x) 5 2x 2 3 2x reaches 0). This means that f(x) is asymptotic to the x-axis. (See Figure 8.62(a).) 6x 1 1 6x < 5 2. So as x S 6`, g (x) S 2 (but never 3x 1 5 3x reaches 2). This means that g(x) is asymptotic to the horizontal line y 2. (See Figure 8.62(b).) 23x2 23x2 1 2x 2 4 < 5 23x. So as x S 6`, c. As x S 6`, h(x) 5 x21 x b. As x S 6`, g(x) 5
h (x) S 23x. This implies that h(x) gets closer to the line y 5 23x, but never touches it. So h(x) is asymptotic to the line y 5 23x. (See Figure 8.62(c).) y
y
y 10
5
x 5
–5
x –10
10
–5 (a) Graph of f (x) =
30
–10 1 2x–3
6x + 1 (b) Graph of g(x) = 3x + 5
–10
10
–30 (c) Graph of h(x) =
3x2 + 2x – 4 x– 1
Figure 8.62 Graphs of three rational functions.
When the end behavior is a horizontal line, we say the rational function has a horizontal asymptote. When the end behavior is a polynomial, we say it has an oblique asymptote.
Graphs of Rational Functions p(x) , where p and q are polynomials with no q(x) common factors and q(x) 2 0, then: For any rational function R(x) 5
The horizontal intercepts of R are the zeros of the numerator p; that is, the values of x where p(x) 5 0. The vertical asymptotes of R occur at the zeros of the denominator q; that is, the values of x where q(x) 5 0. The end behavior of R can be approximated by the ratio of the leading terms of p and q and is either a horizontal or an oblique asymptote.
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Algebra Aerobics 8.7 Graphing program optional for Problems 5 and 6, and required for Problems 9 and 10. 1. Let ƒ(x) 5 x 3, g(x) 5 2x 2 1, and h(x) 5 1x . Evaluate each of the following. a. ƒ(2) d. (h ? g)(2) b. g(2) e. (ƒ 1 g)(2) 2.
3.
4.
5.
c. h(2) f. AhgB s2d Let Q(t) 5 5 1 2t and P(t) 5 3t 3. Evaluate each of the following. a. Q(1) 1 P(1) c. P(21) ? Q(21) b. Q(2) 2 P(2) d. Q(3)/P(3) Let ƒ(t) 5 3 2 2t and h(t) 5 t 2 2 1. Find general expressions for each of the following. a. ƒ(t) 2 h(t) c. ƒ(t) ? h(t) b. ƒ(t) 1 h(t) d. h(t)/ƒ(t) 1 2 Given ƒ(x) 5 x 1 2x 2 3 and g(x) 5 x 2 1 , find: a. ƒ(x 1 1) c. g(x 1 1) b. ƒ(x) 1 1 d. g(x) 1 1 (Graphing program optional.) Identify any horizontal intercepts and any vertical asymptotes for the following functions. If possible, check your answers by graphing the functions with technology. a. ƒsxd 5
sx 2 1d sx 1 5d sx 1 3d
b. gsxd 5
s3x 1 2d sx 1 1d sx 2 3d
6. (Graphing program optional.) Let h(r) 5 r2 2 4r 2 12 and k(r) 5 r 2 2 4r 1 3. Construct the functions f (r) and g(r) and identify their horizontal intercepts and vertical asymptotes, if any exist. If
possible, check your answers by using a graphing program. h(r) k(r) a. f(r) 5 b. g(r) 5 k(r) h(r) 1 7. Using the function f (x) 5 2 , x a. Create a new function g(x) 5 2f (x 1 3) 1 1. b. Show that g(x) is a rational function of the form p(x) g(x) 5 . q(x) c. Identify the horizontal intercepts and vertical asymptotes of g(x), if they exist. p(x) 8. If f (x) 5 q(x) is a rational function, what would you be looking for if you: a. Set f (x) 5 0? c. Set p(x) 0? b. Evaluated f(0)? d. Set q(x) 0? (Graphing program required.) For each function in Problems 9 and 10: i. Identify any horizontal and vertical intercepts. ii. Find any vertical asymptotes. iii. Describe the end behavior of the graph (and any other asymptotes). iv. Sketch the graph, labeling the intercepts. 2x 1 6 9. f (x) 5 x23 10. g (x) 5
x2 1 2x 2 3 3x 2 1
Exercises for Section 8.7 Some exercises recommend or require a graphing program. 1. Given ƒ(t) 5 3t 2 1 4t 2 5 and g(t) 5 6t 1 1, find: a. ƒ(t) 1 g(t)
b. g(t) 2 ƒ(t)
c. ƒ(t) ? g(t)
d.
ƒ(t) g(t)
23m 3 and g(m) 5 , find: m24 2m 2 5 f a. ƒ(m) 1 g(m) c. (ƒ ? g)(m) e. a b (m) g g(m) b. (ƒ 2 g)(m) d. ƒ(m)
2. Given ƒ(m) 5
3. Let ƒ(x) 5 3x5 1 x and g(x) 5 x2 2 1. a. Construct the following functions. j(x) 5 ƒ(x) 1 g(x), k(x) 5 ƒ(x) 2 g(x), l(x) 5 ƒ(x) ? g(x) b. Evaluate j(2), k(3), and l(21). 4. If h(x) 5 ƒ(x) ? g(x) 5 2x2 23x 1 4, what are possible equations for ƒ(x) and g(x)?
5. You own a theater company and you have an upcoming event. a. You decide to charge $25 per ticket. Construct a basic ticket revenue function R(n) (in dollars), where n is the number of tickets sold. b. You need to pay $500 to keep the box office open for ticket sales. Modify R(n) to reflect this. c. You decide to give 30 free tickets to the patrons of your company. Modify your function in part (b) to reflect this. 6. Many colleges around the country are finding they need to buy more computers every year, not only to replace broken or outmoded computers, but also because of the increasing use of computers in classrooms, labs, and studios. A college administrator is preparing a 5-year budget plan. She anticipates that her college, which now has 120 computers, will have to increase that amount by 40 per year for the next 5 years. She currently pays $1000 per computer, but she expects the costs will go up by 3% per year because of inflation.
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a. Construct a function N(t) for the number of computers each year as a function of time t (in years since the present). b. Construct a function C(t) for the individual cost of a computer purchased in year t. c. Construct a function that will describe the total cost of the computers each year. d. For year 5, how much money should the budget allow for computers?
529
11. Using the accompanying graph of f(x) and g(x), find estimates for the missing values in the following table. 10
y
f(x)
7. A worker gets $20/hour for a normal work week of 40 hours and time-and-a-half for overtime. Assuming he works at least 40 hours a week, construct a function describing his weekly paycheck as a function of the number of hours worked.
x f(x) x g(x)
0
1
2
3
4
5
23
21
5
15
29
47
0
1
2
3
4
5
23
25
211
221
235
253
g d. a b (1) f 9. Use the table in Exercise 8 to create a new table for the functions a. h(x) 5 (f 1 g)(x) c. k(x) 5 (f ? g)(x) b. j(x) 5 (g 2 f )(x) a. (f 1 g)(2)
b. (g 2 f )(0)
y 3
3 g(x)
–4
23
x f (x) g(x) f(x) g(x) f(x) g(x) f(x) ? g(x) g(x)/f(x)
c. (f ? g)(3)
10. One method of graphing functions is called “addition of ordinates.” For example, to graph y 5 x 1 1x using this method, we would first graph y1 5 x. On the same coordinate plane, we would then graph y2 5 1x . Then we would estimate the y-coordinates (called ordinates) for several selected x-coordinates by adding geometrically on the graph itself the values of y1 and y2 rather than by substituting numerically. This technique is often used in graphing the sum or difference of two different types of functions by hand, without the use of a calculator. a. Use this technique to sketch the graph of the sum of the two functions graphed in the accompanying figure.
x
–3
8. (Graphing program required.) Using the accompanying table, evaluate the following expressions in parts (a)2(d).
22
21
0
1
2
3
12. From the graph and your results in Exercise 11, find the equations for: a. f(x) c. ( f 1 g)(x) e. ( f ? g)(x) g f. a b (x) f 13. The Richland Banquet Hall charges $500 to rent its facility and $40 per person for dinner. The hall holds a minimum of 25 people and a maximum of 100. A sorority decides to hold its formal there, splitting all the costs among the attendees. Let n be the number of people attending the formal. a. Create a function C(n) for the total cost of renting the hall and serving dinner. b. Create a function P(n) for the cost per person for the event. c. What is P(25)? P(100)? What do these numbers represent? d. ( f 2 g)(x)
b. g(x)
14. If ƒsxd 5 x12 , find 2ƒ(x) 1 2ƒ(x 2 3); then find a common denominator and combine into one rational expression. 15. For each rational function graphed below, estimate the equation for any vertical or horizontal asymptote(s).
y1
10
y
x –3
y2
3 5
y f (x)
g(x)
3
b. Use this technique to sketch the graph of y 5 2x2 1 x3 for 22 # x # 2. Then use a graphing tool (if available) and compare. c. Use this technique to graph y 5 2x 2 x2 for 22 # x # 5.
x
x
–3 –5
y
5
–5
5
x –5
5
h (x) –5
–5
Graph B
Graph C
–7 Graph A
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For each of the functions in Exercises 16–18, identify any horizontal intercepts and vertical asymptotes. Then, if possible, use technology to graph each function and verify your results. 16. ƒ(x) 5
3x 2 13 x24
17. g(x) 5 2 18. h(x) 5
23. Without using technology, match each function with its graph. x2 2 9 2 12 a. f (x) 5 2 c. h(x) 5 3(x 1 2) 5x 2 20 3x 1 5 b. g(x) 5 2 x 24
2 (x 1 3) 2
y
1 22 (x 1 3)(x 2 1)
y
y
10
10
10
x 10 –10
–10
x 10
x 10 –10
19. (Graphing program required.) x ? x 11 b. Does the function have any horizontal intercepts? Any vertical asymptotes? c. What is its end behavior? d. Graph the function. The graph is one of a set of curves called “serpentine” by Isaac Newton. Why would that name be appropriate? a. What is the domain of the rational function S(x) 5
20. (Graphing program required.) Let g(x) 5
–10
2
18 2 2. (x 2 1) 2
a. We can think of g(x) as being created from transformations of the function f(x) 5 x12. Describe the transformations and then write g(x) as a function of f(x) . p(x) b. Show that g(x) is a rational function of the form q(x) . (Hint: Find a common denominator.) c. Identify any horizontal intercepts and any vertical asymptotes. d. What is its end behavior? e. Use graphing technology to confirm your answers and estimate the horizontal asymptote.
22. The function f(x) 5 1/x was transformed into the function g(x) plotted on the accompanying graph. Construct g(x) in terms of f(x) and then write g(x) in rational function form p(x) g(x) 5 q(x) .
Graph C
24. (Graphing program required for part (c).) The rational 11 function gsxd 5 4xx 2 2 3 can be decomposed into a sum by using the following method: write as sum of a fraction and a constant
4x 2 11 A 5 1B x23 x23
find the common denominator
4x 2 11 A B(x 2 3) 5 1 x23 x23 x23 A 1 Bx 2 3B 4x 2 11 5 x23 x23 4x 2 11 5 Bx 1 A 2 3B
multiply and simplify Set numerators equal
4x 5 Bx
Set x values equal
B54
So
211 5 A 2 3B 211 5 A 2 3(4)
Set the constants equal substitute 4 for B
A51
So 21. (Graphing program required.) Construct a rational function f(x) that has horizontal intercepts at (23, 0) and (4, 0) and vertical asymptotes at the lines x 5 1 and x 5 25. Use technology to sketch the graph of f(x).
–10
–10 Graph B
Graph A
gsxd 5
Therefore
4x 2 11 1 5 14 x23 x23
which is the graph of ƒ(x) 5 1x shifted to the right by three units, then shifted up by four units. See the accompanying graph.
10
y
y 10
x –5
10
g(x) –5 x 10
–10
–10
22 a. Use the preceding method to decompose gsxd 5 5xx 1 2 3 . b. Describe the transformation of the function ƒsxd 5 1x into 22 gsxd 5 5xx 1 2 3 . c. Using technology, plot the graphs of ƒ(x) and g(x) to verify that the transformation described in part (b) is correct.
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8.8 Composition and Inverse Functions Composing Two Functions Sometimes it is useful to use the output from one function as the input for another function. As an example, let’s look at some simple “parent” functions. Assuming one’s mother is a unique person (ignoring complexities such as adoption or cloning), we can define a “mother” function M(p) as the mother of p. Similarly, we can define a “father” function F(p) as the (unique) father of p. Then what would the expression F(M(p)) mean? We read the expression from the inside out. Starting with a person p, we apply the mother function to p and then apply the father function to the output. For example, if you are the person p, then F(M(you)) 5 F(your mother) 5 the father of your mother In other words, this is your grandfather on your mother’s side. When we apply M and then F as above, we call it the composition of F and M, a new function denoted by F M. We define s
(F M)(p) 5 F(M(p)) s
Warning: Be careful not to confuse the product of two functions ƒ ? g and the composition ƒ g. The product (ƒ ? g)(x) 5 ƒ(x) ? g(x) means to evaluate ƒ and g both at x and then multiply the results. The composition (ƒ g)(x) 5 ƒ(g(x)) means to evaluate g at x, and then evaluate ƒ at g(x). s
s
EXAMPLE
1
Genealogy Using the parent functions, who is: a. (M F)(you)? b. (M M)(you)? s
SOLUTION
c. Your grandfather on your father’s side?
s
a. (M F)(you) 5 M(F(you)) 5 M(your father) 5 the mother of your father, or your grandmother on your father’s side. b. (M M)(you) 5 M(M(you)) 5 M(your mother) 5 the mother of your mother, or your grandmother on your mother’s side. c. Your grandfather on your father’s side is the father of your father, that is, F(F(you)). s
s
The Composition of Two Functions If ƒ(x) and g(x) are two functions, then the function ƒ g, called the composition oƒ ƒ and g, is defined by (ƒ g)(x) 5 ƒ(g(x)) s
s
EXAMPLE
2
The order of composition matters 1 Let ƒ(x) 5 x 2 and g(x) 5 x 1 1. a. Evaluate (ƒ g)(3). b. Determine a general expression for (ƒ g)(x). c. Does (ƒ g)(x) 5 (g ƒ)(x)? Explain your answer. s
s
s
SOLUTION
a. We have:
s
by definition
(ƒ g)(3) 5 ƒ(g(3)) s
evaluate g(x) when x 5 3
5 ƒ A3
simplify
5 ƒ A14B
evaluate ƒ(x) when x 5 14
5 A14B 2
or
1 5 16
1 1 1B
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(ƒ g)(x) 5 ƒ(g(x))
b. By definition
s
substitute for g(x)
1 5 ƒAx 1 1B
evaluate ƒ
1 2 5 Ax 1 1B
(g ƒ)(x) 5 g(ƒ(x)) 5 g(x 2) 5 21
c. By definition
s
substitute for ƒ(x) evaluate g
x 1 1
1 1 Since A x 1 1 B 2 x2 1 1 (try evaluating both sides when x 5 1), then ( f g)(x) 2 (g f )(x). So the order of composition does matter. 2
s
EXAMPLE
s
3
Toxic plumes For two weeks during April 2004 the Pentagon, in its efforts to fight terrorism, conducted a study that involved the release of simulated airborne toxins at a fixed starting point. Scientists then measured the dispersion path and spread rate of the pseudo-toxic plume, using sensors placed in concentric circles from the point of release. The circles had radii ranging from a few feet to several thousand feet. a. Assume that wind speed is not only the major, but the only dispersion factor. If the wind speed is 10 mph, or about 0.17 miles per minute (from any direction), construct an equation that calculates the circular area in square miles (within which the plume would lie) as a function of time in minutes. b. What circular area would contain the “toxic” plume after 10 minutes? After 30 minutes? c. If the wind speed doubles, will the area double?
SOLUTION
a. The area A(r) (in square miles) of the “toxic” circle is a function of its radius r (in miles), A(r) 5 pr 2 In this case the radius r is not fixed but is a function R(t) of time t (in minutes). So r 5 R(t) The wind is traveling at about 0.17 mile per minute. So if we start at the release point (when t 0), then the radius R(t) of the “toxic” circle is R(t) 0.17t where t is in minutes and R(t) is in miles. Substituting R(t) for r in the equation for A(r) gives us A(R(t)) 5 A(0.17t) 5 p(0.17t)2 < 0.09t2 which represents the circular area (in square miles) at minute t that encloses the “toxic” plume. b. Assuming the wind continues to blow at 10 mph (or 0.17 mile per minute), then after 10 minutes the circular area would be A(R(10)) < 0.09(10)2 9 square miles. After 30 minutes, the circular area would be A(R(30)) < 0.09(30)2 81 square miles, which is larger than all of Washington, D.C. c. If the wind doubles from 10 to 20 mph, the radius at any point in time will be twice as large. Our new radius function will now be Rnew(t) 5 2 ? 0.17t 5 0.34t. But the area would be four times as large, since now A(Rnew(t)) 5 A(0.34t) 5 p(0.34t)2 < 0.36t 2 5 4 ? (0.09t 2) 5 4 ? A(R(t)) or four times the area when the wind was at 10 mph. This makes sense if you think of the composite function (A R) as a direct proportionality with an exponent of 2. s
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533
Composing More Than Two Functions We can think of the transformations of a single function (see Section 8.6) as the result of the composition of two or more functions. For example, the transformation of f (x) 5 x12 to 2f(x 2 1) 5 (x 22 1) 2 can be thought of as the composition of three functions: 1 f (x) 5 2 g(x) 5 2x and h(x) 5 x 2 1 x Since (g f h)(x) 5 g( f(h(x))) 5 g( f (x 2 1)) 5 g a s
s
1 2 b 5 (x 2 1) 2 (x 2 1) 2
It is easy to construct the three functions if you just think of the steps taken to transform x12 into (x 22 1) 2. Starting with x, first subtract 1 (applying h), then square and place the result into the denominator (applying f ), and finally multiply by 2 (applying g). EXAMPLE
4
SOLUTION
Given f (x) 5 x13, rewrite A34B f (x 1 5) 5
3 4(x 1 5) 3
as the composition of three functions.
Letting f (x) 5 x13, g(x) 5 A34B x, and h(x) 5 x 1 5, the composition (g f h)(x) 5 g( f(h(x))) 5 g( f (x 1 5)) 5 ga s
s
1 3 b 5 (x 1 5) 3 4(x 1 5) 3
Algebra Aerobics 8.8a 1. Given ƒ(x) 5 2x 1 3 and g(x) 5 x 2 2 4, find: a. ƒ(g(2)) d. ƒ(ƒ(3)) b. g(ƒ(2)) e. (ƒ g)(x) c. ƒ(g(3)) f. (g ƒ)(x) 1 2. Given P(t) 5 t and Q(t) 5 3t 2 5, find: a. (P Q)(2) d. (P Q)(t) b. (Q P)(2) e. (Q P)(t) c. (Q Q)(3) 2 3. Given F(x) 5 x 2 1 and G(x) 5 3x 2 5, find: a. (F G)(x) b. (G F)(x) c. Are the composite functions in parts (a) and (b) equal? s s
s
s
s
s
s
s
s
Problems 4, 5, and 6 refer to the accompanying graph of f (x) and g(x) .
s
s
s
s
s
s
5
f (x)
g(x)
x 5
0 3 5 2
1 21 3 2
2 1 2 3
3 3 1 4
4 5 0 5
5 7 21 6
a. (h f g)(4) b. (f h g)(1) 8. Use the given functions f, g, and h to evaluate the compositions. 1 f (x) 5 , g(x) 5 "x 1 1, h(x) 5 3 x25 a. (h f g)(3) b. (f g h)(100) s
s
s
s
s
–5
s
s
x f(x) g(x) h(x)
y
–5
4. Using the graphs of f(x) and g(x), determine the values of: a. f (22) e. (g f )(22) b. g(22) f. ( f g)(22) c. f(0) g. (g f )(0) d. g(0) h. ( f g)(0) 5. Using your knowledge of quadratic and linear functions, create algebraic expressions for the following: a. f(x) b. g(x) c. (g f )(x) d. ( f g)(x) 6. Using your functions from Problem 5, evaluate (g f )(22) and ( f g)(22) . Do they agree with your answers in Problem 4, parts (e) and (f)? 7. From the table, determine the values of the following compositions in parts (a) and (b).
s
s
s
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Inverse Functions: Returning the Original Value Sometimes when we compose two functions, the input is equal to the output. For example, if f(x) 5 log x and g(x) 10x, then (f g)(x) 5 f (g(x)) 5 f(10x ) 5 log(10x ) 5 x s
and (g f )(x) 5 g(f(x)) 5 g(log x) 5 10log x 5 x s
What one function “does,” the other “undoes.” We say that f and g are inverse functions of each other. We denote this as f 21 5 g and g21 5 f. The notation f 21 is read as “f inverse” or “the inverse of f.” A function that has an inverse is called invertible.
Inverse Functions The functions f(x) and g(x) are inverse functions of each other if (ƒ g)(x) 5 x s
and
(g ƒ)(x) 5 x s
We write ƒ21 5 g and g21 5 ƒ.
Warning: The “21” in the notation f 21 can be confusing. It is not an exponent, so f 21 does not mean 1/f. The symbol “21” simply indicates that f 21 is the inverse function of f. EXAMPLE
5
SOLUTION
Show that f (x) 5 ln x and g(x) 5 e x are inverses of each other. Since (f g)(x) 5 f (g(x)) 5 f (ex ) 5 ln(ex ) 5 x and (g f )(x) 5 g (f(x)) 5 g(ln x) 5 eln x 5 x(where x . 0) the functions are inverses of each other. So f 21 5 g (and equivalently g21 5 f ). s
s
EXAMPLE
6
SOLUTION
Show that each pair of functions contains a function and its inverse. 3 a. f (x) 5 2x 2 3 and g(x) 5 x 1 2 b. h (x) 5 x 3 1 1 and j(x) 5 (x 2 1) 1/3 3 a. (f g)(x) 5 f (g(x)) 5 f ( (x 1 3)/2) 5 2 A x 1 2 B 2 3 5 (x 1 3) 2 3 5 x (2x 2 3) 1 3 (g f )(x) 5 g(f(x)) 5 g(2x 2 3) 5 5x 2 s
s
b. (h j)(x) 5 h ( j(x)) 5 h(x 2 1) 1/3 5 [(x 2 1) 1/3]3 1 1 5 (x 2 1) 1 1 5 x ( j h)(x) 5 j(h(x)) 5 j(x3 1 1) 5 [(x3 1 1) 2 1]1/3 5 (x3 ) 1/3 5 x s
s
Changing perspectives To get a better understanding of the relationship between a function and its inverse, let’s return to the sales tax example from Chapter 1. There T 0.06P was the function that calculated the 6% sales tax (T) on the price (P). Here P is treated as the input and T the output; that is, given the price we can compute the tax. If we call the function rule F, we have F(P) 5 T 5 0.06P But we also solved for P in terms of T, to get P 5 T/0.06. Now we have changed our perspective. We are thinking of T as the input and P the output; that is, given the tax paid, we can determine the original price. This function is the inverse of F(i.e., F21). Table 8.4 and Figure 8.63 are the table and graph for the function F(P) 5 T 0.06P, where P is the input and T is the output
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Input, P
Output, T
0 2 4 6 8 10
0.00 0.12 0.24 0.36 0.48 0.60
Table 8.4 Table for T 5 0.06P.
T (output in $)
0.60
0.40 (4, 0.24) 0.20
0.00 0
4
2
6
8
10
P (input in $)
Figure 8.63 Graph of T 5 0.06P.
Table 8.5 and Figure 8.64 are the table and graph of the inverse function: T F 21 (T) 5 P 5 0.06 ,
where T is the input and P is the output 10
Output, P
0.00 0.12 0.24 0.36 0.48 0.60
0 2 4 6 8 10
T . Table 8.5 Table for P 5 0.06
8
P (input in $)
Input, T
6
4 (0.24, 4)
2
0 0.00
0.20
0.40
0.60
T (output in $)
T Figure 8.64 Graph of P 5 0.06 .
The function F and its inverse F 21 represent exactly the same information, but from different viewpoints. Comparing Tables 8.4 and 8.5, we see that the columns have been swapped, indicating the reversal of the roles for the input (or independent variable) and the output (the dependent variable). Comparing the graphs (Figures 8.63 and 8.64), we see that the axes have been swapped. So, for example, while the point (4, 0.24) lies on the first graph (Figure 8.63), the point (0.24, 4) lies on the second graph (Figure 8.64). We could write this more formally as F(P) 5 T
means that
F21(T) 5 P
F(4) 0.24
means that
F21(0.24) 4
so
EXAMPLE
7
Evaluating a function and its inverse T Given F(P) 0.06P and F21(T) 5 0.06 , evaluate and interpret F(6) and F21(6).
SOLUTION
F(6) 0.06 ? 6 5 $0.36, so if the price is $6, the sales tax is $0.36. F21(6) 6/0.06 5 $100, so if the sales tax is $6, the price is $100.
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EXAMPLE
8
Finding the inverse from a table If the following table represents a function mapping certain individuals to their cell phone numbers, construct the table of the inverse function. Individual
Cell Phone Number
Janet Davidson John Harbison Elaine Woo Jon Stewart
SOLUTION
810 547-1832 919 287-3557 202 555-6911 212 376-1234
Cell Phone Number
Individual
810 547-1832 919 287-3557 202 555-6911 212 376-1234
Janet Davidson John Harbison Elaine Woo Jon Stewart
Does every function have an inverse? (Answer: No) Not every function is invertible. EXAMPLE
9
SOLUTION
EXAMPLE
10
SOLUTION
Does the function y 5 x 2 have an inverse? If we try to solve y 5 x2 for x in terms of y, we get the equation x 5 6 !y. But for any positive value of y, there will be two values for x, namely, 1 !y and 2!y. For instance, if the input is y 100, then the output could be either x 5 210 or x 5 10. So this new equation is not a function, and hence the original function, y 5 x 2 , has no inverse.
Does the mother function, M(p), from the previous section have an inverse? Recall that M( p) is the mother of a person p. Each person has only one mother (according to our definition), but each mother could have several children. For example, if both Fred and Jenny have the same mother, Sarah, then M(Fred) 5 Sarah 5 M(Jenny). (See Figure 8.65.) People
Fred
Mothers M Sarah
Jenny
Figure 8.65 The mother function, M, showing that Fred
and Jenny have the same mother, Sarah.
There cannot be an inverse function for M, since if the mother, Sarah, were now the input, her output would be both Fred and Jenny. Functions that have an inverse must be “one-to-one” In the last two examples (9 and 10), the original function had two different input values associated with the same output value, making it impossible to construct an inverse
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537
function. In order to be a function, each input for an inverse function must have a single output. But the inputs for an inverse function are the outputs of the original function. So for a function f to have an inverse function f 21 , then f must have only one input for each output value. Technically, that means if x1 2 x2, then f(x1 ) 2 f(x2 ). We call such functions one-to-one, which can abbreviated as 1-1. A good example of a 1-1 function is Social Security numbers. Each American is assigned a unique Social Security number. (See Figure 8.66.) We can think of this process as a function S that assigns to each person p a unique Social Security number n 5 S(p). We can define an inverse function S21 (n) 5 p that maps any existing Social Security number back to its unique owner, p. (See Figure 8.67.) S: The Social Security function is 1-1
David Wright
987654321
Esther Davis
876543219
Kristian Joyce
765432198
Yikwan Chu
654321987
Kathy Knight
543219876
Alfred Ramerez
432198765
Americans
Social Security numbers
Figure 8.66 Mapping people to their Social Security
numbers.
S -1: The inverse of the Social Security function
987654321
David Wright
876543219
Esther Davis
765432198
Kristian Joyce
654321987
Yikwan Chu
543219876
Kathy Knight
432198765
Alfred Ramerez
Figure 8.67 Mapping Social Security numbers to people.
The sales tax function is also 1-1 and has an inverse. But we’ve just seen that the function y 5 x2 and the mother function, M, are not 1-1 and hence don’t have inverses.
A function f (x) is called one-to-one (abbreviated as 1-1) if no two distinct input values are mapped to the same output value; that is, whenever x1 2 x2, then f(x1 ) 2 f(x2 ). Any one-to-one function f has an inverse function f 21 (restricted to an appropriate domain).
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Finding the equation for an inverse function If an invertible function is defined by a simple equation, it can be easy to find its inverse.
EXAMPLE
11
SOLUTION
The circumference C of a sphere is a function of the radius r, C 2pr. But in realworld situations, it is often easier to measure the circumference than the radius of a sphere. a. Find the inverse function by solving for r in terms of C. b. If the circumference of Earth is roughly 25,000 miles, estimate Earth’s radius. C a. C is a 1-1 function of r, so solving for r gives us the inverse function r 5 2p . 25,000 12,500 b. Letting C 25,000 miles, we have r 5 2p 5 p < 4000 miles.
It is a little more complicated when we use function notation. To find the inverse of the 1-1 function f (x) 1 log x, we need to name the output variable, say y, to get y 1 log x. We can then solve this equation for x in terms of y. y 5 1 1 log x(where x . 0)
Given subtract 1 from each side
y 2 1 5 log x
write as a power of 10
10 y21 5 10 log x
simplify and switch sides
x 5 10 y21
So we could write f 21(y) 10 y 2 1 or, more conventionally, since the variables are abstract (and unitless) we could use x as the input to get f 21 (x) 5 10 x21 EXAMPLE
12
SOLUTION
1 a. What is the domain of the function f(x) x 1 1 ? Is f one-to-one? 21 b. If so, find its inverse f and specify its domain.
a. The domain of f is all real numbers except 21. The function f (x) is 1-1 since 1 1 if f (x1 ) 5 f(x2 ) 1 5 1 x2 1 5 x1 1 1 x2 5 x1. So no two x1 1 1 x2 1 1 distinct input values can be mapped to the same output value. b. Thus f has an inverse function that we can determine by naming the output value y 5 f(x) and solving the following formula for x in terms of y. 1 Given y5 (where x 2 21) x11 1 cross-multiply x115 y 1 subtract 1 from each side x5 21 y So the inverse function is f 21(y) 5 A1y B 2 1, where the domain is all real numbers except 0. Since this is an abstract function, we can use any variable name for the input. So following the standard convention of using x as the input, we could rewrite f 21 as f 21 (x) 5
1 2 1(where x 2 0) x
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SOMETHING TO THINK ABOUT
How to tell if graph represents a 1-1 function: The horizontal line test It is easy to tell from its graph whether a function is 1-1 and hence has an inverse (on an appropriate domain). The graph must pass the horizontal line test, which means that no horizontal line can cross the graph twice. (Why?) Compare Figures 8.68 and 8.69. Figure 8.68 shows the graph (from Section 8.3) of the height h(d) of a projectile fired from a battleship gun as a function of d, its distance from the ship. Height of projectile (feet in 000s)
Why must the graph of a function pass the horizontal line test in order to have an inverse?
$60,000 Amount in account ($)
?
539
60
30
d1
$30,000
$0
d2
0
d, distance from ship (feet)
x1
25
Years
Figure 8.68 Graph of height vs. distance
Figure 8.69 Amount in an account
for a projectile fired from a battleship.
over time.
Choose any output height value, say 30 (thousand feet), that is below the maximum height. Then there will be two corresponding inputs, that is, two different distances from the battleship, d1 and d2 , for which the projectile will be at 30 (thousand feet). So d1 2 d2, but h(d1) 5 h(d2) 30,000. (See Figure 8.68.) So the horizontal line at h(d) 30 (thousand feet) crosses the graph twice. Hence the function h is not 1-1 and cannot have an inverse. Figure 8.69 shows the amount A(x) at year x, in an account that started with $10,000 and grew by 7% a year. If we choose any output value (which must be $$10,000), there will be associated with it a unique input value. For example, if we choose an output of $30,000, we can see from Figure 8.69 that it will be associated with one and only one input value, labeled x1. So the function A is 1-1, and there is an inverse function A21 whose domain is restricted to values $$10,000. Recall that a function must pass the vertical line test; that is, no vertical line crosses its graph more than once. A one-to-one function must pass the horizontal line test; that is, no horizontal line crosses its graph more than once.
EXAMPLE
13
Which of the functions in the accompanying graphs have inverses? y y a. b. c. y
x
x
x
Graph A
Graph B Graph C
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SOLUTION
EXAMPLE
The functions in Graphs A and C would each have an inverse, since both function graphs pass the horizontal line test. The function in Graph B would not have an inverse, since we could construct a horizontal line (many, in fact) that would cross the graph twice.
14
Visualizing their graphs, identify which functions from the following function families have inverses. a. Linear functions (of the form y 5 b mx) b. Exponential functions (of the form y 5 Ca x) c. Logarithmic functions (of the form y 5 a log (bx) or y 5 a ln(bx)) d. Power functions (of the form y 5 kx a)
SOLUTION
a. Every linear function, except those representing horizontal lines (where m 0), has an inverse. b, c. Every exponential and logarithmic function has an inverse. d. Odd integer power functions have inverses, but even integer power functions (such as y 5 ax2 or y 5 ax22) do not.
A Final Example The following example combines many of the ideas covered in the course— exponential, log, and power functions as well as compositions and inverses. So it’s a lengthier example than most, but it involves a reasonable model for an important issue—breast cancer.
EXAMPLE
15
Modeling breast cancer tumor growth a. Depending on the aggressiveness of a breast tumor, its volume could double in weeks or months. On average, the volume of a breast tumor doubles every 100 days.5 What is the daily growth factor? The yearly growth factor? b. Describe the tumor volume as a function of time. Use an initial volume of 0.06 cubic cm (the minimum tumor volume detectable by a mammogram) and measure time in years after the tumor reached that volume.6 c. Breast tumors are roughly spherical. So their size is usually reported in terms of the diameter, which is easier to measure than the volume. Construct a function that describes the tumor diameter as a function of its volume. d. Now compose the functions in parts (b) and (c) to create a new function that gives the tumor diameter as a function of time. What is the minimum tumor diameter a mammogram can detect? e. Using technology, graph the function in part (d). What do the points where t , 0 represent? f. Estimate and then calculate how many years it would take the tumor to reach a diameter of 2 cm, the smallest size detectable by touch. This is also the maximum size for a stage I breast cancer and is often used as the decision point for recommending a lumpectomy vs. a mastectomy.
5
Breast cancer data are from the afterword by Dr. Susan M. Love, MD, of Joyce Wadler’s My Breast: One Woman’s Cancer Story (Reading, MA: Addison-Wesley, 1992). 6 Note: The tumor may have been growing for 10 years before being detectable by a mammogram.
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SOLUTION
a. Growth factors. Since the tumor volume has a fixed doubling time, its growth is exponential. Since the tumor volume doubles every 100 days, if a is the daily growth factor, then 2 5 a100 21/100 5 a
1 taking the 100 th root
?
541
The yearly growth factor is the daily growth factor, 21/100, applied 365 times, or
SOMETHING TO THINK ABOUT
(21/100)365 5 2365/100 5 23.65 < 12.6
What is the yearly percentage increase in tumor volume?
So the yearly growth factor is 12.6. (Recall this means a growth rate of 11.6 (in decimal form) or 1,160% per year!) b. Tumor volume as a function of time. Assuming an initial tumor volume of 0.06 cubic cm and a yearly growth factor of 12.6, the tumor volume V(t) is a function of time t given by V(t) 5 0.06(12.6)t
(1)
where V(t) is measured in cubic centimeters and t in number of years since the tumor reached 0.06 cubic cm in volume. c. Tumor diameter as a function of volume. The volume v of a sphere is a function of the radius r, where v 5 A43B pr 3. Since we are interested in the diameter d (twice the radius), we can substitute r 5 d2 into the equation to get 4 4 d 3 4pd 3 pd 3 v 5 pr 3 5 pa b 5 5 3 3 2 3?8 6
(2)
We now have the volume as a function of the diameter. We need the inverse of that function, one that describes the diameter as a function of the volume. We can solve Equation (2) for d in terms of v: v5
Given
pd 3 6
6v 5 d3 p
solving for d taking the cube root
d5
3
6v 5
Åp
3
6
Åp
3 " v < 1.2v1/3
So using function notation, setting d 5 D(v), we have a function that describes the tumor diameter D(v) as a function of its volume v: D(v) 5 1.2v 1/3
(3)
where v is in cubic centimeters and D(v) is in centimeters. d. Tumor diameter as a function of time. Since the volume v can be also be written as a function of time as V(t), then the tumor diameter as a function of time is given by the composition of Equations (1) and (3): (D V)(t) 5 D(V(t)) 5 D(0.06 ? 12.6 t) 5 1.2(0.06 ? 12.6t)1/3 5 1.2 ? (0.06)1/3 ? (12.61/3)t < 0.5 ? 2.3t s
where (D V)(t) is the diameter of the tumor at year t. When t 5 0, then (D V)(t) < 0.5 cm. So the minimum tumor diameter detectable by mammography is about 0.5 cm, less than a quarter of an inch. s
s
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e. Visualizing tumor diameter over time. Figure 8.70 shows the growth of a breast tumor over time.
Diameter of tumor (cm) 10
5
t –3
–2
–1
0
1
2
3
Time (in years)
Figure 8.70 Graph of
breast tumor diameter over time, where t 5 0 corresponds to the minimum tumor diameter detectable by a mammogram.
When t , 0, then (D V)(t) gives the size of the tumor before it was detectable by mammogram. For example, when t 5 25 (5 years before the tumor had a mammogram-detectable size), the tumor diameter was s
(D V)(25) 5 0.5 ? 2.325 < 0.008 cm (about three-thousandths of an inch) s
f. Estimating and calculating tumor diameter growth. Estimating from the graph in Figure 8.70, it would take about 2 years for the diameter to grow from 0.5 cm (when t 5 0) to 2 cm. To get a more exact number, we can substitute 2 cm for (D V)(t) to get the equation s
2 5 0.5 ? 2.3t divide by 0.5 simplify
2 5 2.3t 0.5 4 5 2.3t
take the log of both sides
log(4) 5 log(2.3t)
use rules for logs
log(4) 5 t log(2.3) log(4) t5 log(2.3)
solve for t use a calculator
t < 1 23 years
So it could take almost 2 years for a tumor detectable by a mammogram to grow to a size detectable by touch.
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543
Algebra Aerobics 8.8b 9. Determine from the accompanying function graphs which functions are 1-1.
Graphing program required in Problem 10. t 1. Let g(t) 5 5 2 2t and h(t) 5 5 2 2 . a. Complete the following tables. t
g(t)
t
h(t)
21 1 3 5
0 1 2 3
x x
Graph A
Graph C
b. Find (g h)(3) and (h g)(3). c. Is (g h)(t) 5 (h g)(t) for all t? d. What is the relationship between g and h? s
s
s
s
x
In Problems 2 and 3, verify that ƒ and g are inverse functions by showing that (ƒ g)(x) 5 x and (g ƒ)(x) 5 x. 1 2. ƒ(x) 5 2x 1 1 and g(x) 5 x 2 2 3 3. ƒ(x) 5 " and g(x) 5 x3 2 1 x11 x 1 4. Verify that ƒ(x) 5 x 2 1 and ƒ21(x) 5 1 1 x are inverse functions of one another. s
x
s
Find the inverse (if it exists) for each of the functions in Problems 5, 6, and 7. 5. f(x) 5 3x 1 5 6. g(x) 5 (x22)2/3 7. h(x) 5 5x3 2 4 8. State the inverse action for each of the actions described below. a. Saying “yes” b. Going to class and then taking the bus home c. Unlocking the door, opening the door, entering the room, and turning on the light d. Subtracting 3 from x and multiplying the result by 5 e. Multiplying z by 23 and adding 2
Graph B
Graph D
10. (Graphing program required.) a. Sketch a graph of the function f (x) (x 2)2. Does f have an inverse? b. If not, restrict the domain of f so that f 21 does exist. c. Given the restricted domain of f, find the equation for f 21 and then graph f 21(x). 11. An oil spill is spreading in a roughly circular shape. The radius, r, is growing by 10 feet per hour. The area A(r) (in square feet) of the spill is a function of the radius r (in feet), given by A(r) 5 pr 2. a. Construct a function R(t) that represents the radius r as a function of time t (in hours since the oil spill). b. If the oil spill has been spreading for 2 hours, what is the area of the spill? c. How could you compose the functions A and R to give the area in terms of time t?
Exercises for Section 8.8 1. From the accompanying table, find: a. ƒ(g(1)) c. ƒ(g(0)) b. g(ƒ(1)) d. g(ƒ(0))
e. ƒ(ƒ(2))
2. From the accompanying table, find: a. ƒ(g(1)) c. ƒ(g(0)) b. g(ƒ(1)) d. g(ƒ(0))
e. ƒ(ƒ(2))
x
ƒ(x)
g(x)
x
ƒ(x)
g(x)
0 1 2 3
2 1 3 0
1 0 3 2
21 0 1 2
22 1 2 0
1 2 21 21
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CHAPTER 8 QUADRATICS, POLYNOMIALS, AND BEYOND
3. Using the accompanying graphs, find: a. g(ƒ(2)) b. ƒ(g(21)) c. g(ƒ(0))
d. g(ƒ(1))
y
y
5
4
–3
3
x
–4
4
x
b. Construct a function G that converts Canadian dollars, c, to U.S. dollars. c. What would the function F s G do? Would its input be U.S. or Canadian dollars (i.e., d or c)? Construct a formula for F s G. d. What would the function G s F do? Would its input be U.S. or Canadian dollars (i.e., d or c)? Construct a formula for G s F. 9. A stone is dropped into a pond, causing a circular ripple that is expanding at a rate of 13 ft/sec. Describe the area of the circle as a function of time.
–5
–4
Graph of f(x)
Graph of g (x)
4. Using the accompanying graphs, find: a. g(ƒ(22)) b. ƒ(g(1)) c. g(ƒ(0))
d. g(ƒ(1))
y
y
3
3
x –3
3
x –3
–3
–3
Graph of f(x)
5. Given F(x) 5 2x 1 1 and Gsxd 5 a. F(G(1)) b. G(F(22)) c. F(G(2))
3
Graph of g (x)
x21 , find: x12
d. F(F(0)) e. (F s G)(x) f. (G s F)(x)
6. Given ƒ(x) 5 3x 2 2 and g(x) 5 (x 1 1)2, find: a. ƒ(g(1)) d. ƒ(ƒ(2)) b. g(ƒ(1)) e. (ƒ s g)(x) c. ƒ(g(2)) f. (g s ƒ)(x) 7. The winds are calm, allowing a forest fire to spread in a circular fashion at 5 feet per minute. a. Construct a function A(r) for the circular area burned, where r is the radius. Identify the units for the input and the output of A(r). b. Construct a function for the radius r 5 R(t) for the increase in the fire radius as a function of time t. What are the units now for the input and the output for R(t)? c. Construct a composite function that gives the burnt area as a function of time. What are the units now for the input and the output? d. How much forest area is burned after 10 minutes? One hour? 8. The exchange rate a bank gave for Canadian dollars on March 2, 2007, was 1.18 Canadian dollars for 1 U.S. dollar. The bank also charges a constant fee of 3 U.S. dollars per transaction. a. Construct a function F that converts U.S. dollars, d, to Canadian dollars.
10. The wind chill temperature is the apparent temperature caused by the extra cooling from the wind. A rule of thumb for estimating the wind chill temperature for an actual temperature t that is above 08Fahrenheit is W(t) 5 t 2 1.5S0, where S0 is any given wind speed in miles per hour. a. If the wind speed is 25 mph and the actual temperature is 108F, what is the wind chill temperature? We know how to convert Celsius to Fahrenheit; that is, we can write t 5 F(x), where F(x) 5 32 1 95x , with x the number of degrees Celsius and F(x) the equivalent in degrees Fahrenheit. b. Construct a function that will give the wind chill temperature as a function of degrees Celsius. c. If the wind speed is 40 mph and the actual temperature is 2108C, what is the wind chill temperature? 11. Salt is applied to roads to decrease the temperature at which icing occurs. Assume that with no salt, icing occurs at 328F, and that each unit increase in the density of salt applied decreases the icing temperature by 58F. a. Construct a formula for icing temperature, T, as a function of salt density, s. Trucks spread salt on the road, but they do not necessarily spread it uniformly across the road surface. If the edges of the road get half as much salt as the middle, we can describe salt density S(x) as a function of the distance, x, from the center of the road by Ssxd 5 [1 2 12 A xk B 2]Sd , where k is the distance from the centerline to the road edges and Sd is the salt density applied in the middle of the road. b. What will the expression for S(x) be if the road is 40 feet wide? c. What will the value for x be at the middle of the 40-footwide road? At the edge of the road? Verify that at the middle of the road the value of the salt density S(x) is Sd and that at the edge the value of S(x) is 12Sd. d. Construct a function that describes the icing temperature, T, as a function of x, the distance from the center of the 40-foot-wide road. e. What is the icing temperature at the middle of the 40-foot-wide road? At the edge? 12. Using the given functions f, g, and h where
f(x) x 1
g(x) 5 ex
h(x) x 2
a. Create the function k(x) (f s g s h)(x). b. Describe the transformation from x to k(x).
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8.8 Composition and Inverse Functions
13. Using the given functions J, K, and L, where
J(x) x3
L(x) 1x
K(x) log(x)
a. Create the function M(x) 5 (L s J s K)(x). b. Describe the transformation from x to M(x). In Exercises 14 and 15, rewrite j(x) as the composition of three functions, f, g, and h. 14. j(x) 5
2 (x 2 1) 3
15. j(x) 4ex – 1 In Exercises 16–22, show that the two functions are inverses of each other. x22 16. ƒ(x) 5 3x 1 2 and g(x) 5 3 17. ƒ(x) 5 "x 2 1 (where x . 1) and g(x) 5 x2 1 1 (where x . 0) x11 18. ƒ(x) 5 2x 2 1 and g(x) 5 2
26. On March 2, 2007, the conversion rate from U.S. dollars to euros was 0.749; that is, on that day you could change $1 for 0.749 euros, the currency of the European Union. a. Was a U.S. dollar worth more or less than 1 euro? b. Using the March 2 exchange rate, construct a function C1(d) that converts d dollars to euros. What is C1(1)? C1(25)? c. Now construct a second function C2(r) that converts r euros back to dollars. What is C2(1)? C2(100)? d. Show that C1 and C2 are inverses of each other. e. Reread the beginning of Exercise 8, which describes a conversion process between Canadian and U.S. dollars. In that process the two formulas are not inverses of each other. Why not? 27. Given the accompanying graph of f(x), answer the following. y 5
x3 2 5 4 20. ƒ(x) 5 10 x/ 2 and g(x) 5 log(x 2)
4
3 19. ƒ(x) 5 " 4x 1 5 and g(x) 5
3 f (x)
2
21. F(t) 5 e and G(t) 5 ln(t ) 3t
545
1/3
1 2
22. H(r) 5 ln r and J(r) 5 e
1
2r
x –5
–4
–3
–2
In Exercises 23 and 24, create a table of values for the inverse of the function ƒ(x).
–1
1
2
3
4
5
–1 –2
23.
24.
x
ƒ(x)
22 21 0 1
5 1 2 4
x
ƒ(x)
0 1 22 4
5 3 2 27
–3 –4 –5
a. b. c. d.
28. Determine which of the accompanying graphs show functions that are one-to-one.
25. Cryptology (the creation and deciphering of codes) is based on 1-1 functions. After you code a message using a 1-1 function, the decoder needs the inverse function in order to retrieve the original message. The following table matches each letter of the alphabet with its coded numerical form. A
B
C
Does f(x) have an inverse? Please explain. What is the domain of f(x)? Estimate the range of f(x)? From the graph, determine f(4), f(0), and f(5). Determine f 1(0), f 1(2), and f 1(3).
D
E
F
G
H
I
J
K
L
a. Graph of f (x)
c. Graph of h(x)
x
x
M
26 25 24 23 22 21 20 19 18 17 16 15 14 b. Graph of g(x)
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
13 12 11 10
9
8
7
6
5
4
3
2
1
a. Does this code represent a 1-1 function? Is there an inverse function? If so, what is its domain? b. Decode the message “14 26 7 19 9 6 15 22 8.”
x
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In Exercises 29–32, for each function Q find Q21, if it exists. For those functions with inverses, find Q(3) and Q1(3). 29. Q(x) 5
2 x25 3
30. Q(x) 5 5e0.03x 31. Q(x) 5
x13 x
32. Use the graph of f(x) to evaluate each expression.
[Note: More information can be found in the article by M. P. Pari and F. P. Paloucek, “The origin of the ‘ideal’ body weight equations,” Annals of Pharmacology 34 (9), 2000: 10 :1066–69.]
y 10
f (x)
x
–10
10
–10
b. f 1(2)
a. f(2)
c. f1(4)
d. (f
s
f 1)(8)
33. The following tables represent a function f that converts cups to quarts and a function g that converts quarts to gallons (all measurements are for fluids). a. Fill in the missing values in the chart. (Hint: One quart contains 4 cups, and one gallon contains 4 quarts.) x (cups) f(x) (quarts)
4
8
16
32
x (quarts) g(x) (gallons)
2
4
8
16
b. Now evaluate each of the following and identify the units of the results. i. (g f)(8) iii. ( f 1 g1)(1) iv. ( f 1 g1)(2) ii. g1(2) c. Explain the significance of ( f 1 g1)(x) in terms of cups, quarts, and gallons. s
a. Write the functions for IBW (in kg) for men and women, Wmen(h) and Wwomen(h), where h is a person’s height in inches. Give a reasonable domain for each. b. Evaluate Wmen(70) and Wwomen(66). Describe your results in terms of height and weight. c. Evaluate W1men(77.6). What does this tell you? d. Given that 1 lb 0.4356 kg, alter the functions to create Wnewmen(h) and Wnewwomen(h) so that the weight is given in pounds rather than kilograms. e. Use your functions in part (d) to find W1newwomen(125). What does this tell you?
s
s
s
34. Let f (x) 5 mx b. a. Does f(x) always have an inverse? Explain. b. If f(x) has an inverse, find f 1(x). c. Using the formula for f 1(x), explain in words how, given any linear equation (under certain constraints), you can find the inverse function knowing the slope m and y-intercept b. 35. If you do an Internet search on formulas for “ideal body weight” (IBW), one that comes up frequently was created by Dr. B. J. Devine. His formula states IBW for men (in kilograms) 50 (2.3 kg per inch over 5 feet) IBW for women (in kilograms) 45.5 (2.3 kg per inch over 5 feet)
36. The formula for the volume of a cone is V 5 A13B r2h. Assume you are holding a 6-inch-high sugar cone for ice cream. a. Construct a function V(r) for the volume as a function of r. Why don’t you need the variable h in this case? Find V(1.5) and explain what have you found (using appropriate units). b. Evaluate V1(25). Describe your results. What are the units attached to the number 25? c. When dealing with abstract functions where f (x) y, we have sometimes used the convention of using x (rather than y) as the input to the inverse function f 1(x). Explain why it does not make sense to interchange V and r here to find the inverse function. 37. In Chapter 6 we learned that a logarithm can be constructed using any positive number (except 1) as a base: loga x 5 y means that ay 5 x. Show that F(x) 5 a x and G(x) 5 log a x are inverse functions. The software “E10: Inverse Functions y 5 a x and y 5 log a x” in Exponential and Log Functions can help you visualize the relationship between the two functions. 38. The Texas Cancer Center website, www.texascancercenter.com, notes that the 5-year survival rate for stage I breast cancer (when the tumor diameter is #2 cm) is about 85%. The 5-year survival rate for stage II breast cancer (when the tumor diameter is #5 cm and the cancer has not spread to the lymph nodes) is about 65%. a. Using the equations in Example 15 (the final one in the text) referring to a tumor that doubles in volume every 100 days, how long would it take such a tumor to grow from 0.5 cm in diameter to 5 cm? (Recall that 0.5 cm in diameter corresponds to an initial volume of 0.06 cubic cm, the minimum tumor size detectable by a mammogram.) b. If the tumor were more aggressive, doubling in volume every 50 days, what would the yearly growth factor be? Use this to construct a new function to reflect the volume growth of this more aggressive tumor over time (again using 0.06 cubic cm as the initial volume). c. Using your function from part (b), construct another function to represent the diameter growth over time. d. How many years would it take the aggressive tumor to grow from 0.5 cm to 5 cm in diameter?
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Chapter Summary
547
C H A P T E R S U M M A RY Quadratic Functions and Their Graphs
y
y
y
A quadratic function can be written in standard form as ƒ(x) 5 ax 2 1 bx 1 c (where a 2 0)
x
x
x
Its graph • has a distinctive ∪-shape called a parabola • is symmetric across its axis of symmetry • has a minimum or a maximum point called its vertex • is concave up if a . 0 and concave down if a , 0 • becomes narrower as k a k increases 1 • has a focal point @ 4a @ units above (or below) the vertex on the axis of symmetry y
(a) No x-intercepts No real zeros
(b) One x-intercept One real zero
The Quadratic Formula Setting the quadratic function ƒ(x) 5 ax 2 1 bx 1 c equal to 0 and solving for x using the quadratic formula, gives
x5 Axis of symmetry
(c) Two x-intercepts Two real zeros
2b 6 "b 2 2 4ac 2a
The term b 2 2 4ac is called the discriminant and can be used to predict the number of horizontal intercepts (or real zeros) of ƒ(x).
Focal point
x
0
Vertex
Finding the Vertex Any quadratic function ƒ(x) 5 ax 2 1 bx 1 c can also be written in vertex or a-h-k form as ƒ(x) 5 a(x 2 h)2 1 k b b where the vertex is at (h, k) 5 A22a , ƒ A22a BB . The vertex “anchors” the graph, and the coefficient a determines the shape of the parabola.
Finding the Horizontal Intercepts Every quadratic function ƒ(x) has two, one, or no horizontal intercepts x. To find the x-intercepts, we set ƒ(x) 5 0 and solve for x. The solutions are called the zeros of the function. The Factor Theorem says that any quadratic function ƒ(x) 5 ax 2 1 bx 1 c can be written in factored form as ƒ(x) 5 a(x 2 r1)(x 2 r2) where r1 and r2 are the zeros of ƒ(x). If the zeros, r1 and r2, are real numbers, they are the horizontal intercept(s) of the quadratic function ƒ(x).
If the discriminant . 0, there are two distinct real roots and hence two x-intercepts. If the discriminant 5 0, there is only one distinct real root and hence only one x-intercept. If the discriminant , 0, then the "b 2 2 4ac is not a real number and hence there are no x-intercepts. These zeros are complex numbers of the form a 1 bi, where a and b are real numbers (b 2 0) and i 5 "21. The Average Rate of Change of a Quadratic Function Given a quadratic function ƒ(x) 5 ax 2 1 bx 1 c, the average rate of change between two points on the parabola approaches 2ax 1 b over very small intervals. We can think of the linear function g(x) 5 2ax 1 b as representing the average rate of change of ƒ(x) with respect to x.
Polynomial Functions and Their Graphs A polynomial function of degree n is of the form ƒ(x) 5 an x n 1 an21x n21 1 c 1 a1x 1 a0 where n is a nonnegative integer and an 2 0. The graph of a polynomial function of degree n will • Have at most n 2 1 turning points • Cross the horizontal axis at most n times
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For large values for x, the graph of ƒ(x) 5 a n x n 1 a n21 x n21 1 c 1 a1x 1 a0 will resemble the graph of the power function g(x) 5 an xn. If r1, r2, . . . , rn are zeros of a polynomial ƒ(x) 5 anx n 1 an21x n21 1 c 1 a1x 1 a0, then ƒ(x) can be written in factored form as ƒ(x) 5 an(x 2 r1)(x 2 r2) ? c ? (x 2 rn) Any real zero is an x-intercept for ƒ(x).
The Algebra of Functions If ƒ(x) and g(x) are two functions with the same domain, we can define new functions by Adding to get ƒ(x) 1 g(x) 5 (ƒ 1 g)(x) Subtracting to get ƒ(x) 2 g(x) 5 (ƒ 2 g)(x) Multiplying to get ƒ(x) ? g(x) 5 (ƒ ? g)(x) Dividing to get
Creating New Functions from Old Ways to Transform the Graph of f(x)
ƒsxd f 5 a b sxd (where g(x) 2 0) gsxd g
A rational function R(x) is the quotient of two polynomial p(x) functions. If R(x) 5 q(x) (where p(x) and q(x) have no common terms and q(x) 2 0) then
To stretch or compress:
Multiply the output of ƒ(x) by a constant to get aƒ(x).
To shift vertically:
Add a constant to the output of ƒ(x) to get ƒ(x) 1 k.
The end behavior can be approximated by the ratio of the leading terms of p and q, and is either a horizontal or an oblique asymptote.
To shift horizontally:
Subtract a constant from the input of ƒ(x) to get ƒ(x 2 h).
Composition and Inverses
To reflect across the x-axis:
Multiply the output of ƒ(x) by 21 to get 2ƒ(x).
To reflect across the y-axis:
Multiply the input of ƒ(x) by 21 to get ƒ(2x).
Symmetry If ƒ(x) 5 ƒ(2x), then ƒ is symmetric across the y-axis. If ƒ(x) 5 2ƒ(2x), then ƒ is symmetric about the origin.
Set p(x) 5 0 to find horizontal intercepts. Set q(x) 5 0 to find vertical asymptotes.
If ƒ(x) and g(x) are two functions, then the function ƒ g, called the composition of ƒ and g, is defined by s
(ƒ g)(x) 5 ƒ(g(x)) s
If both (ƒ g)(x) 5 x and (g ƒ)(x) 5 x, then the functions ƒ and g are inverses of each other, written as ƒ21 5 g and g21 5 ƒ. A function f(x) is called one-to-one (or 1-1) if no two distinct input values are mapped to the same output value. A 1-1 function must pass the horizontal line test. A function that has an inverse must be one-to-one. s
s
C H E C K Y O U R U N D E R S TA N D I N G I. Is each of the statements in Problems 1221 true or false? Give an explanation for your answer. 1. If ƒ(t) 5 2(t 2 1)2, then ƒ(0) 5 22. 2. If the vertical axis is the axis of symmetry for a quadratic function g(x), then g(22) 5 g(2).
4. The function y 5 3(x 2 2)2 1 5 has a focal point at 1 A2 12 , 5B . 5. The graph of the quadratic function y 5 2x 2 2 3x 1 1 is steeper than the graph of y 5 3x 2 2 3x 1 1. 6. The graph of y 5 5 2 x 2 is concave down.
3. The polynomial function in the accompanying figure has a minimum degree of 5.
7. The graph of y 5 x 2 1 2x 1 3 is three units higher than the graph of y 5 x 2 1 2x.
y
8. The graph of y 5 (x 1 4)2 lies four units to the right of the graph of y 5 x 2. 9. A quadratic function that passes through the points (1, 5) and (7, 5) will have an axis of symmetry at the vertical line x 5 4. 10. If ƒ(x) 5 x 2 2 3x 2 4, then ƒ(4) 5 0. x
11. The function ƒ(x) 5 (x 1 2)(x 1 5) has zeros at 2 and 5.
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12. In the accompanying figure it appears that g(x) 5 22ƒ(x). y 10
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25. Two more polynomial functions with horizontal intercepts at 21, 3, and 4. 26. A quadratic function concave down with a vertex at (1, 3).
f
27. A quadratic function concave up with its axis of symmetry at the line x 5 3. x 3
–3
28. A quadratic function concave down with vertical intercept at 2 and zeros at 22 and 2. 29. A quadratic function whose graph will be exactly the same shape as the graph of the function r 5 s 2 2 s but five units higher.
g –10
13. The quadratic function ƒ(x) 5 2x 2 2 3x 2 1 has a discriminant with a value of 1. 14. The function ƒ(x) 5 3(x 2 1)2 1 2 has an axis of symmetry at x 5 1. 15. The function ƒ(x) 5 22(x 1 3)2 2 1 has a vertex at (3, 21). 16. There is only one quadratic function ƒ(x) with x-intercepts at 3 and 0. 17. The function h(t) 5 t 2 2 3t 1 2 has a zero at t 5 2 because h(0) 5 2.
30. A quadratic function G(x) whose graph will be exactly the same shape as the function F(x) 5 x 2 1 2x but two units to the left. 31. A quadratic function whose graph will be the reflection across the t-axis of the graph of h(t) 5 (t 2 2) 2. 32. Two distinct quadratic functions that intersect at the point (1, 1). 33. A quadratic function with one zero at x 5 24. 34. A polynomial function h(t) that could describe the function in the accompanying figure. 5
h(t)
18. In the accompanying figure it appears that g(x) 5 2ƒ(x 2 1) 1 3.
t –3
4
–2
y 6
35. Two functions ƒ and g such that ƒ(x) 2 g(x) 5 x 3 1 2x 2 2 5x 1 2.
f
36. Two functions Q and H such that (Q
s
H)(t) 5 "3t 1 1.
37. A rational function with a horizontal intercept at (2, 0) and two vertical asymptotes at x 5 0 and x 5 23. x –3
g
4
–2
19. Assume the height of a ball thrown vertically upward is modeled by the function h(t) 5 24.9t 2 1 38t 1 55, (where t is time in seconds, and h(t) is the height in meters). Then the ball will hit the ground after approximately 9 seconds. 20. If revenue R (in dollars) from an item sold at price p (in dollars) is modeled by the function R 5 p(100 2 5p), the revenue will be at a maximum when the price is $10. 3s 21. The functions mssd 5 2s 11 3 and nssd 5 1 2 are 2s inverses of each other.
II. For Problems 22237 give an example of a function or functions with the specified properties. Express your answer using equations. 22. A polynomial function that does not intersect the horizontal axis. 23. A quadratic function with vertex at the point (0, 0) and with focal point (0, 21). 24. A polynomial function with horizontal intercepts at 21, 3, and 4.
III. Is each of the statements in Problems 38258 true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample. 38. All polynomial functions are power functions. 39. All linear functions are polynomial functions. 40. All power functions are polynomial functions. 41. All quadratic functions are polynomial functions. 42. The quadratic function with vertex at the origin, (0, 0) and focal point at (0, 1) will be narrower than the quadratic function with the same vertex but with focal point at (0, 4). 43. A polynomial function of degree 4 will always have three turning points. 44. A polynomial function of degree 4 will cross the horizontal axis exactly four times. 45. A polynomial function of odd degree must cross the horizontal axis at least one time. 46. Quadratic functions ƒ(x) 5 ax 2 1 bx 1 c always have two distinct zeros because the equation ax 2 1 bx 1 c 5 0 2 2 4ac . always has two roots, x 5 2b 6 "b 2a
47. If the discriminant is 0, then the associated quadratic function has no horizontal intercept.
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48. The leading term determines the global shape of the graph of a polynomial function.
50. Quadratic functions that open upward have a minimum value at the vertex.
49. The three polynomial functions in the accompanying figure are all of even degree.
51. (ƒ g)(x) 5 (g ƒ)(x) for any functions ƒ and g.
y
A
C
s
s
52. The functions ƒ(x) 5 ln x and g(x) 5 ex are inverses of each other. 1 53. ƒ21 (x) 5 ƒ(x) for any function ƒ.
54. If ƒ(x) 5 1x and g(x) 5 2x2 1 1, then (ƒ g)(x) 5 (2x2 11 1) (2x2 1 1) and ƒ(x) ? g(x) 5 . x s
55. If ƒ(x) 5 ƒ(2x), the graph of ƒ is symmetric across the x-axis.
x
56. If ƒ(x) 5 2ƒ(2x), the graph of ƒ is symmetric about the origin. 57. A function that passes the vertical line test has an inverse. 58. Every function is 1-1.
B
CHAPTER 8 REVIEW: PUTTING IT ALL TOGETHER Some problems are identified as needing a calculator that calculates roots, a graphing program, or technology to generate a best-fit polynomial. 1. For each of the accompanying parabolas, identify the graph as concave up or down, and then estimate the minimum (or maximum) point, the axis of symmetry, and any horizontal intercepts. y
1
Q
40
3. (Requires graphing program for parts (e) and (f).) A wood craftsman has created a design for a parquet floor. The pattern for an individual tile is shown in the accompanying image. The square center (x inches wide) of the tile is made from white oak hardwood and is surrounded by 1-inch strips of maple hardwood.
10
30
t –6
–4
–2
2
4
6
20
x
–10 10
x –4
–2
4
–10 Graph A
6
–20
1
8 –30
Graph B
2. An electric heater is being designed as a parabolic reflector 6 deep. To prevent accidental burns, the center of the heating element is placed at the focus, which is set 1.5 from the vertex of the reflector. a. What equation describes the shape of the reflector? b. How wide will the reflector be? c. Sketch an image of the parabolic reflector with the vertex at its origin. Put a circle at the focal point, and label the depth and width of the reflector.
a. What is the area of the interior white oak square (in terms of x)? The area of each of the maple 1-inch strips? b. White oak costs $2.39 per square foot; maple costs $4.49 per square foot. Calculate the cost per square inch for white oak and for maple. c. What is the cost for the white oak in one tile of the parquet? What is the cost of the four maple strips in one tile? d. The approximate labor cost to make each tile is $5.00. Create a cost function C(x) (in dollars) for making one parquet tile. What type of function is this?
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e. Use technology to graph the function C(x), where x is the width of the inner white oak square. Use a domain of 0 # x # 15 inches. f. From your graph, estimate the size of a parquet tile if the total cost (including labor) is to be $7.00 or less per tile.
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ii. In the crop development growth stage, the canopy is between 10% and 75%. Estimate the crop coefficient when the canopy is at 75% and then calculate it. What does this number mean? 5. Construct a function for each parabola g(x) and h(x)in the accompanying graph.
4. California produces nearly 95% of the processing tomatoes grown in the United States. Therefore, managing irrigation water for tomatoes is a major issue. Agricultural researchers have been able to quantify the relationship between C, the canopy coverage, and KC, the crop coefficient.7 Canopy cover is the percentage of the total plot covered by shade produced by the leaves of the plants. The crop coefficient is a measure of the water needed by the plants. (Technically, it is the ratio of the amount of water plants need divided by the amount used for an equivalent area of well-watered grass.) The following graph shows the relationship between the crop canopy and the crop coefficient.
8
g(x)
x –2
1.2
2
1.1 h(x)
1.0
–2
Crop coefficient
0.9 0.8 0.7 0.6 2004 2003 2003 2002 2001 Regression
0.5 0.4 0.3 0.2 0.1 0 0
10
20
30
40
50
60
70
80
90
100
Canopy coverage (%)
a. As the plants grow, will the canopy cover increase or decrease? Why? b. The graph suggests that as the plant grows, the crop coefficient increases—but at a decreasing rate. Why would that be true? c. The best-fit function to the data is a quadratic: KC 0.126 (0.0172)C – (0.0000776)C2 where 0 # C # 100% i. The initial growth stage has 10% canopy coverage. Estimate the corresponding crop coefficient from the graph, then calculate it using the quadratic model. What does this number mean in terms of well-watered lawn grass?
7
B. R. Hanson and D. M. May, “New crop coefficients developed for highyield processing tomatoes,” California Agriculture 60(2), April–June
2006.
6. (Graphing program optional.) On Earth, the distance dEarth a freely falling object has traveled is a function of time t. It can be modeled by the equation dEarth 16t2, where t is in seconds and dEarth is in feet. On Mars, the comparable equation is dMars 6.1t2. a. Will an object fall faster on Earth or on Mars? b. Plot the two functions on the same graph for 0 # t # 2.5 seconds. c. Calculate the number of seconds on Earth it would take a freely falling object to fall 100 feet. Does your graph confirm your answer? d. How far would a freely falling object fall on Mars during the same time (as in part (c))? e. On Jupiter, the equation for the distance a freely falling object has traveled (in feet) is given by dJupiter 40.65t2 (where t is in seconds). Will an object fall faster or more slowly on Jupiter than on Earth? How would its graph compare with those of Earth and Mars? f. If you double the time, what happens to the distance an object has fallen on Earth? On Mars? On Jupiter? 7. a. Identify the coordinates of the vertex for each the following quadratic functions. F(x) x2, G(x) x2 5, H(x) (x 2)2 , and J(x) (x 1)2 5 b. Without using technology, draw a rough sketch on the same grid of all the functions for 4 # x # 4. c. Describe how the graph of F was transformed into the graphs of G, H, and J, respectively.
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8. Find any horizontal intercepts for the following functions. a. y (x 3)(2x 1) c. Q(t) 2t2 t 1 2 b. G(z) 2z z 3 9. a. Construct a quadratic function Q(t) that is concave up and has horizontal intercepts at t 4 and t 2. Write it in both factored and standard form. Find its vertex. b. Construct a second function M(t) that has the same horizontal intercepts as Q(t) but is steeper. Write M(t) in both factored and standard form. Do the two functions have the same vertex? c. Add a term to Q(t) to create a function P(t) that has no horizontal intercepts. 10. Explain why you could (or couldn’t) construct a parabola through any three points. Problems 11, 12, and 13 refer to the accompanying diagram of the cross section of a swimming pool with a reflective parabolic roof. 100' d
32'
72'
11. Find an equation for the cross section of the parabolic roof of the swimming pool in the diagram. (Hint: Place the origin of your coordinate system at the vertex and identify two other points on the parabola in terms of d.) 12. (Requires results of Problem 11.) The pool designer wants to mount a light source at the parabolic focus so that it sheds light evenly on the water surface below. How many feet down from the vertex must that be? 13. A diver jumps up off the high board, which is 25 feet above the surface of the water. Her height, H(t), in feet above the water at t seconds, can be modeled by the function H(t) 25 12t 16t2. a. What will be the highest point above the water of her dive? b. When will she hit the water? 14. In the United States a “heat wave” is a period of three or more consecutive days at or above 90F. A heat wave is often accompanied by high humidity, making the air feel even hotter. The following formula combines an air temperature of 90F with relative humidity, H, to give the apparent temperature, A, the perceived level of heat: A 86.61 0.132H 0.0059H
2
This formula uses relative humidity as a percentage (e.g., 70% relative humidity appears in the formula as H 70). Remember, this formula applies only for an air temperature of 90F. a. If the relative humidity is 0%, what is the apparent temperature for an air temperature of 90F? Does the apparent temperature feel lower or higher than the air temperature of 90F? b. On a 90F day, if the relative humidity is 60%, what is the apparent temperature? How much hotter do you feel? c. An apparent temperature of 105F or above is considered dangerous, especially for children and elders. At what relative humidity on a 90F day is an apparent temperature of 105F reached? 15. a. Complete the y x2 4x.
x
y
1 0 1 2 3 4 5
5 0 3 4
following
table
for
the
function
Average Rate of Change
Average Rate of Change of Average Rate of Change
n.a. 5 3
n.a. n.a. [3 (5)]/(1 0) 2
b. If you plotted the points with coordinates of the form (x, average rate of change) and connected adjacent points, what type of function would you get? c. Does the fourth column verify your result in part (b)? Why or why not? 16. Using the three accompanying graphs of polynomial functions, determine whether the degree of the polynomial is odd or even, identify its minimum possible degree, and estimate any visible horizontal intercepts of the function.
100
y
x –10
10
–100 Graph A
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10
y x
–10
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19. On the same grid, hand-draw rough sketches of the three functions f(x) ex, g(x) 4ex, and h(x) 0.5ex for 0 # x # 5. Describe the relationships among the graphs.
10
20. a. Given the following graph of the function f(x), sketch: y
f (x)
–50 Graph B
1000
x
y
x 10
–10
i. g(x) f(x) ii. h(x) f(x) iii. j(x) | f(x) | b. Describe each function in relation to f(x). 21. (Graphing program required.)
–2000 Graph C
17. (Graphing technology that can generate a best-fit polynomial is required.) Commercial beekeepers rent out their bees to farmers. The table below gives the average price paid by California almond farmers for each hive of bees (according to Lance Sundberg, who owns and operates the honey business Sunshine Apiary in Columbus, Montana). 2000 2001 2002 2003 2004 2005 2006 $/hive $51.00 $51.50 $52.00 $52.50 $53.00 $82.50 $128.50
a. Plot the data. What sort of function do you think might best fit the data? Use technology to find that best-fit function P(t), where t years since 2000. b. Use your model to predict the average price per hive a beekeeper would get in 2010. c. On average, there are 2.5 hives per acre of almonds. Predict the amount a beekeeper might get per acre in 2010. d. Unfortunately, in 2007, for some unknown reason, bees throughout the country have not been returning to their hives after being released. Do you think that the price per hive will go up or down? Why? 18 a. Generate two different polynomials, M(z) and N(z), that have horizontal intercepts at z 2, 0, and 3. b. Generate a third polynomial, P(z), with the same horizontal intercepts but a higher degree.
a. If g(x) 5 3(x 22 4) 2 1 and f(x) 1x , describe the transformation of the graph of f(x) into the graph of g(x). b. Rewrite g(x) as a ratio of two polynomials. c. What is the domain of g(x)? Sketch g(x). d. What are its horizontal and vertical intercepts (if any)? e. Does g(x) have a vertical asymptote? f. What is its end behavior? 22. Global warming melts glaciers and polar ice, so scientists predict that the sea level will rise, flooding coastal areas. a. Assuming Earth is a sphere with radius r and that roughly three-quarters of Earth’s surface is ocean (with or without melt water), develop a formula to estimate the volume of melt water necessary to raise the sea level 1 foot. (Note: The volume of a sphere is 43pr 3.)
r+1 r
b. Given the Earth’s radius is currently about 3959 miles and 1 cubic foot 7.481 gallons, how many gallons of melt water does your estimate predict? (Recall: 1 mile 5280 feet.) 23. Retirement fund counselors often recommend a mixed portfolio of investments, including some higher-risk investments, which offer higher interest rates, and some more secure investments, with lower interest rates. A woman wants to put half of her $10,000 savings in a safe 4% fund, and the other half in a riskier
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10% fund. She expects to retire in 30 years but would like to know how much she can expect to get if she retires earlier. a. Create three functions where t is the number of years since the start of the investments and S(t) is the amount of money in the 4% account, R(t) is the amount in the 10% account, and T(t) is the total amount invested in both accounts. b. On one graph show how the 4% fund, the 10% fund, and the combined fund total accumulate over 30 years. c. In the worst-case scenario, if she loses all the money in the 10% fund, how much will she be left with in 30 years? d. You might think that if she is getting 4% on $5000 plus 10% on another $5000, this is the same as getting 14% on $5000. Is it? If not, why not? You can explain your answer using a table and/or a graph. 24. Given the functions f(x) x2, g(x) x 1, and h(x) 3x, evaluate each of the following compositions. a. (f g)(x) and (g f )(x) b. (h g)(x) and (g h)(x) c. (f g h)(x) and (h g f)(x) s
s
s
s
s
s
s
s
25. Does the function f(x) (x 2)3 1 have an inverse? If not, explain why. If so, what is it? 26. Which of these functions has an inverse? If there is one, what is it? y 300
2000
y
x –20
20
x
–15
15 –100
a. Graph of y 3x2 1
–2000
b. Graph of y 2x3
27. When lightning strikes, you seem to see it right away, but the associated thunder often comes a few seconds later. One rule of thumb is that each second of delay represents 1000 feet; that is, if you hear the thunder 3 seconds after the lightning strike, the strike was about 3000 feet away from you. a. Light travels about 186,000 miles per second, so the light created from a lightning strike a few thousand feet away is seen virtually simultaneously with the strike. However, sound travels much more slowly, at about 761 mph at sea level. i. Convert 761 mph into feet/second. ii. Now construct a function that gives the distance D(t) (in feet) that the thunder has traveled from the strike site in t seconds. iii. Does the rule of thumb seem reasonable? b. The sound travels in all directions, creating expanding “sound circles” that radiate out from the lightning strike. i. Create a function A(r) (in square feet) that gives the area of a sound circle with radius r (in feet). ii. We can think of the radius r of the sound circle as D(t), the distance thunder has traveled (in any direction) in t seconds. Substituting r for D(t), construct a composite function A(D(t)) to describe the circular area at time t within which the thunder can be detected. What is the circular area (in square feet) within which thunder can be heard 4 seconds after the lightning strike? What is the area in square miles? iii. When the time doubles, what happens to the distance the thunder has traveled? What happens to the area within which it can be heard?
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E X P L O R AT I O N 8 . 1 How Fast Are You? Using a Ruler to Make a Reaction Timer1 Objective • learn about the properties of freely falling bodies and your own reaction time Materials/Equipment • several 12s rulers • narrow strips of paper and tape • calculators Procedure General Description Work in groups of two or three. Each group has a 12s ruler and will attach a 12s paper strip to the ruler, adding some marks (specified below). One student drops the ruler between the thumb and forefinger of a second student. The second student tries to catch the ruler as quickly as possible (see image). The reaction time of the second student can be measured by how far the ruler falls before it is caught.
Mathematical Background Near the surface of Earth, and neglecting air resistance, gravity causes dropped objects to fall approximately according to the formula d 5 16t2 1
This exploration was developed by Karl Schaffer, Mathematics Department, De Anza College, Cupertino, CA. 555
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where d 5 distance fallen in feet and t 5 time of fall in seconds. 1. Thus an object in free fall for 1 second will fall 16 feet, and an object in free fall for 0.5 second will fall ________ feet. (Calculate.) 2. Can you give an intuitive explanation for why the object falling for 0.5 second does not fall half as far as the object that fell for 1 second? 3. What assumptions must we make about the shape of the dropped object for it to fall according to this formula? (Hint: Will a sheet of paper fall 16 feet in 1 second?) 4. Do you think a heavy object will fall faster or slower or at the same rate as a light object? Explain your reasoning. 5. For each of the following times, use the formula to calculate how far a dropped object will fall.
Time, t
Distance in feet, d
Distance in inches
0.05 second 0.10 second 0.15 second 0.20 second 0.25 second
__________ __________ __________ __________ __________
__________ __________ __________ __________ __________
6. Use tape to attach a strip of paper along the length of the ruler. Think of the ruler as measuring distance fallen. For each distance (in inches) in your previous table, put a mark on the paper that indicates the time corresponding to that distance. So you’ll have unevenly spaced marks for the times 0.05 sec, 0.10 sec, up to 0.25 sec. Now the ruler is a reaction timer. 7. One member of your group holds the ruler just above the outstretched thumb and first finger of a second person. The “dropper” suddenly drops the ruler and the “catcher” tries to catch it. Use the time marks on the strip of paper to get an estimate for the reaction time of the catcher. Use the actual number of inches at the point where the catcher caught the ruler to calculate the reaction time. Record the reaction time, and average several tries. If you like, measure the reaction time for someone else in your group.
Person
Distance in inches
Distance in feet, d
Reaction time, t
__________ __________ __________ __________ __________ __________ __________
__________ __________ __________ __________ __________ __________ __________
__________ __________ __________ __________ __________ __________ __________
__________ __________ __________ __________ __________ __________ __________
What was your average reaction time? How did yours compare with that of others in your group? Further Investigations 1. A popular party trick has one person drop a dollar bill between the fingers of a second person. Usually the bill will fall through the second person’s fingers without being caught. How long is a dollar bill, and what must the second person’s reaction time be for the bill to be caught? Does the use of money speed up the reaction times you measured with the ruler? Does the bill fall with only negligible air resistance? 2. How do medications or drugs affect our reaction times? Test the reaction times of someone who is taking cold or flu medication or aspirin, or has just drunk a cup of coffee or glass of alcohol (outside of class, of course). Is his or her reaction time impaired? Is there a correlation between the amount of alcohol consumed and reaction time that might enable you to use your ruler
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3.
4.
5. 6.
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reaction timer as a portable tester to determine whether someone who has consumed alcohol should not drive? Do reaction times measured in this activity improve with practice? Why or why not? (Try it!) Do you think the catcher learns to detect subtle indications of the dropper that she or he is about to drop the ruler? How might these biases be removed from this experiment? Jugglers, athletes, and dancers need to understand, either intuitively or objectively, their own reaction times. In what other occupations is reaction time important? Using the Internet, can you find the reaction times necessary in any of these areas? The Guinness Book of World Records lists the fastest times for drawing and firing a gun. Look this up. Are these times consistent with your results? What else might you investigate about reaction times and your ability to measure them?
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AN EXTENDED EXPLORATION
THE MATHEMATICS OF MOTION OVERVIEW In this extended exploration we use the laboratory methods of modern physicists to collect and analyze data about freely falling bodies and then examine the questions asked by Galileo about bodies in motion. After conducting this exploration, you should be able to • understand the importance of the scientific method • describe the relationship between distance and time for freely falling bodies • derive equations describing the velocity and acceleration of a freely falling body
559
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The Scientific Method Today we take for granted that scientists study physical phenomena in laboratories using sophisticated equipment. But in the early 1600s, when Galileo did his experiments on motion, the concept of laboratory experiments was unknown. In his attempts to understand nature, Galileo asked questions that could be tested directly in experiments. His use of observation and direct experimentation and his discovery that aspects of nature were subject to quantitative laws were of decisive importance, not only in science but in the broad history of human ideas. Ancient Greeks and medieval thinkers believed that basic truths existed within the human mind and that these truths could be uncovered through reasoning, not empirical experimentation. Their scientific method has been described as a “qualitative study of nature.” Greek and medieval scientists were interested in why objects fall. They believed that a heavier object fell faster than a lighter one because “it has weight and it falls to the Earth because it, like every object, seeks its natural place, and the natural place of heavy bodies is the center of the Earth. The natural place of a light body, such as fire, is in the heavens, hence fire rises.”1 Galileo changed the question from why things fall to how things fall. This question suggested other questions that could be tested directly by experiment: “By alternating questions and experiments, Galileo was able to identify details in motion no one had previously noticed or tried to observe.”2 His quantitative descriptions of objects in motion led not only to new ways of thinking about motion, but also to new ways of thinking about science. His process of careful observation and testing began the critical transformation of science from a qualitative to a quantitative study of nature.3 Galileo’s decision to search for quantitative descriptions “was the most profound and the most fruitful thought that anyone has had about scientific methodology.”4 This approach became known as the scientific method.
The Free Fall Experiment Instructions for conducting the free fall experiment are in the last section.
In this extended exploration, you will conduct a modern version of Galileo’s free fall experiment. This classic experiment records the distance that a freely falling object falls during each fraction of a second. The experiment can be performed either with a graphing calculator connected to a motion sensor or in a physics laboratory with an apparatus that drops a heavy weight and records its position on a tape. In this experiment, Galileo sought to answer the following questions: How can we describe mathematically the distance an object falls over time? Do freely falling objects fall at a constant speed? If the speed of freely falling objects is not constant, is it increasing at a constant rate?
The software “Q11: Freely Falling Objects” in Quadratic Functions provides a simulation of the free fall experiment.
You can try to find answers to these questions by collecting and analyzing your own data or by using the data provided as both Excel and graph link files. Instructions for using technology to collect and analyze data are provided in the last section. The following discussion will help you analyze your results and provide answers to Galileo’s questions.
1
M. Kline, Mathematics for the Nonmathematician (New York: Dover, 1967), p. 287. E. Cavicchi, “Watching Galileo’s Learning,” in the Anthology of Readings on the course website. 3 Galileo’s scientific work was revolutionary in terms not only of science but also of the politics of the time; his work was condemned by the ruling authorities, and he was arrested. 4 M. Kline, op. cit., p. 288. 2
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Interpreting Data from a Free Fall Experiment DATA
See the Excel or graph link file FREEFALL, which contains the data in Table 1.
t=0 d=0
The sketch of a tape given in Figure 1 gives data collected by a group of students from a falling-object experiment. Each dot represents how far the object fell in each succeeding 1/60 of a second. Since the first few dots are too close together to get accurate measurements, we start measurements at the sixth dot, which we call dot0. At this point, the object is already in motion. This dot is considered to be the starting point, and the time, t, at dot0 is set at 0 seconds. The next dot represents the position of the object 1/60 of a second later. Time increases by 1/60 of a second for each successive dot. In addition to assigning a time to each point, we also measure the total distance fallen, d (in cm), from the point designated dot0. For every dot we have two values: the time, t, and the distance fallen, d. At dot0, we have t 5 0 and d 5 0. The time and distance measurements from the tape are recorded in Table 1 and plotted on the graph in Figure 2. Time, t, is the independent variable, and distance, d, is the dependent variable. The graph gives a representation of the data collected on distance fallen over time, not a picture of the physical motion of the object. The graph of the data looks more like a curve than a straight line, so we expect the average rates of change between different pairs of points to be different. We know how to calculate the average rate of change between two points and that it represents the slope of a line segment connecting the two points: change in distance 5 slope of line segment change in time
t =1/60 d = 1.72
average rate of change 5
t = 2/60 d = 3.75 t = 3/60 d = 6.10 t = 4/60 d = 8.67 t = 5/60 d = 11.58
Figure 1 Tape from a
free fall experiment.
0.0000 0.0167 0.0333 0.0500 0.0667 0.0833 0.1000 0.1167 0.1333 0.1500 0.1667 0.1833 0.2000 0.2167 0.2333 0.2500 0.2667 0.2833 0.3000 0.3167 0.3333 0.3500 Table 1
Total Distance Fallen (cm) 0.00 1.72 3.75 6.10 8.67 11.58 14.71 18.10 21.77 25.71 29.90 34.45 39.22 44.22 49.58 55.15 60.99 67.11 73.48 80.10 87.05 94.23
100
d
90
80
70
Distance (cm)
Time (sec)
60
50
40
30
20
10
t 0.0
0.1
0.2 Time (sec)
Figure 2 Free fall: distance versus time.
0.3
0.4
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Table 2 and Figure 3 show the increase in the average rate of change over time for three different pairs of points. The time interval nearest the start of the fall shows a relatively small change in the distance per time step and therefore a relatively gentle slope of 188 cm/sec. The time interval farthest from the start of the fall shows a greater change of distance per time step and a much steeper slope of 367 cm/sec. t
d
Average Rate of Change
0.0500 0.1333
6.10 21.77
21.77 2 6.10 < 188 cm/sec 0.1333 2 0.0500
0.0833 0.2333
11.58 49.58
49.58 2 11.58 < 253 cm/sec 0.2333 2 0.0833
0.2167 0.3333
44.22 87.05
87.05 2 44.22 < 367 cm/sec 0.3333 2 0.2167
Table 2
100
d
90 80 70 Distance (cm)
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Slope = 367 cm/sec
60 50 40 Slope = 253 cm/sec
30 20 10
Slope = 188 cm/sec
t 0.0
0.1
0.2 Time (sec)
0.3
0.4
Figure 3 Slopes (or average velocities)
between three pairs of end points.
In this experiment the average rate of change has an additional important meaning. For objects in motion, the change in distance divided by the change in time is also called the average velocity for that time period. For example, in the calculations in Table 2, the average rate of change of 188 cm/sec represents the average velocity of the falling object between 0.0500 and 0.1333 second.
average velocity 5
change in distance change in time
Important Questions Do objects fall at a constant speed?5 The rate-of-change calculations and the graph in Figure 3 indicate that the average rate of change of position with respect to time—the
5
In everyday usage, “speed” and “velocity” are used interchangeably. In physics, “velocity” gives the direction of motion by the sign of the number—positive for forward, negative for backward. “Speed” means the absolute value, or magnitude, of the velocity. So speed is never negative, whereas velocity can be positive or negative.
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velocity—of the falling object is not constant. Moreover, the average velocity appears to be increasing over time. In other words, as the object falls, it is moving faster and faster. Our calculations agree with Galileo’s observations. He was the first person to show that the velocity of a freely falling object is not constant. This finding prompted Galileo to ask more questions. One of these questions was: If the velocity of freely falling bodies is not constant, is it increasing at a constant rate? Galileo discovered that the velocity of freely falling objects does increase at a constant rate. If the rate of change of velocity with respect to time is constant, then the graph of velocity versus time is a straight line. The slope of that line is constant and equals the rate of change of velocity with respect to time. A theory of gravity has been built around Galileo’s discovery of a constant rate of change for the velocity of a freely falling body. This constant of nature, the gravitational constant of Earth, is denoted by g and is approximately 980 cm/sec2.
Deriving an Equation Relating Distance and Time Galileo wanted to describe mathematically the distance an object falls over time. Using mathematical and technological tools not available in Galileo’s time, we can describe the distance fallen over time in the free fall experiment using a “best-fit” function for our data. Galileo had to describe his finding in words. Galileo described the free fall motion first by direct measurement and then abstractly with a time-squared rule. “This discovery was revolutionary, the first evidence that motion on Earth was subject to mathematical laws.”6 Using Galileo’s finding that distance is related to time by a time-squared rule, we use technology to find the following best-fit quadratic function for the free fall data in Table 1: d 5 487.8t 2 1 98.73t 2 0.0528
d 5 488t 2 1 99t 2 0 5 488t 2 1 99t
(1)
We now have a mathematical model for our free fall data.
100
d
90 80 Distance (cm)
If you are interested in learning more about how Galileo made his discoveries, read Elizabeth Cavicchi’s “Watching Galileo’s Learning.”
Figure 4 shows a plot of the data and the function. If your curve-fitting program does not provide a measure of closeness of fit, such as the correlation coefficient for regression lines, you may have to rely on a visual judgment. Rounding the coefficients to the nearest unit, we obtain the equation
70 60 50 40 30 20 10
t 0.0
0.1
0.2 0.3 Time (sec)
0.4
Figure 4 Best-fit function for
distance versus time.
6
E. Cavicchi, “Watching Galileo’s Learning,” in the Anthology of Readings on the course website.
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What are the units for each term of the equation? Since d is in centimeters, each term on the right-hand side of Equation (1) must also be in centimeters. Since t is in seconds, the coefficient, 488, of t 2 must be in centimeters per second squared: cm sec2 ? 5 cm sec2 1 The coefficient, 99, of t must be in centimeters per second, and the constant term, 0, in centimeters. If we ran the experiment again, how would the results compare? In one class, four small groups did the free fall experiment, plotted the data, and found a corresponding bestfit second-degree polynomial. The functions are listed below, along with Equation (1). In each case we have rounded the coefficients to the nearest unit. All of the constant terms rounded to 0. d 5 488t 2 1 99t d 5 486t 2 1 72t d 5 484t 2 1 173t d 5 486t 2 1 73t d 5 495t 2 1 97t
(1) (2) (3) (4) (5)
Examine the coefficients of each of the terms in these equations. All the functions have similar coefficients for the t 2 term, very different coefficients for the t term, and zero for the constant term. Why is this the case? Using concepts from physics, we can describe what each of the coefficients represents. The coefficients of the t 2 term found in Equations (1) to (5) are all close to one-half of 980 cm/sec2, or half of g, Earth’s gravitational constant. The data from this simple experiment give very good estimates for 12 g. The coefficient of the t term represents the initial velocity, v0, of the object when t 5 0. In Equation (1), v0 5 99 cm/sec. Recall that we didn’t start to take measurements until the sixth dot, the dot we called dot0. So at dot0, where we set t 5 0, the object was already in motion with a velocity of approximately 99 cm/sec. The initial velocities, or v0 values, in Equations (2) to (5) range from 72 to 173 cm/sec. Each v0 represents approximately how fast the object was moving when t 5 0, the point chosen to begin recording data in each of the various experiments. The constant term rounded to zero in each of Equations (1) to (5). On the tape where we set t 5 0, we set d 5 0. So we expect that in all our best-fit equations the constant terms, which represent the distance at time zero, are approximately zero. If we substitute zero for t in Equations (1) to (5), the value for d is indeed zero. If we looked at additional experimental results, we might encounter some variation in the constant term, but all should have values of approximately zero. Galileo’s discoveries are the basis for the following equations relating distance and time:
The general equation of motion of freely falling bodies that relates distance fallen, d, to time, t, is 1 d 5 gt 2 1 v0 t 2 where v0 is the initial velocity and g is the acceleration due to gravity on Earth.
For example, in our original model, d 5 488t 2 1 99t, the coefficient 488 approximates 1 2 g (in centimeters per second squared) and 99 approximates the initial velocity (in centimeters per second).
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Returning to Galileo’s Question If the velocity for freely falling bodies is not constant, is it increasing at a constant rate? Galileo discovered that the rate of change of the velocity of a freely falling object is constant. In this section we confirm his finding with data from the free fall experiment.
Velocity: Change in Distance over Time If the rate of change of velocity is constant, then the graph of velocity vs. time should be a straight line. Previously we calculated the average rates of change of distance with respect to time (or average velocities) for three arbitrarily chosen pairs of points. Now, in Table 3 we calculate the average rates of change for all the pairs of adjacent points in our free fall data. The results are in column 4. Since each computed velocity is the average over an interval, for increased precision we associate each velocity with the midpoint time of the interval instead of one of the end points. In Figure 5, we plot velocity from the fourth column against the midpoint times from the third column. The graph is strikingly linear.
Time, t Distance Midpoint (sec) Fallen, d (cm) Time, t (sec) 0.0000 0.0167 0.0333 0.0500 0.0667 0.0833 0.1000 0.1167 0.1333 0.1500 0.1667 0.1833 0.2000 0.2167 0.2333 0.2500 0.2667 0.2833 0.3000 0.3167 0.3333 0.3500 Table 3
0.00 1.72 3.75 6.10 8.67 11.58 14.71 18.10 21.77 25.71 29.90 34.45 39.22 44.22 49.58 55.15 60.99 67.11 73.48 80.10 87.05 94.23
0.0083 0.0250 0.0417 0.0583 0.0750 0.0917 0.1083 0.1250 0.1417 0.1583 0.1750 0.1917 0.2083 0.2250 0.2417 0.2583 0.2750 0.2917 0.3083 0.3250 0.3417
Velocity, v (cm/sec)
450
103.2 121.8 141.0 154.2 174.6 187.8 203.4 220.2 236.4 251.4 273.0 286.2 300.0 321.6 334.2 350.4 367.2 382.2 397.2 417.0 430.8
400
v
350
300 Velocity (cm/sec)
See “C3: Average Velocity and Distance” in Rates of Change.
250
200
150
100
50
t 0.0
0.1
0.2
0.3
0.4
Time (sec)
Figure 5 Best-fit linear function for average velocity versus time.
Generating a best-fit linear function and rounding to the nearest unit, we obtain the equation average velocity 5 977t 1 98 where the average velocity, v, is in centimeters per second, and time t, is in seconds. The graph of this function appears in Figure 5. The slope of the line is constant and equals the rate of change of velocity with respect to time. So although the velocity is not constant, its rate of change with respect to time is constant.
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The coefficient of t, 977, is the slope of the line and in physical terms represents g, the acceleration due to gravity. The conventional value for g is 980 cm/sec2. So the velocity of the freely falling object increases by about 980 cm/sec during each second of free fall. With this equation we can estimate the velocity at any given time t. When t 5 0, then v 5 98 cm/sec. This means that the object was already moving at about 98 cm/sec when we set t 5 0. In our experiment, the velocity when t 5 0 depends on where we choose to start measuring our dots. If we had chosen a dot closer to the beginning of the free fall, we would have had an initial velocity lower than 98 cm/sec. If we had chosen a dot farther away from the start, we would have had an initial velocity higher than 98 cm/sec. Note that 98 cm/sec closely matches the value of 99 cm/sec in our best-fit quadratic function (Equation 1).
The general equation that relates, v, the velocity of a freely falling body, to t, time, is v 5 gt 1 v0 where v0 5 initial velocity (velocity at time t 5 0) and g is the acceleration due to gravity.
Acceleration: Change in Velocity over Time Acceleration means a change in velocity or speed. If you push the accelerator pedal in a car down just a bit, the speed of the car increases slowly. If you floor the pedal, the speed increases rapidly. The rate of change of velocity with respect to time is called acceleration. Calculating the average rate of change of velocity with respect to time gives an estimate of acceleration. For example, if a car is traveling at 20 mph and 1 hour later the car has accelerated to 60 mph, then s60 2 20d mph change in velocity 5 5 (40 mph)/hr 5 40 mi/hr 2 change in time 1 hr In 1 hour, the velocity of the car changed from 20 to 60 mph, so its average acceleration was 40 mph/hr, or 40 mi/hr 2.
average acceleration 5
change in velocity change in time
Table 4 uses the average velocity data and midpoint time from Table 3 to calculate average accelerations. Figure 6 shows the plot of average acceleration in centimeters per second squared (the third column) versus time in seconds (the first column). The data lie along a roughly horizontal line. The average acceleration values vary between a low of 756 cm/sec2 and a high of 1296 cm/sec2 with a mean of 982.8. Rounding off, we have acceleration < 980 cm/sec2 This expression confirms that for each additional second of free fall, the velocity of the falling object increases by approximately 980 cm/sec. The longer it falls, the faster it goes. We have verified a characteristic feature of gravity near the surface of Earth: It causes objects to fall at a velocity that increases every second by about 980 cm/sec. We say that the acceleration due to gravity near Earth’s surface is 980 cm/sec2.
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Midpoint Time, t (sec) 0.0083 0.0250 0.0417 0.0583 0.0750 0.0917 0.1083 0.1250 0.1417 0.1583 0.1750 0.1917 0.2083 0.2250 0.2417 0.2583 0.2750 0.2917 0.3083 0.3250 0.3417
Average Velocity, v (cm/sec) 103.2 121.8 141.0 154.2 174.6 187.8 203.4 220.2 236.4 251.4 273.0 286.2 300.0 321.6 334.2 350.4 367.2 382.2 397.2 417.0 430.8
567
Average Acceleration (cm/sec2) n.a. 1116 1152 792 1224 792 936 1008 972 900 1296 792 828 1296 756 972 1008 900 900 1188 828
1500 Acceleration (cm/sec2)
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1000
500
0.0
0.1
0.2 0.3 Time (sec)
0.4
Figure 6 Average acceleration for free fall data.
Table 4
In order to express g in feet per second squared, we need to convert 980 centimeters to feet. We start with the fact that 1 ft 5 30.48 cm. So the conversion factor for centimeters to feet is (1 ft)/(30.48 cm) 5 1. If we multiply 980 cm by (1 ft)/(30.48 cm) to convert centimeters to feet, we get 980 cm 5 s980 cmd a
1 ft b < 32.15 ft < 32 ft 30.48 cm
So a value of 980 cm/sec2 for g is equivalent to approximately 32 ft/sec2. The numerical value used for the constant g depends on the units being used for the distance, d, and the time, t. The exact value of g also depends on where it is measured.7 The conventional values for g, the acceleration due to gravity near the surface of Earth, are g 5 32 ft/sec2 or equivalently
7
g 5 980 cm/sec2 5 9.8 m/sec2
Because Earth is rotating, is not a perfect sphere, and is not uniformly dense, there are variations in g according to the latitude and elevation. The following are a few examples of local values for g. Location Panama Canal Jamaica Denver, CO Pittsburgh, PA Cambridge, MA Greenland
North Latitude (deg) 9 18 40 40.5 42 70
Elevation (m) 0 0 1638 235 0 0
g (cm/sec2) 978.243 978.591 979.609 980.118 980.398 982.534
Source: H. D. Young, University Physics, Vol. I, 8th ed. (Reading, MA: Addison-Wesley, 1992), p. 336.
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Deriving an Equation for the Height of an Object in Free Fall Assume we have the following motion equation relating distance fallen, d (in centimeters), and time, t (in seconds): d 5 490t 2 1 45t Also assume that when t 5 0, the height, h, of the object was 110 cm above the ground. Until now, we have considered the distance from the point the object was dropped, a value that increases as the object falls. How can we describe a different distance, the height above ground of an object, as a function of time, a value that decreases as the object falls? At time zero, the distance fallen is zero and the height above the ground is 110 centimeters. After 0.05 second, the object has fallen about 3.5 centimeters, so its height would be 110 2 3.5 5 106.5 cm. For an arbitrary distance d, we have h 5 110 2 d. Table 5 gives associated values for time, t, distance fallen, d, and height above ground, h. The graphs in Figure 7 show distance versus time and height versus time.
Time, t (sec)
Distance Fallen, d (cm) (d 5 490 t 2 1 45t)
Height above Ground, h (cm) (h 5 110 2 d)
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
0.0 3.5 9.4 17.8 28.6 41.9 57.6 75.8 96.4
110.0 106.5 100.6 92.2 81.4 68.1 52.4 34.2 13.6
Table 5
120
d
120 Height above ground (cm)
100 Distance fallen (cm)
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80 60 40 20
t 0.0
0.1
0.2 0.3 Time (sec)
0.4
h
100 80 60 40 20
0.0
(a) Distance versus time
0.1
0.2 0.3 Time (sec)
t
0.4
(b) Height versus time
Figure 7 Representations of free fall data.
How can we convert the equation d 5 490t 2 1 45t, relating distance fallen and time, to an equation relating height above ground and time? We know that the relationship between height and distance is h 5 110 2 d. We can substitute the expression for d into the height equation: h 5 110 2 d 5 110 2 (490t 2 1 45t) 5 110 2 490t 2 2 45t
(6)
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Switching the order of the terms, we could rewrite this equation as h 5 2490t 2 2 45t 1 110. The constant term, here 110 cm, represents the initial height when t 5 0. By placing the constant term first as in Equation (6), we emphasize 110 cm as the initial or starting value. Height equations often appear in the form h 5 c 1 bt 1 at 2 to emphasize the constant term c as the starting height. This is similar to writing linear equations in the form y 5 b 1 mx to emphasize the constant term b as the base, or starting, value. Note that in Equation (6) for height, the coefficients of both t and t 2 are negative. If we consider what happens to the height of an object in free fall, this makes sense. As time increases, the height decreases. (See Table 5 and Figure 7.) When we were measuring the increasing distance an object fell, we did not take into account the direction in which it was going (up or down). We cared only about the magnitudes (the absolute values) of distance and velocity, which were positive. But when we are measuring a decreasing height or distance, we have to worry about direction. In this case we define downward motion to be negative and upward motion to be positive. In the height equation h 5 110 2 45t 2 490t 2, the constant term, the initial height, is 110 cm. The change in height resulting from the initial velocity, 245t, is negative because the object was moving down when we started to measure it. The change in height caused by acceleration, 2490t 2, is also negative because gravity pulls objects downward in what we are now considering as a negative direction, reducing the height of a falling object. Once we have introduced the notion that downward motion is negative and upward motion is positive, we can also deal with situations in which the initial velocity is upward and the acceleration is downward. The velocity equation for this situation is v 5 2gt 1 v0 where v0 could be either positive or negative, depending on whether the object is thrown upward or downward, and the sign for the g term is negative because gravity accelerates downward in the negative direction.
If we treat upward motion as positive and downward motion as negative, then the acceleration due to gravity is negative. So the general equations of motion of freely falling bodies that relate height, h, and velocity, v, to time, t, are h 5 h 0 1 v0t 2 12gt 2 v 5 2gt 1 v0 where h0 5 initial height, g 5 acceleration due to gravity, and v0 5 initial velocity (which can be positive or negative).
Working with an Initial Upward Velocity If we want to use the general equation to describe the height of a thrown object, we need to understand the meaning of each of the coefficients. Suppose a ball is thrown upward with an initial velocity of 97 cm/sec from a height of 87 cm above the ground. Describe the relationship between the height of the ball and time with an equation. The initial height of the ball is 87 cm when t 5 0, so the constant term is 87 cm. The coefficient of t, or the initial-velocity term, is 197 cm/sec since the initial motion is upward. The coefficient of t2, the gravity term, is 2490 cm/sec2, since gravity causes objects to fall down. Substituting these values into the equation for height, we get h 5 87 1 97t 2 490t 2
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Table 6 gives a series of values for heights corresponding to various times. Figure 8 plots height above ground (cm) vs. time (sec.). 100
h
90
h (cm)
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
87.00 90.63 91.80 90.53 86.80 80.63 72.00 60.93 47.40 31.43 13.00
Table 6
80 70 Height (cm)
t (sec)
60 50 40 30 20 10
t 0.0
0.1
0.2 0.3 Time (sec)
0.4
0.5
Figure 8 Height of a thrown ball.
The graph of the heights at each time in Figure 8 should not be confused with the trajectory of a thrown object. The actual motion we are talking about is purely vertical— straight up and straight down. The graph shows that the object travels up for a while before it starts to fall. This corresponds with what we all know from practical experience throwing balls. The upward (positive) velocity is decreased by the pull of gravity until the object stops moving upward and begins to fall. The downward (negative) velocity is then increased by the pull of gravity until the object strikes the ground.
Collecting and Analyzing Data from a Free Fall Experiment Objective • to describe mathematically how objects fall
Equipment/Materials • graphing calculator with best-fit function capabilities or computer with spreadsheet and function graphing program • notebook for recording measurements and results (sample Lab Book on course website) DATA
If using precollected data, see the Excel or graph link file FREEFALL.
Equipment needed for collecting data in physics laboratory: a. Free fall apparatus b. Meter sticks 2 meters long c. Masking tape Equipment needed for collecting data with CBL® (Calculator-Based Laboratory System®): a. b. c. d.
CBL® unit with AC-9201 power adapter Vernier CBL® ultrasonic motion detector Graphing calculator Extension cord and some object to drop, such as a pillow or rubber ball
Preparation If collecting data in a physics laboratory, schedule a time for doing the experiment and have the laboratory assistant available to set up the equipment and assist with the
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experiment. If collecting data with a CBL® unit with graphing calculator, instructions for using a CBL® unit are in the Instructor’s Manual.
Procedure The following procedures can be used for collecting data in a physics laboratory.8 If you are collecting data with a CBL®, collect the data and go to the Results section. If you are using the precollected data in the file FREEFALL, go directly to the Results section. Collecting the data Since the falling times are too short to record with a stopwatch, we use a free fall apparatus. Every sixtieth of a second a spark jumps between the falling object or “bob” and the vertical metal pole supporting the tape. Each spark burns a small dot on the fixed tape, recording the bob’s position. The procedure is to: 1. 2. 3. 4.
Position the bob at the top of the column in its holder. Pull the tape down the column so that a fresh tape is ready to receive spark dots. Be sure that the bob is motionless before you turn on the apparatus. Turn on the spark switch and bob release switch as demonstrated by the laboratory assistant. 5. Tear off the length of tape recording the fall of the bob. Obtaining and recording measurements from the tapes The tape is a record of the distance fallen by the bob between each sixtieth-of-a-second spark dot. Each pair of students should measure and record the distance between the dots on the tape. Let d 5 the distance fallen in centimeters and t 5 time in seconds. 1. Fasten the tape to the table using masking tape. 2. Inspect the tape for missing dots. Caution: The sparking apparatus sometimes misses a spark. If this happens, take proper account of it in numbering the dots. 3. Position the 2-meter stick on its edge along the dots on the tape. Use masking tape to fasten the meter stick to the table, making sure that the spots line up in front of the bottom edge of the meter stick so you can read their positions off of the stick. 4. Beginning with the sixth visible dot, mark the time for each spot on the tape; that is, write t 5 0/60 sec by the sixth dot, t 5 1/60 by the next dot, t 5 2/60 by the next dot, and so on, until you reach the end of the tape. Note: The first five dots are ignored in order to increase accuracy of measurements. One cannot be sure that the object is released exactly at the time of the spark, instead of between sparks, and the first few dots are too close together to get accurate measurements. When the body passes the sixth dot, it already has some velocity, which we call v0, and this point is arbitrarily taken as the initial time, t 5 0.
t = 0/60 d0 = 0
1/60
2/60
3/60
4/60
d1 =
d2 =
d3 =
d4 =
5. Measure the distances (accurate to a fraction of a millimeter) from the sixth dot to each of the other dots. Record each distance by the appropriate dot on the tape.
8
These procedures are adapted from “Laboratory Notes for Experiment 2: The Kinematics of Free Fall,” University of Massachusetts, Boston, Elementary Physics 181.
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6. Recheck your measurements. 7. Clean your work area.
Results Use your notebook to keep a record of your data, observations, graphs, and analysis of the data. a. Record the data obtained from your measurements on the tape or from using a CBL® unit. If you are entering your data into a function graphing program or a spreadsheet, you can use a printout of the data and staple it into your laboratory notework. Your data should include time, t, and distance fallen, d, as in the following table: t (sec)
d (cm)
0/60 1/60 2/60 ....... To last record
0 – – – –
This table assumes regular time intervals of one-sixtieth of a second. Check your equipment to see whether it uses a different interval size. b. Note at which dot on the tape you started to make your measurements. Analysis of data 1. By hand: a. Graph your data, using the vertical axis for distance fallen, d, in centimeters and the horizontal axis for time, t, in seconds. What does your graph suggest about the average rate of change of distance with respect to time? b. Calculate the average rate of change for distance, d, with respect to time, t, for three pairs of points from your data table: average rate of change 5
change in distance ⌬d 5 change in time ⌬t
Show your work. This average rate of change is called the average velocity of the falling object between these two points. Do your calculations support your answer in part (a)? c. Jot down your observations from your graph and calculations in your notebook. Staple your graph into your notebook. 2. With graphing calculators or computers: a. Use technology to graph your data for the free fall experiment. Plot time, t, on the horizontal axis and distance fallen, d, on the vertical axis. b. Find a best-fit function for distance fallen versus time. c. Use your spreadsheet or graphing calculator to calculate the average rate of change in distance over each of the small time intervals. This average rate of change is the average velocity over each of these time intervals. d. Plot average velocity versus time, with time on the horizontal axis and average velocity on the vertical axis. e. Jot down your observations from your graphs and calculations in your notebook. Be sure to specify the units for any numbers you recorded.
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Conclusions Summarize your conclusions from the experiment: • Describe what you found out from your graph of distance vs. time and your calculations for the average rate of change of distance with respect to time. Is the average rate of change of distance with respect to time the same for each small time interval? • What does your graph of the average velocity vs. time tell you about the average velocity of the freely falling body? Is the average rate of change in velocity from one interval to the next roughly constant? • In light of the readings and class discussion, interpret your graphs for distance and average velocity and interpret the coefficients in the equation you found for distance. In his own version of this experiment, Galileo sought to answer the following questions: How can we describe mathematically the distance an object falls over time? Do freely falling objects fall at a constant speed? If the velocity of a freely falling object is not constant, is it increasing at a constant rate? Use your results to answer these questions.
EXERCISES (A graphing program is optional for many exercises and required for Exercise 30.) 1. Complete the accompanying table. What happens to the average velocity of the object as it falls?
Time (sec) 0.0000
Distance Fallen (cm)
Average Velocity (average rate of change for the previous 1/30 of a second)
0.00
0.0333
3.75
0.0667 0.1000 0.1333 0.1667
8.67 14.71 21.77 29.90
n.a. 3.75 0.0333 8.67 0.0667
2 2 2 2
0.00 0.0000 3.75 0.0333
< 113 cm/sec < 147 cm/sec
2. The essay “Watching Galileo’s Learning” examines the learning process that Galileo went through to come to some of the most remarkable conclusions in the history of science. Write a summary of one of Galileo’s conclusions about motion. Include in your summary the process by which Galileo made this discovery and some aspect of your own learning or understanding of Galileo’s discovery. 3. (Graphing program optional.) The equation d 5 490t 2 1 50t describes the relationship between distance fallen, d, in centimeters, and time, t, in seconds, for a particular freely falling object. a. Interpret each of the coefficients and specify its units of measurement. b. Generate a table for a few values of t between 0 and 0.3 second. c. Graph distance versus time by hand. Check your graph using a computer or graphing calculator if available.
4. A freely falling body has an initial velocity of 125 cm/sec. Assume that g 5 980 cm/sec2. a. Write an equation that relates d, distance fallen in centimeters, to t, time in seconds. b. How far has the body fallen after 1 second? After 3 seconds? c. If the initial velocity were 75 cm/sec, how would your equation in part (a) change? 5. (Graphing program optional.) The equation d 5 4.9t 2 1 1.7t describes the relationship between distance fallen, d, in meters, and time, t, in seconds, for a particular freely falling object. a. Interpret each of the coefficients and specify its units of measurement. b. Generate a table for a few values of t between 0 and 0.3 second. c. Graph distance versus time by hand. Check your graph using a computer or graphing calculator if available. d. Relate your answers to earlier results in this chapter. 1
6. In the equation of motion d 5 2 gt 2 1 v0 t, we specified that distance was measured in centimeters, velocity in centimeters per second, and time in seconds. Rewrite this as an equation that shows only units of measure. Verify that you get centimeters 5 centimeters. 1
7. The equation d 5 2 gt 2 1 v0 t could also be written using distances measured in meters. Rewrite the equation showing only units of measure and verify that you get meters 5 meters. 1
8. The equation d 5 2 gt 2 1 v 0 t could be written using distance measured in feet. Rewrite the equation showing only units of measure and verify that you get feet 5 feet.
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9. A freely falling object has an initial velocity of 50 cm/sec. a. Write two motion equations, one relating distance and time and the other relating velocity and time. b. How far has the object fallen and what is its velocity after 1 second? After 2.5 seconds? Be sure to identify units in your answers. 10. A freely falling object has an initial velocity of 20 ft/sec. a. Write one equation relating distance fallen (in feet) and time (in seconds) and a second equation relating velocity (in feet per second) and time. b. How many feet has the object fallen and what is its velocity after 0.5 second? After 2 seconds? 11. (Graphing program optional.) A freely falling object has an initial velocity of 12 ft/sec. a. Construct an equation relating distance fallen and time. b. Generate a table by hand for a few values of the distance fallen between 0 and 5 seconds. c. Graph distance vs. time by hand. Check your graph using a computer or graphing calculator if available. 12. (Graphing program optional.) Use the information in Exercise 11 to do the following: a. Construct an equation relating velocity and time. b. Generate a table by hand for a few values of velocity between 0 and 5 seconds. c. Graph velocity versus time by hand. If possible, check your graph using a computer or graphing calculator. 13. If the equation d 5 4.9t 2 1 11t represents the relationship between distance and time for a freely falling body, in what units is distance now being measured? How do you know? 14. The distance that a freely falling object with no initial velocity falls can be modeled by the quadratic function d 5 16t 2 , where t is measured in seconds and d in feet. There is a closely related function v 5 32t that gives the velocity, v, in feet per second at time t, for the same freely falling body. a. Fill in the missing values in the following table: Time, t (sec)
Distance, d (ft)
Velocity, v (ft/sec)
1 1.5 2 80 144 b. When t 5 3, describe the associated values of d and v and what they tell you about the object at that time. c. Sketch both functions, distance versus time and velocity versus time, on two different graphs. Label the points from part (b) on the curves. d. You are standing on a bridge looking down at a river. How could you use a pebble to estimate how far you are above the water?
One screen in “Q11: Freely Falling Objects” in Quadratic Functions simulates this activity.
15. (This exercise requires a free fall data tape created using a spark timer.) a. Make a graph from your tape: Cut the tape with scissors crosswise at each spark dot, so you have a set of strips of paper that are the actual lengths of the distances fallen by the object during each time interval. Arrange them evenly spaced in increasing order, with the bottom of each strip on a horizontal line. The end result should look like a series of steps. You could paste or tape them down on a big piece of paper or newspaper. b. Use a straight edge to draw a line that passes through the center of the top of each strip. Is the line a good fit? Each separate strip represents the distance the object fell during a fixed time interval, so we can think of the strips as representing change in distance over time, or average velocity. Interpret the graph of the line you have constructed in terms of the free fall experiment. 16. In the Anthology Reading “Watching Galileo’s Learning,” Cavicchi notes that Galileo generated a sequence of odd integers from his study of falling bodies. Show that in general the odd integers can be constructed from the difference of the squares of successive integers, that is, that the terms (n 1 1)2 2 n2 (where n 5 0, 1, 2, 3, . . .) generate a sequence of all the positive odd integers. 17. The data from a free fall tape generate the following equation relating distance fallen in centimeters and time in seconds: d 5 485.7t2 1 7.6t a. Give a physical interpretation of each of the coefficients along with its appropriate units of measurement. b. How far has the object fallen after 0.05 second? 0.10 second? 0.30 second? 18. What would the free fall equation d 5 490t 2 1 90t become if d were measured in feet instead of centimeters? 19. In the equation d 5 4.9t 2 1 500t, time is measured in seconds and distance in meters. What does the number 500 represent? 20. In the height equation h 5 300 1 50t 2 4.9t 2, time is measured in seconds and height in meters. a. What does the number 300 represent? b. What does the number 50 represent? What does the fact that 50 is positive tell you? 21. (Graphing program optional.) The height of an object that was projected vertically from the ground with initial velocity of 200 m/sec is given by the equation h 5 200t 2 4.9t 2, where t is in seconds. a. Find the height of the object after 0.1, 2, and 10 seconds. b. Sketch a graph of height vs. time. c. Use the graph to determine the maximum height of the projectile and the approximate number of seconds that the object traveled before hitting the ground.
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22. (Graphing program optional.) The height of an object that was shot downward from a 200-meter platform with an initial velocity of 50 m/sec is given by the equation h 5 24.9t 2 2 50t 1 200, where t is in seconds and h is in meters. Sketch the graph of height versus time. Use the graph to determine the approximate number of seconds that the object traveled before hitting the ground. 23. (Graphing program optional.) Let h 5 85 2 490t 2 be a motion equation describing height, h, in centimeters and time, t, in seconds. a. Interpret each of the coefficients and specify its units of measurement. b. What is the initial velocity? c. Generate a table for a few values of t between 0 and 0.3 second. d. Graph height versus time by hand. Check your graph using a computer or graphing calculator if possible. 24. (Graphing program optional.) Let h 5 85 1 20t 2 490t2 be a motion equation describing height, h, in centimeters and time, t, in seconds. a. Interpret each of the coefficients and specify its units of measurement. b. Generate a table for a few values of t between 0 and 0.3 second. c. Graph height versus time by hand. Check your graph using a computer or graphing calculator if possible. 25. At t 5 0, a ball is thrown upward at a velocity of 10 ft/sec from the top of a building 50 feet high. The ball’s height is measured in feet above the ground. a. Is the initial velocity positive or negative? Why? b. Write the motion equation that describes height, h, at time, t.
Weight
Growth rate
Acceleration of growth
26. The concepts of velocity and acceleration are useful in the study of human childhood development. The accompanying figure shows (a) a standard growth curve of weight over time, (b) the rate of change of weight over time (the growth rate or velocity), and (c) the rate of change of the growth rate over time (or acceleration). Describe in your own words what each of the graphs shows about a child’s growth.
Time (a)
Time ( b)
0
Time (c)
Source: Adapted from B. Bogin, “The Evolution of Human Childhood,” BioScience, Vol. 40, p. 16.
27. The relationship between the velocity of a freely falling object and time is given by v 5 2gt 2 66
575
where g is the acceleration due to gravity and the units for velocity are centimeters per second. a. What value for g should be used in the equation? b. Generate a table of values for t and v, letting t range from 0 to 4 seconds. c. Graph velocity vs. time by hand and interpret your graph. d. What was the initial condition? Was the object dropped or thrown? Explain your reasoning. 28. A certain baseball is at height h 5 4 1 64t 2 16t2 feet at time t in seconds. Compute the average velocity over each of the following time intervals and indicate for which intervals the baseball is rising and for which it is falling. In which interval was the average velocity the greatest? a. t 5 0 to t 5 0.5 e. t 5 2 to t 5 3 b. t 5 0 to t 5 0.1 f. t 5 1 to t 5 3 c. t 5 0 to t 5 1 g. t 5 4 to t 5 4.01 d. t 5 1 to t 5 2 29. At t 5 0, an object is in free fall 150 cm above the ground, falling at a rate of 25 cm/sec. Its height, h, is measured in centimeters above the ground. a. Is its velocity positive or negative? Why? b. Construct an equation that describes its height, h, at time t. c. What is the average velocity from t 5 0 to t 5 21 ? How does it compare with the initial velocity? 30. (Graphing program required.) The force of acceleration on other planets. We have seen that the function d 5 12 gt 2 1 v0 t (where g is the acceleration due to Earth’s gravity and v0 is the object’s initial velocity) is a mathematical model for the relationship between time and distance fallen by freely falling bodies near Earth’s surface. This relationship also holds for freely falling bodies near the surfaces of other planets. We just replace g, the acceleration of Earth’s gravitational field, with the acceleration for the planet under consideration. The following table gives the acceleration due to gravity for planets in our solar system: Acceleration Due to Gravity m/sec2 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune
3.7 8.9 9.8 3.7 24.8 10.4 8.5 11.6
ft/sec2 12.1 29.1 32.1 12.1 81.3 34.1 27.9 38.1
Note: Pluto is no longer considered a planet. Source: The Astronomical Almanac, U.S. Naval Observatory, 1981.
a. Choose units of measurement (meters or feet) and three of the planets (other than Earth). For each of these planets, find an equation for the relationship between the distance an object falls and time. Construct a table as shown at the top of the next page. Assume for the moment that the initial velocity of the freely falling object is 0.
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Name of Planet
Function Relating Distance and Time (sec)
Units for Distance
b. Using a graphing program, plot the three functions, with time on the horizontal axis and distance on the vertical axis. What domain makes sense for your models? Why? c. On which of your planets will an object fall the farthest in a given time? On which will it fall the least distance in a given time? d. Examine the graphs and think about the similarities that they share. Describe their general shape. What happens to d as the value for t increases? e. Think about the differences among the three curves. What effect does the coefficient of the t2 term have on the shape of the graph; that is, when the coefficient gets larger (or smaller), how is the shape of the curve affected? Which graph shows d increasing the fastest compared with t? 31. Suppose an object is moving with constant acceleration, a, and its motion is initially observed at a moment when its velocity is v0. We set time, t, equal to 0, at this point when velocity equals v0. Then its velocity t seconds after the initial observation is V(t) 5 at 1 v0. (Note that the product of acceleration and time is velocity.) Now suppose we want to find its average velocity between time 0 and time t. The average velocity can be measured in two ways. First, we can find the average of the initial and final velocities by calculating a numerical average or mean; that is, we add the two velocities and divide by 2. So, between time 0 and time t, average velocity 5
v0 1 Vstd 2
(1)
We can also find the average velocity by dividing the change in distance by the change in time. Thus, between time 0 and time t, average velocity 5
d20 d ⌬distance 5 5 ⌬time t20 t
(2)
If we substitute the expression for average velocity (from time 0 to time t) given by Equation (1) into Equation (2), we get d v 1 Vstd 5 0 t 2
(3)
We know that V(t) 5 at 1 v0. Substitute this expression for V(t) in Equation (3) and solve for d. Interpret your results. 32. In 1974 in Anaheim, California, Nolan Ryan threw a baseball at just over 100 mph. If he had thrown the ball straight upward at this speed, it would have risen to a height of over 335 feet and taken just over 9 seconds to fall back to
Earth. Choose another planet and see what would have happened if he had been able to throw a baseball straight up at 100 mph on that planet. In your computations, use the table for the acceleration due to gravity on other planets from Exercise 30. 33. An object that is moving horizontally along the ground is observed to have (an initial) velocity of 60 cm/sec and to be accelerating at a constant rate of 10 cm/sec2. a. Determine its velocity after 5 seconds, after 60 seconds, and after t seconds. b. Find the average velocity for the object between 0 and 5 seconds. 34. (Requires results from Exercise 33.) Find the distance traveled by the object described in the previous exercise after 5 seconds by using two different methods. a. Use the formula distance 5 rate ? time. For the rate, use the average velocity found in Exercise 33(b). For time, use 5 seconds. b. Write an equation of motion d 5 12 at2 1 v0 t using a 5 10 cm/sec2 and v0 5 60 cm/sec and evaluate when t 5 5. Does your answer agree with part (a)? 35. An object is observed to have an initial velocity of 200 m/sec and to be accelerating at 60 m/sec2. a. Write an equation for its velocity after t seconds. b. Write an equation for the distance traveled after t seconds. 36. You may have noticed that when a basketball player or dancer jumps straight up in the air, in the middle of a blurred impression of vertical movement, the jumper appears to “hang” for an instant at the top of the jump. a. If a player jumps 3 feet straight up, generate equations that describe his height above ground and velocity during his jump. What initial upward velocity must the player have to achieve a 3-foot-high jump? b. How long does the total jump take from takeoff to landing? What is the player’s downward velocity at landing? c. How much vertical distance is traveled in the first third of the total time that the jump takes? In the middle third? In the last third? d. Now explain in words why it is that the jumper appears suspended in space at the top of the jump. 37. Old Faithful, the most famous geyser at Yellowstone National Park, regularly shoots up a jet of water 120 feet high. a. At what speed must the stream of water be traveling out of the ground to go that high? b. How long does it take to reach its maximum height? 38. A vehicle trip is composed of the following parts: i. Accelerate from 0 to 30 mph in 1 minute. ii. Travel at 30 mph for 12 minutes. iii. Accelerate from 30 to 50 mph in 21 minute. iv. Travel at 50 mph for 6 minutes. v. Decelerate from 50 to 0 mph in 12 minute.
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a. Sketch a graph of speed versus time for the trip. b. What are the average velocities for parts (i), (iii), and (v) of the trip? c. How much distance is covered in each part of the trip, and what is the total trip distance? 39. In general, for straight motion of a vehicle with constant acceleration, a, the velocity, v, at any time, t, is the original velocity, v0, plus acceleration multiplied by time: v 5 v0 1 at. The distance traveled in time t is d 5 v0 t 1 12 at 2. a. A criminal going at speed vc passes a police car and immediately accelerates with constant acceleration ac. If the police car has constant acceleration ap . ac , starting from 0 mph, how long will it take to pass the criminal? Give t in terms of vc, ap, and ac. b. At what time are the police and the criminal traveling at the same speed? If they are traveling at the same speed, does it mean the police have caught up with the criminal? Explain.
577
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Office of Institutional Research and Policy Studies (OIRP)
1:44 PM
STATISTICAL PORTRAIT FALL 2006
10/11/07
The University of Massachusetts Boston
Tables for Exploration 2.1 (p. 139) from:
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2,740 2,347 1,541 65.7% 743 57 48.2% 806 393 2,890 2,510 2,234 89.0% 1,371 121 61.4% 276 380 5,630 4,857 3,775 77.7% 2,114 178 56.0% 1,082 773
TRANSFERS Applied Decision Ready Admitted Admit Rate Enrolled Admitted but Deferred Yield Rate Not Admitted: Denied Inc. Applications
TOTAL UNDERGRADUATES Applied Decision Ready Admitted Admit Rate Enrolled Admitted but Deferred Yield Rate Not Admitted: Denied Inc. Applications
1996
FRESHMEN Applied Decision Ready Admitted Admit Rate Enrolled Admitted but Deferred Yield Rate Not Admitted: Denied Inc. Applications
Undergraduate Admissions Summary
5,662 4,881 3,665 75.1% 1,984 146 54.1% 1,216 781
6,492 5,338 3,939 73.8% 2,248 222 57.1% 1,399 1,154
7,251 5,688 4,295 75.5% 2,379 119 55.4% 1,393 1,563
3,790 2,964 2,601 87.8% 1,590 89 61.1% 363 826
3,461 2,724 1,694 62.2% 789 30 46.6% 1,030 737
1999
7,486 5,705 4,193 73.5% 2,262 145 53.9% 1,512 1,781
4,008 3,038 2,631 86.6% 1,556 113 59.1% 407 970
3,478 2,667 1,562 58.6% 706 32 45.2% 1,105 811
2000
8,074 5,568 4,103 73.7% 2,243 59 54.7% 1,465 2,506
4,172 2,916 2,564 87.9% 1,542 45 60.1% 352 1,256
3,902 2,652 1,539 58.0% 701 14 45.5% 1,113 1,250
2001
7,901 5,596 3,856 68.9% 1,958 115 50.8% 1,740 2,305
4,063 2,892 2,378 82.2% 1,382 92 58.1% 514 1,171
3,838 2,704 1,478 54.7% 576 23 39.0% 1,226 1,134
2002
8,263 5,613 3,921 69.9% 1,949 33 49.7% 1,692 2,650
4,076 2,779 2,360 84.9% 1,339 20 56.7% 419 1,297
4,187 2,834 1,561 55.1% 610 13 39.1% 1,273 1,353
2003
6,920 5,600 3,678 65.7% 1,758 97 47.8% 1,922 1,320
3,390 2,697 2,125 78.8% 1,193 67 56.1% 572 693
3,530 2,903 1,553 53.5% 565 30 36.4% 1,350 627
2004
7,187 5,813 4,009 69.0% 2,107 66 52.6% 1,804 1,374
3,317 2,639 2,089 79.2% 1,326 51 63.5% 550 678
3,870 3,174 1,920 60.5% 781 15 40.7% 1,254 696
2005
1:44 PM
3,515 2,872 2,479 86.3% 1,574 161 63.5% 393 643
2,977 2,466 1,460 59.2% 674 61 46.2% 1,006 511
1998
10/11/07
2,994 2.576 2,224 86.3% 1,347 102 60.6% 352 418
2,668 2,305 1,441 62.5% 637 44 44.2% 864 363
1997
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0.6% 11.9% 15.3% 7.0% 0.9% 35.7% 55.7% 8.7% [1,509]
2001 0.8% 11.9% 14.0% 8.8% 1.0% 36.5% 58.4% 5.1% [1,244]
2002 0.4% 12.9% 13.3% 8.0% 1.8% 36.4% 60.1% 3.4% [960]
2003 0.6% 12.8% 15.3% 7.1% 1.9% 37.7% 58.6% 3.7% [885]
2004 0.6% 12.5% 12.7% 11.8% 1.4% 39.0% 59.1% 2.0% [1034]
2005
Trend in New Student Race/Ethnicity in the College of Liberal Arts
Native American Asian/Pacific Islander Black Hispanic Cape Verdean Total minority White Non-resident alien Known race [N]
SAT Scores of New Freshmen by College/Program
College of Liberal Arts
SATVerbal SATMath Combined [N]
510 541 1,051 [34]
518 500 1,018 [288]
1996
450 521 971 [51]
502 504 1,006 [244]
1997
468 546 1,014 [54]
513 511 1,024 [328]
1998
481 537 1,018 [43]
521 513 1,034 [418]
1999
516 548 1,064 [42]
515 529 1,044 [363]
2000
511 556 1,067 [34]
524 536 1,060 [369]
2001
515 546 1,061 [37]
520 532 1,052 [317]
2002
492 542 1,034 [36]
522 519 1,041 [224]
2003
487 523 1,010 [35]
528 517 1,045 [201]
2004
511 554 1,065 [59]
540 539 1,079 [278]
2005
SAT Scores of New Freshmen by College/Program, 10-Year Trend (Excluding the DSP Program, Learning Disabled and Foreign Students)
College of Management
SATVerbal SATMath Combined [N]
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SOLUTIONS 8. Table for the histogram in Figure 1.6
CHAPTER 1 Section 1.1 Algebra Aerobics 1.1 Fraction
Decimal
Percent
7 12 1 40 1 50 1 200 7 20
0.583
58.3%
0.025
2.5%
0.02
2%
0.005
0.5%
0.35
35%
0.008
0.8%
(0.20)(1352) < 270 (0.35)(1352) < 473 (0.30)(1352) < 406 (0.15)(1352) < 203 1352
9. a. sum 5 $8750, so mean 5 $8750/9 < $972.22; median 5 $300 b. sum 5 4.7, so mean 5 4.7 4 8 < 0.59; median 5 (0.4 1 0.5)/2 5 0.45
Exercises for Section 1.1
Relative Frequency (%)
52 35 28 22 137
Table 1.3
38 26 20 16 100
5. a. 45 b. Quantitative data 3 < 6.7% c. 45 d.
Salaries Distribution for Graduates 25 20 15 10 5 0 46–50
Frequency Count
3. a. Paying off bills/debts. The other categories add up to 62%, not 66%. b. They do not add up to 100%. Also, the pie chart is 3D and tipped. Thus the front slices are disproportionately larger than the other slices.
41–45
5. a. The numbers (in millions) for the Hispanic population. 35.3 42.7 5 282.4 million; 2005: 0.144 < 296.5 million b. 2000: 0.125 6. a. Frequency Count (FC) for Age 1–20 interval is 38% of total: 38% of 137 5 0.38(137) < 52. FC for Age 61–80 interval is total FC minus all the others: 137 2 (52 1 35 1 28) 5 137 2 115 5 22. Relative Frequency (RF) for each interval is its FC divided by total: 35 RF of (21–40) interval is: 137 < 0.255 < 26% 28 RF of (41–60) interval is: 137 < 0.204 < 20% 22 RF of (61–80) interval is: 137 < 0.161 < 16%
1. a. 34.7% of 13- to 17-year-old females spent at least 3 hours per day watching TV in 2006. b. 39.9% c. The total numbers of male and female teenagers in that age range.
36–40
4. mean GPA < 2.06; median GPA 5 2.0
31–35
3. mean 5 $25,040; median 5 $20,000
1–20 21–40 41–60 61–80 Total
20 35 30 15 100
10. One of the values ($6,000) is much higher than the others which forces a high value for the mean. In cases like this, the median is generally a better choice for measuring central tendency.
2. a. 500 people b. 38.94 or 39 students c. 37.5%
Age
1–20 21–40 41–60 61–80 Total
26–30
125
Frequency Count
21–25
1
Relative Frequency (%)
Counts
1.
Age
Ranges of Salaries in $1000's
b. 20% 1 16% 5 36% 7.
Distribution of Ages (in years) 40% 20% 0%
1-20 21-40 41-60 61-80
Distribution of Ages (in years) 16% 1–20 21–40 38% 20% 41–60 61–80 26%
7. a. In housing; 36.16% b. 0.1438 ? $35,000 5 $5033 c. Answers will vary. Americans spend over 36% of their income for housing. 9. a. Mean 5 386/7 5 55.14; median 5 46. b. Changing any entry in the list that is greater than the median to something still higher will not change the median of the list but will increase the mean. The same effect can be had if an entry less than the median is increased to a value that is still less than or equal to the median. 583
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11. The mean annual salary is $24,700 and the median annual salary is $18,000. The mean is heavily weighted by the two high salaries. The mean salary is more attractive but is not likely to be an accurate indicator. 13. No answer is given here. (In general, when answers from students can vary quite a bit, either a typical answer or none is given.) 15. The mean age in the United States is slightly higher than the median age since there are a lot of older Americans (including the baby boomers), which pulls the mean age higher than the median. In developing countries, the mean age will be less than the median age since there are a lot of younger people and this pulls the mean lower than the median. Answers will vary by State and will depend on how the ages are distributed. 17. He is correct, provided the person leaving state A has an IQ that is below the average IQ of people in state A and above the average IQ of the people in state B. 19. The mean net worth of a group, e.g., American families, is heavily biased upward by the very high incomes of a relatively small subset of the group. The median net worth of a group such as this is not as biased. The two measures would be the same if net worths were distributed symmetrically about the mean. 5
21. a. a g xi b ^5 i51
n
b. a g ti b ^ n
5
g 2k 5 30
c.
i51
k51
23. Here is a progressive table of the work needed. The mean age of 36.6 years was obtained by dividing total people-years by the total number of persons. Total people-years was obtained by adding up the products of midpoint ages by the population counts in each age group.
Age (years)
Population (thousands)
Midpoint Age
Product of Midpoint Age and Population (people-years)
Under 10 10 to 19 20 to 29 30 to 39 40 to 49 50 to 59 60 to 74 75 to 84 85 and over
39,677 41,875 40,532 41,523 45,179 35,986 31,052 12,971 4,860
4.5 14.5 24.5 34.5 44.5 54.5 67 79.5 92.5
178,546 607,187 993,034 1,432,543 2,010,465 1,961,237 2,080,484 1,031,194 449,550
Total
293,655
n.a. Mean age
10,744,242 36.6
25. If we assume an estimated allowance of $5 for all students with an allowance in the $0–9 category, an estimate of $15 for all students with an allowance in the $10–19 category, etc., then the mean < ($5 ? 8 1 $15 ? 6 1 $25 ? 12 1 $35 ? 14 1 $45 ? 9)/49 5 1325/49 < $27. The median lies somewhere between $20 and $29, probably closer to $29.
27. Answer is omitted. Student answers will vary. 29. Some factors to note are given below. You may find others. It is important in giving your answer to cite some population numbers and specific age brackets when making specific comparisons. 1) In Ghana there is a steady decline in the number in each age group from 5 to 9 up to over 80 (from approx. 2.7 million to 0.1 million), whereas in the United States the number of people in each age category is generally the same or more than the number of people 0 to 4 years of age (approx. 19 million) up to the ages 45 to 49 (approx. 23 million). 2) At almost all age levels there seems to be an even distribution of males and females in Ghana; it is somewhat the same in the United States except that from age 60 onward there is a dominance of females in each age category. 3) There is a small bulge in population in the teens as compared to those younger and those just older in the United States (by about 2 million) as compared to a steady decline in Ghana in the population in all age groups once past the 5 to 9 age bracket.
Section 1.2 Algebra Aerobics 1.2a 1. a. The median net worth of households, after decreasing from 1988 to 1993, increased from 1993 to 2000. It reached a low of about $43,600 in 1993 and increased to $55,000 in 2000. b. Many factors would be useful, including size of households and number of wage earners in household. 2. a. 1975 b. 2030 c. 55 years 3. a. b. c. d.
approximately 11 billion approximately 1 billion approximately 11 2 1 5 10 billion The total world population increased rapidly from 1950 to 2000, and is projected to continue to increase but at a decreasing rate, reaching approximately 11 billion in the year 2150. pop 2000
6 5 4. The population of the world in year 4. a. pop 1900 < 1.5 2000 was approximately 4 times greater than in 1900. The difference < 4.5 billion. pop 2100
b. pop 2000 < 10.5 6 5 1.75, so the population in 2100 was approximately 1.75 times greater than in 2000. The difference < 4.5 billion. c. The population from 2000 to 2100 is expected to grow by 4.5 billion people, which is the same as the 4.5 billion increase from 1900 to 2000. The world population in 2000 was approximately 4 times greater than in 1900 and it is projected to be about 1.75 times greater in 2100 than in 2000. While the population continues to increase, the rate of increase is slowing down.
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Algebra Aerobics 1.2b 1. a. Square the value of x, then multiply that result by 3, then subtract the value of x and add 11. b. (0, 1) is the only one of those ordered pairs that is a solution. c. x 23 22 21 0 1 2 3 15
5
1
3
11
14,000
16
9
x
24
22
21
0
1
2
4
y1
16
10
7
4
1
22
28
y2
215
3
6
5
0
29
239
3.
4
1
0
1
4
y
y 20
y1
x 5
–5 –20 x 5
–5
–40
–15
c. Yes for y1; no for y2. d. No for y1; yes for y2. e. No, since 4 2 3(23) 5 13, 22(23)2 2 3(23) 1 5 5 24 4.
y 3 y = x 2 –3x+2 2 1 x –2
–1
0
1
2
12,000 10,000 8000 6000 4000 Personal crimes 2000
a. & b. 25
Personal Crimes from 1995 to 2003
25
2. a. Subtract 1 from the value of x, then square the result. b. (0, 1) and (1, 0) are the only ones of those ordered pairs that are solutions. c. x 23 22 21 0 1 2 3 y
3. a.
3
–1
a. y 5 21/4 b. Many answers: (0, 0) and (22, 5) are examples
Exercises for Section 1.2 1. a. Answers may vary. 1) AIDS cases in the United States have nearly halved from 1993 to 2004, since they dropped from an alltime high of 79,879 in 1993 to 42,514 in 2004. 2) AIDS cases increased slightly from 2001 to 2004. 3) AIDS cases reached a low of about 39,200 in 2001. b. Student answers will vary. Possible topic sentences: AIDS cases in the United States went down from 1993 to 2001 but then started to rise very slowly. The three points
0 1995 1997 1999 2001 2003 Year
b. In 1995 there were approximately 6.7 times more property crimes than violent crimes. In 2003 there were approximately 7.6 times more property crimes than violent crimes. c. Answers may vary. Some observations could be the following. Property crime, from 1995 to 2003, was over six times as great as personal crime. Both property crimes and personal crimes decreased from 1995 to 2003, but from 1995 to 2003 the ratio of property to personal crimes actually increased. 5. a. As men grow older, the risk of cancer goes up dramatically—for example, from 1 in 25,000 at age 45, to 1 in 25 at age 65, to 1 in 6 at age 80. b. Prostate Cancer Risk vs. Age Risk at Percent
31
mentioned in part (a) could form the rest of the paragraph. A concluding sentence might be: “It is worrisome that the number of cases has been on the rise from 2001 to 2004. The number of cases has increased by a little over 3000 from 2001 to 2004.”
Count in thousands
y
585
18.00 16.00 14.00 12.00 10.00 8.00 6.00 4.00 2.00 0.00 40 45 50 55 60 65 70 75 80 85 Age in Years
c. The ratio of risks for a 50- vs. a 45-year-old man is 0.21%/0.004% 5 52.5. This means that a 50-year-old man is 52.5 times more likely to have had prostate cancer than a 45-year-old man. The ratio of risks for a 55- vs. 50-yearold man is 0.83%/0.21% < 4. So a 55-year-old man is only 4 times as likely to have had prostate cancer than a 50-year-old man. At first this may seem contradictory. But the biggest incremental risk occurs between 45 and 50. So while the absolute risk continues to rise (note that the percent risk is cumulative), the ratios of the percent risk decrease over subsequent 5-year age intervals. d. The medical profession has recommended this test for all men from age 40 up. The insurance companies think that this is too expensive. Your answer may be different. 7. a. Except for a dip in the count for 12- to 19-year-olds in the 1976 to 1980 period there has been a dramatic rise in the percentage of children who are overweight.
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b. It increased in going from every time interval to the next, except in going from the first (1963 to 1970) to the second (1971 to 1974), when the percentage stayed the same. c. From the period of 1963 to 1970 to the period of 1971 to 1974. d. During the 1976–1980 period. e. The goal is reasonable, but the data suggest that the trend in obesity will keep rising. 9. a. True; B is the newer car and it costs more than A. b. True; A is the slower car in cruising speed and it has the larger size. c. False; A is the larger car but it is older than B. d. True; A carries more passengers and it is less expensive. e. Your answers will vary. You may mention that the larger range car also has the larger passenger capacity. f. Your answers will vary. Much depends on what features you value. There are many trade-offs. 11. a. Only (1, 23) satisfies the formula. b. There are many answers: e.g., (0, 2), (21, 7). In general, pick a value of T and plug it into the formula to compute the corresponding value of R. c. Here is the scatter plot for the three points given: 8
(0,2) –1.5
1.5 –4
Section 1.3 Algebra Aerobics 1.3 1. Table A represents a function. For each input value there is one and only one output value. Table B does not represent a function. For the input of 2 there are two outputs, 7 and 8. 2. There are two output values, 5 and 7, for the input of 1. For the table to represent a function, there can be only one output value for each input. If you change 5 to 7 or change 7 to 5, then there will be only one output value for the input of 1. 3. Yes, it passes the vertical line test.
Points that satisfy R = 2 – 5T
(–1, 7)
tools could mean more frequent prevention of death. Lastly, not all cancers are fatal. b. Most of the observations given by William M. London are cited in part (a) but his arguments make claims that are not buttressed clearly enough by data. We need to have data for cancer incidence, cancer treatment, cancer conquest, and cancer fatality, for all age groups by sex and type of cancer before we can make any solid overall claims either way. c. The statements given above in parts (a) and (b) can be used to forge a paragraph.
(1, –3)
d. The plot suggests that solutions could be found on the straight line through these three points by eyeballing the line. Another source would be using the formula with other values of T to generate the corresponding values of R. 13. a. Add 1 to the value of x; divide the result by what one gets by subtracting 1 from the value of x. b. (5, 1.5) c. (2, 3) d. No, the formula is not defined if x 5 1. 15. a. b. c. d.
x 5 0 implies y 5 0 If x . 0 then y , 0. If x , 0 then y , 0. No
17. a. b. c. d.
Only (21, 3) satisfies y 5 2x 1 5. (1, 0) and (2, 3) satisfy y 5 x2 2 1. (21, 3) and (2, 3) satisfy y 5 x2 2 x 1 1 Only (1, 2) satisfies y 5 4/(x 1 1).
19. a. There are far more people now than in 1971, and those who are most susceptible, the elderly, are now a greater proportion of the population than in 1971. Thus, it is not necessarily the case that proportionately twice as many now as then will have cancer. Also, better cancer detection
4. Graph B represents a function, while graphs A and C do not represent functions since they fail the vertical line test. 5. Neither is a function of the other. The graph fails the vertical line test, so weight is not a function of height. At height 51 inches there are two weights (115 and 120 pounds) and at height 56 inches there are two weights (135 and 140 pounds). If we reverse the axes, so that weight is on the horizontal axis and height is on the vertical axis, that graph will also fail the vertical line test, since the weight of 140 pounds has two corresponding heights of 56 inches and 58 inches. 6. a. D is a function of Y, since each value of Y determines a unique value of D. b. Y is not a function of D, since one value of D, $2.70, yields two values for Y, 1993 and 1997. 7.
y
a.
y
b.
4
4
2
2
x –2
0
–2
2
4
x –2
0
2
–2
a. The line containing the points is the graph of a function since for each value of x we have a unique value of y. The line is horizontal. Its equation can be written as y 5 3.
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b. The line containing the points is not the graph of a function since for one value of x we have infinitely many values of y. The line is vertical. Its equation can be written as x 5 22.
Exercises for Section 1.3 1. a. Yes, each date has one and only one temperature. b. No, the temperature 278C goes with two different dates. 3. a. function (all input values are different and thus each input has one and only one output) b. function [same reason as in part (a)] c. not a function [same input values have different output values] d. not a function [same reason as in part (c)] 5. a. Not a function: fails the vertical line test, e.g., look at the y-axis. b. Function: passes the vertical line test. c. Not a function: fails the vertical line test, e.g., look at the y-axis 7. a. The formulas are: y 5 x 1 5; y 5 x 2 1 1; y 5 3 b. All three represent y as a function of x; each input of x has only one output y. 9. a. b. c. d.
D
400 (mg)
8. a. Tip 5 15% of the cost of meal, so T 5 0.15M. Independent variable: M (meal price); Dependent variable: T (tip). The equation is a function since to each value of M there corresponds a unique value of T. b. T 5 (0.15)($8) 5 $1.20 c. T 5 (0.15)($26.42) 5 $3.96. One would probably round that up to $4.00.
500
587
300 200 100
0
2
4
6
8
W 10
(kg)
d. The domain and range are now all the real numbers. The table can now include values of W that are less than or equal to zero and values greater than or equal to 10. The graph will continue indefinitely in both directions.
Section 1.4 Algebra Aerobics 1.4a 1. g(0) 5 0 g(21) 5 23 g(1) 5 3 g(20) 5 60 g(100) 5 300 2. ƒ(0) 5 (0)2 2 5(0) 1 6 5 6, so ƒ(0) 5 6 ƒ(1) 5 (1)2 2 5(1) 1 6 5 2, so ƒ(1) 5 2 ƒ(23) 5 (23)2 – 5(23) 1 6 5 30, so ƒ(23) 5 30 2 3. ƒ(0) 5 0 2 1 5 22, so ƒ(0) 5 22; 2 ƒ(21) 5 (21)21 5 21, so ƒ(21) 5 21; 2 ƒ(1) 5 (1)21 5 20, so ƒ(1) is undefined; 2 5 212 , so ƒ(23) 5 21/2 ƒ(23) 5 (23)21
S1 5 0.90 ? P; $90 S2 5 (0.90)2 ? P; $81 S3 5 (0.90)3 ? P; $72.90 S5 5 (0.90)5 ? P; $59.05; 40.95%
4. 5 2 2t 5 3, so t 5 1; 3t 2 9 5 3, so t 5 4; 5t 2 12 5 3, so t 5 3.
11. y is a function of x in parts (a), (b), and (c) but not in (d). In (d), for example, if x 5 1, then y 5 61. 13. a. Since the dosage depends on the weight, the logical choice for the independent variable is W (expressed in kilograms) and for the dependent variable is D (expressed in milligrams). b. In this formula, each value of W determines a unique dosage D, so D is a function of W. c. The following table and graph are representations of the function. Since the points (0, 0) and (10, 500) are not included in the model, these points are represented with a hollow circle on the graph. W (kg)
D (mg)
0 2 4 6 8
0 100 200 300 400
5. 2(x 2 1) 2 3(y 1 5) 5 10 1 2x 2 2 2 3y 2 15 5 10 1 27 2x 2 17 2 3y 5 10; 2x 2 27 5 3y 1 y 5 2x 2 3 27 So y is a function of x. ƒ(x) 5 2x 2 . 3
6. x 2 1 2x 2 y 1 4 5 0 1 x 2 1 2x 1 4 5 y. So y is a function of x. ƒ(x) 5 x2 1 2x 1 4 7. 7x – 2y 5 5 1 –2y 5 –7x 1 5 1 y 5 (–7x 1 5)/(–2) 1 y 5 7x 22 5 So y is a function of x. ƒ(x) 5 7x 22 5
8. ƒ(24) 5 2; ƒ(21) 5 21; ƒ(0) 5 22; ƒ(3) 5 1. When x 5 22 or x 5 2, then ƒ(x) 5 0. 9. ƒ(0) 5 20, ƒ(20) 5 0, ƒ(x) 5 10 when x 5 10 and x 5 30. It is a function because for every value of x, there is one and only one value of ƒ(x).
Algebra Aerobics 1.4b 1. a. (2, `) b. [4, 20) c. (2`, 0] d (500, 1`)
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CH. 1 Algebra Aerobics Solutions for Section 1.5
2. a. 23 # x , 10 b. 22.5 , x # 6.8 c. x # 5 or x $ 12 3. a. [2.5, 3.6] b. [0.333, 1.000] (Note: The highest possible batting average is 1.000, meaning that the batter had a hit every time he has been at bat. This is commonly known as “batting a thousand.”) c. [35000, 50000]
Miles 0 10 20
Cost ($) 2.00 5.20 8.40
Miles 30 40 50
Cost ($) 11.60 14.80 18.00
The graph is in the accompanying diagram. Some of the table values are marked. C
4. 2(x 1 1) 1 3y 5 5 1 2x 1 2 1 3y 5 5 1
35
3y 5 22x 1 3 1 y5
C = 2.00 + 0.32M
22x 1 3 3
Cost in $
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So y is a function of x. The domain and range are all real numbers. f(x) 5 22x31 3
(50, 18)
(20, 8.4)
5. x 1 2y 5 3x 2 4 1 2y 5 2x 2 4 1 y 5 x 2 2
M 0
So y is a function of x. The domain and range are all real numbers. f(x) 5 x 2 2 6. y 5 !x So y is a function of x. The domain and range are all real numbers $ 0. f(x) 5 !x 6 7. 2xy 5 6 1 y 5 2x 5 3x . So y is a function of x. The domain is all real numbers except 0. The range is all real numbers except 0. f(x) 5 3x y
8. 6 A2x 1 3B 5 6(1) 1 3x 1 2y 5 6 1 2y 5 6 2 3x 1 y 5 6 22 3x So y is a function of x. The domain and range are all real numbers. f(x) 5 6 22 3x 9. f(x) is undefined at x 5 25; domain: (2`, 25) d (25, 1`); range: all real numbers except 0. g(x) is undefined at x 5 21; domain: (2`, 21) d (21, 1`); range: all real numbers except 0. h(x) is undefined at x , 10; domain: [10, 1`); range: [0, 1`).
Miles
7. a. ƒ(2) 5 4 b. ƒ(21) 5 4
100
c. ƒ(0) 5 2 d. ƒ(25) 5 32.
9. a. p(24) 5 0.063, p(5) 5 32 and p(1) 5 2. b. n 5 1 only 11. a. ƒ(22) 5 5, ƒ(21) 5 0, ƒ(0) 5 23, and ƒ(1) 5 24. b. ƒ(x) 5 23 if and only if x 5 0 or 2. c. The range of ƒ is from 24 to 1 ` since we may assume that its arms extend out indefinitely. 13. ƒ(0) 5 1, ƒ(1) 5 1, and ƒ(22) 5 25 15. a. b. c. d.
f(0) 5 1, g(0) 5 1 f(22) 5 2, g(23) 5 10 f(2) 5 0, f(1) 5 0.5 f(3) 5 –0.5, g(3) 5 10
17. a. x 5 22 and 2
b. x 5 2
c. x 5 0 and 2
Section 1.5 Exercises for Section 1.4 1. a. T(0) 5 2, T(21) 5 6, T(1) 5 0, T(25) 5 42 3. a. Tax 5 0.16 ? Income b. Income is the independent variable and Tax is the dependent variable. c. Yes, the formula represents a function: for each input there is only one output. d. As the Income gets closer and closer to $20,000, the Tax gets closer and closer to $3200. In fact the Tax, at some point will round off to $3200.00, even though the Income is not quite $20,000. Its domain is [0, 20000] and its range is [0, 3200]. 5. The equation is C 5 2.00 1 0.32M. It represents a function. The independent variable is M measured in miles. The dependent variable is C measured in dollars. Here is a table of values.
Algebra Aerobics 1.5 1. Possible titles: a. From 2004 to 2006 Prices Decreased 50% for LCD TVs b. iTunes Sales Increase Tenfold From December 2003 to January 2005 2. a. maximum value: approx. $46,000 in 1999; (1999, $46000) b. minimum value: approx. $40,000 in 1993; (1993, $40000) c. Median household income decreased from 1990 to 1993, reaching a low point of about $40,000 in 1993. From 1993 to 1999 it increased steadily, reaching a high point of about $46,000 in 1999, which was followed by a decrease until 2004. 3. Graph B is the best match for the situation. It is the only graph that represents the child stopping at the top of the slide with a speed of zero for a few minutes.
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AIDS cases
Number of U.S. AIDS Cases
e. Concave down approximately over interval (0, 5); concave up approximately over interval (5, 8). f. ƒ(x) 5 4 when x 5 1 and 3. g. ƒ(28) is not defined. 11. a. Graph B seems best. It indicates several stops, rises and falls in speed and, most importantly, it is the only one that ends with a stop. b. Graph E is suitable. The horizontal parts on the graph indicate a time at which the bus stopped. Graphs D and F seem out of place. They would indicate that the bus went backwards.
Time
5. a. Graph that is concave up with minimum at (22, 1) y Decreasing
589
3
13. a. Graph of h(x) 5 x4 y
Increasing 270
–3
3
x
h(x) = x4
b. Graph that is concave down with maximum at (3, 22) y 2
–4
x 4
–5
x
0
b. Graph of k(x) 5 x4 2 24x 1 50 Decreasing
Increasing
5
y
–4
200
Exercises for Section 1.5 1. a. Graph B b. Graph A
3. The price of Rockwell Collins hovered under $30 at the beginning of the week and then rose to $32 on Wednesday and then slowly climbed for the rest of the week. The price of Transkaryotic, by contrast, rose from $16.50 on Friday to $17 on Monday and then dropped to $13 on Tuesday and hovered near $13 for the rest of the week. 5. a. b. c. d. e. f.
Positive over (25, 0) and (5, 1 ` ) Negative over (0, 5) and (2 ` , 25) Decreasing over (23, 3) Increasing over (2 ` , 23) and (3, 1 ` ) There is no minimum. There is no maximum.
7. a. A, domain 5 (2 ` , 1 ` ) and range 5 (2 ` , 2) B, domain 5 [0, 1 ` ) and range 5 [0, 1 ` ) C, domain 5 (2 ` , 1 ` ) and range 5 (0, 1 ` ) D, domain 5 (2 ` , 1 ` ) and range 5 [2, 1 ` ) b. A, (25, 21), B, (0, 1 ` ), C, (2 ` , 1 ` ), D, (2 ` , 1 ` ) B, nowhere c. A, (2 ` , 25) and (21, 1` ) C, nowhere D, nowhere 9. a. b. c. d.
k(x) = x 4 – 24x + 50
c. Graph A d. Graph B
[26, 23) and (5, 11) (23, 5) and (11, 12] (26, 2) and (8, 12) (2, 8)
x –5
0
5
The graphs of h(x) and k(x) are everywhere concave up. 15. a. b. c. d.
1996 to 1997, 1998 to 1999, 2000 to 2001. 1995 to 1996, 1997 to 1998, 1999 to 2000, 2001 to 2003. 17 billion dollars in 1999. 10 billion dollars in 2003.
17. a. Johnsonville’s population goes from 2.4 ? 100,000 5 240,000 to 5.8 ? 100,000 5 580,000. Palm City’s population ranges from 1.8 ? 100,000 5 180,000 to a high of 3.8 ? 100,000 5 380,000. (This notation adheres to what is found in the graph.) b. The population of Palm City increased from 1900 to 1930. c. The population of Palm City decreased from 1930 to 1990. d. The two populations were equal sometime around 1940. 19. a. Yes, P is a function of Y since the inputs are all distinct. b. The domain is the set of years from 1990 to 1995 inclusive; the range is the set of corresponding values, namely {20.5, 0, 1.2, 1.4, 2.3}. c. The maximum P value is 2.3; it occurs when Y 5 1991. d. P is increasing from 1990 to 1991 and from 1993 to 1994. It is decreasing from 1991 to 1993 and 1994 to 1995. e. Y is not a function of P, since the two inputs of 1.4 have different outputs.
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CH. 1 Check Your Understanding
21. a. From 1750 to 2100. b. Yes, from 2100 to 2200. c. Student answers will vary; it is hoped that something is said about the dramatic increase and the slowing down noted in parts (a) and (b). 23. a. Male and female enlisted reserve personnel reached their respective maxima in 1992. b. The maximum for males was approximately 1350 (thousand) and for women it was approximately 215 (thousand). c. For men, there was a steady rise from 1990 to 1992 and then a gradual decline, reaching a low of 800,000 in 2004. For women there was a sharp rise from 1990 to 1992; then a steady decline to 175,000 until 1998; and then a slight rise from 1998 to 2001, when it climbed to 175,000 and a gradual decline since then to 2004, when it reached 170,000. d. Answers will vary.
a. As x S 1 ` y1 approaches 1` , y3 approaches 0; b. As x S 2 ` y1 approaches 2` , y3 approaches 0
y2 approaches 1 ` , y4 approaches 2 y2 approaches 1 ` , y4 approaches 2.
29. Clearly one cannot simulate the session here. But one can give the graphs of ƒ and g (see the accompanying diagram) and note that the graph of g is the mirror image of the graph of ƒ with respect to the x-axis. y 40 y = f(x)
x –5
(–2, 0)
(3, 0)
5
25. y = g(x) Levels of mobility
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Ch. 1: Check Your Understanding 0 2
4
6 12 Weeks gone by
1. True
8. False
15. True
22. True
2. False
9. True
16. False
23. True
3. True
10. False
17. True
24. False
4. True
11. True
18. False
25. True
5. True
12. False
19. False
26. True
6. True
13. False
20. True
27. True
7. False
14. True
21. True
16
Here is one graph. Phrases like “after a while” and “for a while” were loosely interpreted. As for “levels of mobility,” the idea was to give a good picture by means of a graph without being held to numerical values. Time intervals, when explicitly given, were respected. There are discontinuities in the going-to-crutches phases. 27. Here are the graphs asked for: y
28. Possible answer: z4 5 2w 1 3
y
29. Possible answer: y1 = x3
z w
y2 = x2
1 10
2 23
3 9
4 23
x
30. Possible answer:
31. Possible answer:
y
y
x 0.00
y
x
y 7.00
32. Possible answer:
1
y4 = x + 2 x = –3
y3 = 1 x+3
–10.00
–7.00
33. Possible answer: y
y
y=2 x
x 7.00
–7.00
(–2,0) –7.00
x
7.00
–7.00
x
x
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CH. 1 Review: Putting It All Together
34. Possible answer: The number of wolf pups born and the number that have survived in Yellowstone National Park have both increased significantly since 1995. 35. False
37. False
39. False
41. True
36. False
38. True
40. False
42. True
Ch. 1 Review: Putting It All Together 1. a. 25% b. 20% c. 40 lb is a smaller percentage of 200 lb (his weight at age 60) than it is of 160 lb (his weight at age 20). 3. This could be explained by changes in the distribution of incomes. For example, some incomes (originally below the median) could have increased to more than the previous median (to raise the median), while other incomes decreased on either side of the median (to lower the mean). To simultaneously decrease the mean while raising the median, the decrease in incomes would need to be greater than the increase in incomes. For example, for a small set of numbers such as {5, 15, 20, 50, 110}, the median is 20 and a mean is 40. If we change the set such that 5 increases to 25 and 110 decreases to 80, then the new set of {15, 20, 25, 50, 80} would have a median of 25 and a mean of 38.
591
iii. Function iv. Not a function b, c. Answers will vary. 13. a. Approx. 51%. b. In November 2004. c. There was a major increase between September and November of 2004; other lesser increases occurred between June and August of 2004 and between February 2005 and April 2005. d. In June 2005 there was a difference of about 30% (approx. 63% minus 33%). 15. a.
d y
1 8
2 16
3 24
4 32
10 80
20 160
100 800
80 10
100 12.5
b. y 5 8d (d independent, y dependent) c.
y d
8 1
12 1.5
16 2
20 2.5
50 6.25
d. d 5 (1/8)y (y independent, d dependant) 17.
100
f(t)
5. a. Heart disease b.
The Four Leading Causes of Death Worldwide
50
14
Percentage of all deaths
12
t
10
0
5
10
8 6 4 2 0 Heart disease
Strokes
Lower HIV/AIDS respiratory infections
c. Sixty-six percent of the deaths were not accounted for. For example, tuberculosis, malaria, lung cancer, war, and car accidents are not included. No other cause can account for more than 5% of deaths, or else it would supersede HIV/AIDS and be one of the four leading causes. 7. E 5 0.025P 9. Answers will vary. Possible title and summary: “The Popularity of the Name Emma over Time.” Emma was a very popular name in 1880, when it was given to about 9,000 out of every million babies. Its use fell rapidly until it bottomed out in the 1970s, when it was given to about 200 out of every million babies. It has been rising in popularity since the 1980s, although it was only about half as popular in 2004 as it had been in 1880. 11. a. i. Function ii. Not a function
19. a. Yes, it passes the vertical line test. Domain: [1970, 2010]; approx. range: [41, 63]. b. Approx. 1990; approx. 63 years of age. c. Approx. 2003 (projected); approx. 41 years of age. AIDS is a likely candidate. d. Increasing from approx. 1970 to 1990 and projected to increase from 2003 to 2010; decreasing from 1990 to 2003. e. Concave down over interval (1970, 1997); concave up over interval (1997, 2010). f. In 1970 the life expectancy in Botswana was about 52 years, as opposed to about 57 years worldwide. Over the next 20 years this gap gradually closed. In 1990 the life expectancy in Botswana was about 63, and 64 was the worldwide average. However, after 1990 the life expectancy in Botswana began to rapidly decline; this decline was expected to continue until about 2003, at which time it was projected to begin to increase slowly. 21. a. It means that 3% of the worldwide out-of-school population is in Latin America or the Caribbean. b. It does not. It tells us only what percentage of the worldwide out-of-school population is in sub-Saharan Africa.
Ch. 1
CH. 1 Review: Putting It All Together
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CH. 2 Algebra Aerobics Solutions for Section 2.1
Ch. 2
c. This graph does not show the number of children out of school worldwide or within each region. Nor does it show the percentage of children within each region who are out of school. The number of children out of school in a particular region may be relatively small compared to the total worldwide out-of-school population, but quite large compared to the total population of the region. This could be confusing. d. Sub-Saharan Africa has the largest percentage of the world’s out-of-school population. Seventy-three percent of the world’s out-of-school population reside in subSaharan Africa or South Asia. Industrialized nations represent the smallest percentage. e. Answers will vary. One idea is that UNICEF should consider focusing on sub-Saharan Africa and South Asia. This is where almost three-quarters of the children who are out of school can be found. Evidence from the graph should be cited. 23. a. f(0) 5 212,f(21) 5 213,f(2) is undefined b. Domain: all real numbers, x 2 2 Range: all real numbers, f(x) 2 0. 25. a. February has the maximum cost and July the lowest. This makes sense, as heating costs would be less in the warm months and more in the cold ones. b. Rounded off: mean 5 $104.67 and median 5 $86. Since the mean is higher than the median, together the mean and median suggest that there are a few expensive months, which account for most of the cost of gas for the year. c. Number of months
4
$(1,273.2 2 618.4)billion (2006 2 1990)
billion 5 $654.8 < $40.93 billion/yr 16 yr
$(2453.0 2 (281.2) )billion (2006 2 1990)
billion 5 2$371.8 16 yr
< 2$23.24 billion/yr b. While both exports and imports were increasing, imports were increasing at a faster rate, and hence, the trade balance was decreasing. The trade deficit increased by over $23 billion/year from 1990 to 2006. 3. a. b.
(41.8 2 52.1) thousand deaths 20 yrs
deaths < 20.52 thousand , yr or a decrease of 520 deaths per year. (42.6 2 41.8) thousand deaths 4 yrs
deaths 5 0.2 thousand , yr
or an increase of 200 deaths per year. 4. 60 mph/5 sec 5 12 mph per second 5.
($867 2 $689) 6 yrs
< $29.67/yr
6. (978 2 1056)/4 5 278/4 5 219.5. On the average, his performance was declining by 19.5 yards per year. 7. (30,000 2 150,000)/10 5 212,000 elephants per year. On the average, the African elephant population is decreasing by 12,000 elephants per year.
Exercises for Section 2.1 1. a. inches/pound
b. minutes/inch
c. pounds/inch
3. 212 miles/10.8 gal. < 19.6 miles per gallon 7.5 2 9.4 21.9 5. 2000 2 1960 5 40 5 20.0475 pounds per year
13,600,000 2 630,000 2005 2 1985
5
12,970,000 20
5 648,500
computers
per year.
2 1 200– 250
150– 199
100– 149
50 –99
0 –49
Dollars
d. Answers will vary. Some patterns: For one-third of the year, the cost of gas is between $150 and $250. Gas costs peak in February and then start to decrease throughout the spring and summer. In August they begin to increase, continuing throughout the fall and early winter.
CHAPTER 2 Section 2.1 lb 5 1.6 yr
$(820.2 2 537.2)billion (2006 2 1990)
4.0 2 84.1 280.1 b. 2005 students per 2 1985 5 20 5 24.005 < 24 computer per year, or a decrease of 4 students per computer each year.
11. a.
500,000 2 4,000,000 2000 2 1930
5
23,500,000 70
5 250,000;
i.e., on average there were 50,000 fewer elephants per year. b. In 1980 the rate was twice as large as the rate computed in part (a) for 1930 to 2000. That means that either before or after the 1980s, the average rate of decline must have been much smaller. 59.7 85.8 2 26.1 13. a. For whites: 2004 2 1940 5 64 < 0.9 percentage points per year. 80.6 2 7.3 73.3 For blacks: 2004 2 1940 5 64 < 1.1 percentage points per year. 85.0 2 22.6 62.4 For Asian/Pacific Islanders: 2004 2 1940 5 64 < 1.0 percentage point per year. 85.2 2 24.5 60.7 For all: 2004 2 1940 5 64 < 0.9 percentage point per year.
Algebra Aerobics 2.1
2. a. i.
iii.
9. a.
3
0
(143 2 135) lb 5 yr
ii.
504 2 498 6 7. a. Math: 2005 2 2000 5 5 5 1.2 points per year. 505 2 504 1 b. Verbal: 2005 2 2000 5 5 5 0.2 point per year.
Monthly Cost of Gas 5
1.
CH. 2 Exercises Solutions for Section 2.1
5 $28316billion < $17.69 billion/yr yr
b. In 2007: For whites: 85.8 1 3 ? 0.9 5 88.5% will have completed 4 years of high school. For blacks: 80.6 1 3 ? 1.1 5 83.9%.
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15. a. White females had the highest life expectancy in 1900 and in 2005. Black males had the lowest life expectancy in 1900 and in 2005. 75.4 2 46.6 28.8 b. For white males: 2005 2 1900 5 105 < 0.27 years per year. 81.1 2 48.7 32.4 For white females: 2005 2 1900 5 105 < 0.31 years per year. 69.9 2 32.5 37.4 For black males: 2005 2 1900 5 105 < 0.36 years per year. 76.8 2 33.5 43.3 For black females: 2005 2 1900 5 105 < 0.41 years per year. Thus black females had the largest average rate of change in life expectancy between 1900 and 2005. c. In all four groups the average rate of change of life expectancy rose over the 105 years. White males had the smallest increase, 0.27 years per year, and black females had the largest increase, 0.41 years per year.
Section 2.2 Algebra Aerobics 2.2
593
b. Graph of average rate of change over prior 50 yrs. 80 70 60 50 40 30 20 10 0 1850 1900 1950 2000 2050 Year
c. During 1950–2000, average annual rate of change 5 71.4 million/yr was the greatest. d. The average rate of change increased in the time interval 1800 to 2000, yet despite a projected increase in world population from 2000 to 2050, the rate of change is projected to decrease to 59.72 million/yr. 2. From 2003 to 2005 the graph would rise, indicating that profits increased. From 2005 to 2006, the graph would be flat, so there would be no change in profits. From 2006 to 2007, the graph would fall, indicating that profits decreased. (Note, however, that a lower profit does not imply a loss.) 3. a. High School Completers
Year
Number (thousands)
1960 1970 1980 1990 2000 2004
1679 2757 3089 2355 2756 2752
Average Rate of Change (thousands/yr) n.a. 107.8 33.2 273.4 40.1 21.0
Table 2.7
1. a. World Population Year
Total Population (in millions)
1800
980
Average Rate of Change over Prior 50 yrs. n.a.
1850
1260
(1260 2 980) million (1850 2 1800)
1900
1650
(1650 2 1260) million (1990 2 1850)
5 39050million yrs 5 7.8 million/yr
1950
2520
(2520 2 1650) million (1950 2 1900)
5 87050million yrs 5 17.4 million/yr
2000
6090
(6090 2 2520) million (2000 2 1950)
million 5 3570 50 yrs 5 71.4 million/yr
2050
9076
(9076 2 6090) million (2050 2 2000)
million 5 2986 50 yrs 5 59.72 million/yr
5 28050million yrs 5 5.6 million/yr
b. The number of individuals completing high school each year rose between 1960 and 1980, when it peaked at 3,089,000. The 1960s had the greatest rate of change, with an annual increase of 107.8 (thousand) per year. The rate slowed during the 1970s. In the decade from 1980 to 1990, the number of high school completers each year showed a drastic decline on average of 73,400 per year. The trend reversed in the next decade (1990 to 2000), increasing on average by 40,100 per year. Between 2000 and 2004 the numbers remained almost the same, with only a slight decrease of 4000 over the 4 years. c. If the average rate of change is positive, there is an increase in the number of high school completers. An example would be from 1970 to 1980, where the number of high school completers increased from 2,757,000 to 3,089,000. d. If the average rate of change is negative, there is a decrease in the number of high school completers. An example would be from 1980 to 1990, where the number
Ch. 2
For Asian/Pacific Islanders: 85.0 1 3 ? 1.0 5 88.0%. For all: 85.2 1 3 ? 0.9 5 87.9%. c. There was a major increase in the percentage of those who completed 4 years of high school or more in all four categories. Over the 64-year period the increases ranged from 59.7 percentage points for whites to 73.3 percentage points for blacks. Blacks showed the highest average gain: 1.14 percentage points per year. Other comments could be made. d. For x years since 1940: Whites: 100 5 26.1 1 0.9x 1 x < 82.1 years, or in early 2022. Blacks: 100 5 7.3 1 1.1x 1 x < 84.3 years, or in early 2024. Asian Pacific Islanders: 100 5 22.6 1 1.0x 1 x 5 77.4 years, or in mid-2017. All: 100 5 24.5 1 0.9x 1 x 5 83.9, or in late 2023. These 100% predictions may not make much sense since the time is too far into the future to allow a reliable prediction.
CH. 2 Algebra Aerobics Solutions for Section 2.2
World population (millions/yr)
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CH. 2 Exercises Solutions for Section 2.2
CH. 2 Exercises Solutions for Section 2.2
Ch. 2
of high school completers decreased from 3,089,000 to 2,355,000. e. The growth is slowing down.
9. a. Table A
Exercises for Section 2.2 1.
x
ƒ(x)
0 1 2 3 4 5
0 1 8 27 64 125
Avg. Rate of Change n.a. 1 7 19 37 61
Table B
x
y
Average Rate of Change
0 1 2 3 4 5 6
2 5 8 11 14 17 20
n.a. 3 3 3 3 3 3
y
0 1 2 3 4 5 6
0 0.5 2 4.5 8 12.5 18
n.a. 0.5 1.5 2.5 3.5 4.5 5.5
b. The average rate of change is increasing. c. Its graph is concave up.
b. The average rate of change is constant. c. Its graph is a straight line.
a. The function is increasing. b. The average rate of change is also increasing.
x
Average Rate of Change
27,791,000
3. a. Year
Registered Motor Vehicles (millions)
Average Annual Rate of Change (over prior decade)
1960 1970 1980 1990 2000
74 108 156 189 218
n.a. 3.4 4.8 3.3 2.9
11. a. In 1920 there were 106,000,000 5 0.3 copy of newspapers printed per person (about one-third of a copy per person, or equivalently, roughly one copy for every three people). 55,800,000
In 2000 there were 281,400,000 5 0.2 copy of newspapers printed per person (about one-fifth of a copy per person, or equivalently, about one copy for every five people). b.
b. From 1990 to 2000. c. From 1970 to 1980. d. Student paragraphs might include the observation that from 1960 to 2000 there was a continuous rise in the number of registered motor vehicles in the United States (from 74 million to 218 million). The largest average rate of change (4.8 million per year) was in the decade from 1970 to 1980 and the lowest average rate of change (2.9 million per year) was from 1990 to 2000.
Avg. Annual Avg. Annual Rate of Rate of No. of Change No. of Change TV (over prior Newspapers (over prior Year Stations decade) Published decade) 1950 1960 1970 1980 1990 2000
98 515 677 734 1092 1248
n.a. 41.7 16.2 5.7 35.8 15.6
1772 1763 1748 1745 1611 1480
n.a. 20.9 21.5 20.3 213.4 213.1
Average Rates of Change for Number of TV Stations and Newspapers Published
I. x 0 10 20 30 40 50
ƒ(x) 5 25 45 65 85 105
Average Rate of Change
II.
n.a. 2 2 2 2 2
a. ƒ(x) is increasing. b. The average rate of change is constant.
7. a. Graph B
b. Graph A
x
g(x)
Average Rate of Change
0 10 20 30 40 50
270 240 210 180 150 120
n.a. 23 23 23 23 23
a. g(x) is decreasing. b. The average rate of change is constant.
c. Graph C
Average annual rate of change (in percentage points) over prior decade
5. 50 40
TV Newspapers
30 20 10 0 1960 –10
1970
1980
1990
2000
–20 Year
c. From the table in part (b), the average rate of change from 1990 to 2000 was 15.6 stations per year. At this rate, in 2010 there will be 1248 1 15.6 ? 10 5 1404 stations. This could be off, given that the avg. rate per year for each decade vacillates a fair amount.
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CH. 2 Algebra Aerobics Solutions for Section 2.3
CH. 2 Algebra Aerobics Solutions for Section 2.4
Algebra Aerobics 2.3 i. ii. iii. b. i. ii. iii.
7. a. b. c. d.
m 5 75/10 5 7.5 70 2 y 5 24 1 y 5 74 32 5 4(28 2 x) 1 32 5 112 2 4x 1 x 5 20 6 5 0.6(x 2 10) 1 12 5 0.6x 1 x 5 20
9. a. 24 5 (1 2 t)/(22 2 3) 1 20 5 1 2 t 1 t 5 219 b. 2/3 5 (9 2 6)/(t 2 5) 1 2(t 2 5) 5 9 1 t 2 5 5 4.5 1 t 5 9.5
Section 2.3 1. a.
b. The slope of segment F is steepest. c. Segment C has a slope equal to 0.
Ch. 2
d. It seems that news is being more disseminated through TV stations than through newspapers. But there is not enough information to really give an answer here. A great increase in TVs does not necessarily indicate a great increase in getting the news via TV. Many other programs besides the news are watched on TV.
595
m 5 (11 2 1)/(8 2 4) 5 10/4 5 212 m 5 (6 2 6)/[2 2 (23)] 5 0 m 5 [23 2 (21)]/[0 2 (25)] 5225 m 5 (1 2 11)/(4 2 8) 5 (–10)/(–4) 5 212 m 5 (6 2 6)/(23 22) 5 0 m 5 [21 2 (23)]/(25 2 0) 5 2/(25) 5225
11. a. m1 5 (7 2 3)/(4 2 2) 5 4/2 5 2; m2 5 (15 2 7)/(8 2 4) 5 2; collinear b. m1 5 (4 2 1)/(2 1 3) 5 3/5; m2 5 (8 2 1)/(7 1 3) 5 0.7; not collinear 2 0 1 13. a. m 5 22 6 2 0 5 23
7 2 0 7 b. m 5 24 2 0 5 24 y 8
2. Between 1999 and 2002 the slope is approximately zero. 3. positive: 2001–2002, 2004–2005 negative: 1998–2000, 2003–2004 zero: 2000–2001, 2002–2003 y
y 2 (22)
4. x 5 4 1 5 2 3 1 y 5 6. y
54 1
(–4, 7) y = (– 1 )x y 3 2 (0,0) x –6 –2
y 1 2 2
(0,0)
6
x (6, –2)
–6
6
54 1 y1258
y = (– 7 )x 4
(9 1 2h) 2 9
5. x 5 (2 1 h) 2 2 5 2h h 5 2
–8
6. a. Points lie on the graph 1 the coordinates (x, y) satisfy the equation y 5 x 2 since 151 ,
050 , 2
2
2 2 2 2
452 , 2
953 . 2
1 b. P1 and P2: m 5 5 1; P2 and P3: m 5 42 2 2 1 5 3; P3 and P4: m 5 5 5. c. The positive slopes suggest that the function increases between x 5 0 and x 5 3; because the slopes increase in size as we move further to the right, the graph rises at an increasing rate. 1 1 9 3
0 0 4 2
Exercises for Section 2.3 3 2 (26)
1. a. m 5 2 2 (25) 5 97
29 2 6 b. m 5 223 2 (25) 5 7
3. a. Graph A crosses the y-axis at (0, 24) and goes through (24) 2 (24) (1, 24). So the slope is 5 0. 1 2 0 b. Graph B crosses the x-axis at (2, 0) and the y-axis at 2 0 (0, 28). So the slope is 28 0 2 2 5 4. 5. a. slope of segment A 5 [2 2 (26)]/[24 2 (28)] 5 8/4 5 2 slope of segment B 5 [28 2 2]/[0 2 (24)] 5 210/4 5 22.5 slope of segment C 5 [28 2 (28)]/[2 2 0] 5 0/2 5 0 slope of segment D 5 [26 2 (28)]/[6 2 2] 5 2/4 5 0.5 slope of segment E 5 [6 2 (26)]/[10 2 6] 5 12/4 5 3 slope of segment F 5 [216 2 6]/[12 2 10] 5 222/2 5 211
15. a. 0 2 "2 5 21 "2 2 0
0 1 3/2 b. 23/2 2 0 5 21 2 b b c. b0 2 0 5 2b 5 21
d. All slopes are 21, since each pair of points is of the form (0, a) and (a, 0). 17. The points in parts (a) and (c). 19. a. 1/10 b. 1/12 3 3 1 1 c. old: run 5 10 1 run 5 30 ft.; new: run 5 12 1 run 5 36 ft. 21. Student answers will vary.
Section 2.4 Algebra Aerobics 2.4 1. a. (1960, 22.2) and (2000, 11.3) b. (1970, 12.6) and (2005, 12.6) c. (2000, 11.3) and (2005, 12.6) 2. a. Between 2005 and 2006 the stock price surged from $1.02 to $1.12 per share, a 9.8% increase or, equivalently, an increase of $0.10 per share. See Graph A.
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CH. 2 Algebra Aerobics Solutions for Section 2.5
Ch. 2
b. Between 2006 and 2007 the stock price dropped drastically from $1.12 to $1.08 per share, about a 3.6% decrease or, equivalently, a decrease of $0.04 per share. See Graph B.
7. a.
y 5
x
Stock Price
–5
Stock Price
1.12
5 (0, –2)
1.12
Price per share
Price per share
1.10 1.08 1.06 1.04
b, c. The task is impossible. The slope of the line is 0, so it is not possible to make a graph through the given points appear to have a large positive slope.
1.11
1.10
9. “Precipitous,” “dire,” and “catastrophic” come to mind. Students will probably think of others.
1.09
1.02 1.00 2005
1.08 2006 2007 Year
2006 Year
Graph B
Graph A
3. a. You could draw a graph that is cropped and stretched vertically, as shown in Graph A. b. You could show a graph that is not too steep (stretched horizontally or compressed vertically) and emphasize the decline in the number of casualties in week three, as in Graph B.
25 20
13. Graph A gives the appearance that the percentage has declined quite a bit, while Graph B gives the impression that the percentage has not declined much at all. The difference is in the vertical scale: it is greatly magnified in Graph A.
100 75 50 25 0
1
2 Week
11. a. The number of persons with AIDS who are still alive has been steadily growing since 1995 by approximately 425,000 2 200,000 5 25,000 people per year. Also, deaths 2004 2 1995 from AIDS have declined since 1993 from 55,000 to less than 20,000 (using left scale). b. Despite the progress in treating AIDS and education about it, there has been an increase in cases diagnosed since 2000 of about 1000 per year. c. AIDS cases increased until 1993, when new treatments, public awareness, education, and testing for the disease may have caused a decrease in the number of new cases.
American Casualties Number of casualties
Number of casualties
American Casualties 30
15
(5, –2)
–5
3
Graph A
1
2 Week
3
Graph B
4. Some of the strategies are: use of dramatic language (the title “Gold Explodes” in big bold type, followed by “Experts Predict $1,500.00 an Ounce”; cropped vertical axis on graph (it starts at $300, not $0, making the graph look steeper); use of powerful graphics (the arrow that increases in size), suggesting dynamic growth.
Exercises for Section 2.4 1. Graph A: “30-Year Mortgage Rates Steadily Climb” Graph B: “30-Year Mortgage Rates Rise Sharply” Graph C: “30-Year Mortgage Rates Show Little Gain” 3. a. Graph B appears to have the steeper slope. b. The slope of Graph A is 26 and the slope of Graph B is 24. So Graph A actually has the steeper slope. 5. a. From 1984 to 2005 the federal appropriations for the most part rose, but the reading scores stayed the same. b. There is no relationship between the scale on the left (for federal funding) and the scale on the right (for NAEP scores). The NAEP scale was arbitrarily started at 185. Had it started at 0, the NAEP scores would still remain constant, but would lie much higher on the graph.
Section 2.5 Algebra Aerobics 2.5 1. The weight of a 4.5-month-old baby girl appears to be <14 lb. From the equation, the exact weight is W 5 7.0 1 1.5(4.5) 5 13.75 lb. Our estimate is within 0.25 lb of the exact weight. 2. The age appears to be < 2.5 months. Solve 11 5 7 1 1.5A 1 11 2 7 5 1.5A 1 4 5 1.5A 1 A 5 2.67 months. 3. a. The units for 15 are dollars/person. The units for 10 are dollars. b. dollars 5 (dollars/person)(persons) 1 dollars 4. a. The units for 1200 are dollars and for 50 are dollars/month. b. dollars 5 dollars 1 Adollars month B months 5. a. 0.8 million dollars per year is the slope or average rate of change in the sales per year. b. 19 million dollars represents the sales this year. c. In 3 years S 5 0.8(3) 1 19 5 21.4 million dollars d. dollars 5 Adollars year B years 1 dollars, where all values for dollars are in millions. 6. a. The average cost to operate a car is $0.45 per mile, and the units are dollars per mile. b. 0.45(25,000) 5 $11,250 c. dollars 5 Adollars mile B miles
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8. a. b. c. d. e. f. g. h.
m 5 5, b 5 3 m 5 3, b 5 5 m 5 5, b 5 0 m 5 0, b 5 3 m 5 21, b 5 7.0 m 5 211, b 5 10 m 5 22/3, b 5 1 m 5 5, b 5 23 since 2y 5 10x 2 6 1 y 5 5x 2 3
9. a. ƒ(x) is a linear function because it is represented by an equation of the form y 5 mx 1 b, where here b 5 50 and m 5 225. b. ƒ(0) 5 50 2 25(0) 5 50 2 0 5 50 ƒ(2) 5 50 2 25(2) 5 50 2 50 5 0 c. Since the line passes through (0, 50) and (2, 0), the slope m 2 0 is: m 5 50 0 2 2 5 225. 10. a. b. c. d.
linear: m 5 3, b 5 5 linear: m 5 1, b 5 0 not linear linear: m 5 22/3, b 5 4
11. a. y 5 3x 1 4 b. y 5 2x c. y 5 23 12. Graph A: Graph B: Graph C: Graph D:
y 5 4x 1 3 y 5 22x 1 05 22x y 5 0x 1 3 5 3 y5x105x
9. a. C(0) 5 $11.00; C(5) 5 11 1 10.50 ? 5 5 $63.50; C(10) 5 11 1 10.50 ? 10 5 $116.00 b. $11.00 c. $10.50 for every thousand cubic feet. d. C(96) 5 $1019 b. 0; p
11. a. 0; 4
d. 217.78; 5/9
c. 0; 2p
13. a. matches f, since $42.50 seems the most likely cost for producing one text. b. matches g, since $0.30 seems the most likely cost to produce one CD. c. matches e, since $800 seems the most likely cost of producing one computer. 15. a. C(p) 5 1.06 ? p b. C(9.50) 5 1.06 ? 9.50 5 10.07; C(115.25) 5 1.06 ? 115.25 5 122.17 (rounded up); C(1899) 5 1.06 ? 1899 5 2012.94. All function inputs and outputs are measured in dollars. 17. a. hours
b. miles/gallon
c. calories/gram of fat
19. a. Slope 5 0.4, vertical intercept 5 220 b. Slope 5 2200, vertical intercept 5 4000 a.
b.
y
P
20
10,000 (50, 0) P = –200C + 4000
x 100
–100
(0, 4000)
(0, 20)
y = 0.4x – 10
(20, 0)
C 40
–40 –2000
–60
21. The equation is: y 5 3x 2 2 5
Exercises for Section 2.5 1. a. E 5 5000 when n 5 0, E 5 5100 when n 5 1, and E 5 7000 when n 5 20 b. (0, 5000), (1, 5100) and (20, 7000) 3. a. (5000, 0) is not a solution to either equation. b. (15, 24000) is a solution to the second equation but not the first. c. (35, 40000) is a solution to the second equation but not the first.
597
Ch. 2
7. a. $840 b. The beginning mortgage is $302,400. c. After 10 years, B 5 $302,400 2 840(12)(10) 5 $201,600; After 20 years, B 5 $302,400 2 840(12)(20) 5 $100,800; After 30 years, B 5 $302,400 2 840(12)(30) 5 $0.
CH. 2 Exercises Solutions for Section 2.5
y (2, 4)
x
y
0 1 2
22 1 4
(1, 1) x –5
y = 3x – 2
5
(0, –2)
–5
23. The equation is y 5 0 ? x 1 1.5 5 1.5. y
5. a. D 5 3.40, 3.51, 3.62, 3.73, and 3.84, respectively b. 0.11 is the slope; it represents the average rate of change of the average consumer debt per year; it is measured in thousands of dollars per year. c. 3.40 represents the average consumer debt when n 5 0 years. It is measured in thousands of dollars. 7. dollars 5 dollars 1 dollars year ? years
5
x
y
25 0 5
1.5 1.5 1.5
y = 1.5 x –5
5
–5
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CH. 2 Algebra Aerobics Solutions for Section 2.6
CH. 2 Exercises Solutions for Section 2.6
6. Answers will vary. For example, ƒ(x) 5 4 2 2x g(x) 5 4 2 5x h(x) 5 4 2 7x h(x) has the steepest slope since k 27 k . k 25 k . k 22 k
Section 2.6 Algebra Aerobics 2.6 1. 0, k 21 k , k 23 k , 4, k 27 k , 9, k 212 k 2. a. The graph of y 5 6x goes through the origin and is two units above the graph of y 5 6x 2 2. b. The graph of y 5 2 1 6x intersects the y-axis at (0, 2). The lines are parallel, so y 5 2 1 6x is four units above the graph of y 5 6x 2 2. c. The graph of y 5 22 1 3x has the same y-intercept (0, 22) but is less steep than the line y 5 6x 2 2. d. The graph of y 5 22 22x has the same y-intercept (0, 22) but is falling left to right, whereas y 5 6x 2 2 is rising and has a steeper slope.
y 20 g(x) 10 f(x) x –4
0
–2
2
4
–10 h(x) –20
3. For (a), m 5 22, so For (b), m 5 21, so For (c), m 5 23, so For (d), m 5 25, so
kmk kmk kmk kmk
5 k 22 k 5 k 21 k 5 k 23 k 5 k 25 k
52 51 53 55
7. ƒ(x) matches Graph B, g(x) matches Graph D, h(x) matches Graph A, k(x) matches Graph C.
Exercises for Section 2.6
Line d is the steepest. In order from least steep to steepest, we have b, a, c, d. 4. Answers will vary depending on the original ƒ(x). For example, if ƒ(x) 5 2 1 2x, then we could have: a. g(x) 5 5 1 2x b. h(x) 5 22 1 2x c. k(x) 5 2 2 2x See graph below
1. a. Graph B b. Graphs A, B, and D
c. Graph C d. Graphs A and D
3. Answers may vary for (b) and (c). a. y 5 23 b. y 5 x 2 3 c. y 5 23x 1 1 y 8
y = –3x + 1 x y
–3
10
5
g(x) y=–3 f(x)
y=x–3
5
–8
h(x) x –4
–2
0
2
–5
4
5. a. matches Graph B b. matches Graph D
c. matches Graph C d. matches Graph A
k(x)
7. Graphs may vary. a. Same slope
–10
b. Same vertical intercept
y
5. For ƒ(x) 5 3x 2 5, m 5 3. For g(x) 5 7 2 8x, m 5 28. g(x) has a steeper slope than ƒ(x) since k 28 k . k 3 k
y
10
10 y = 3x + 1
y = 3x + 3 y=x+1 x –5
y g(x)
20
5
x –5
5
y = 3x – 3
y = 3x
y = –3x + 1
f(x) 10
–10
x –4
–2
0 –10
–20
2
4
9. a. b. c. d. e.
–10
R(t) 5 5 1 13 2 5t 5 18 2 5t S(t) 5 23 1 13 2 5t 5 10 2 5t T(t) 5 13 1 2t (slope may vary) U(t) 5 12 2 5t (vertical intercept may vary) V(t) 5 15 1 5t (vertical intercept may vary)
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CH. 2 Algebra Aerobics Solutions for Section 2.7
5. a.
y = (5/6)x + 2
y 8
599
16
Ch. 2
11.
CH. 2 Algebra Aerobics Solutions for Section 2.7
y = (1/2)x + 2
Hourly wage in dollars
14
(0, 2) x
–5
5 y = 2x + 2
12 10 8 6 4 2
–8 0
13. a. C b. B c. m3 , m2 , m1 (Note: m3 is negative, while m2 and m1 are both positive and m2 , m1.) d. The steepness is the absolute value of the slope. So although m3 is negative, its absolute value is positive. Compared to the absolute values of the other slopes, we have u m2 u , u m1 u , u m3 u.
Section 2.7 1. a. y 5 1.2x 2 4 b. x y
y 6
27.6 24.0 20.4
2 5.30 3.20 13.30 2 8.50 4.80 b. Yes. 8.50 10 2 8 5 2 5 1.6, 13 2 10 5 3 5 1.6, 13.30 2 5.30 8 5 5 5 1.6. Since the rate of change is 1.6, 13 2 8 the same over all intervals, the points lie on the same line with slope 1.6. c. If we choose the point (8, 5.30) 1 x 5 8, y 5 5.30. From part (b) we have m 5 1.6. Solve: 5.3 5 (1.6)8 1 b 1 5.3 5 12.8 1 b 1 27.5 5 b, so the equation is y 5 1.6x 2 7.5.
8. a. b. c. d.
y = 1.2x – 4
x 6
2. a. y 5 300 2 400x b. x y
B 5 $10,800 2 300P $300 per month B 5 $10,800 2 $300(24) 5 $3600 0 5 $10,800 2 $300P 1 $300P 5 $10,800 1 P 5 36 or 36 months.
9. a. R 5 7.5T b. $7.50 per ticket c. R 5 $7.50(120) 5 $900.00
–6
10. Graph A: y 5 x Graph B: Q 5 3t 1 2 Graph C: y 5 6 2 2x
y 1,500
1100 300 2500
11. a. R 5 15,000 1 12T b. R 5 15,000 1 12(40,000) 5 $495,000 x –2
2 –500
3. The vertical intercept is at (0, 1), so b 5 1. The line passes 1 2 (25) through (0, 1) and (3, 25), so m 5 0 2 3 5 22. The equation is: y 5 22x 1 1. 4. a. S 5 $12,000 1 $3000x Year
$ Salary
0 1 3 5
12,000 15,000 21,000 27,000
12. Any equation of the form y 5 6 1 mx with three different values for m; some examples are y 5 6 1 2x, y 5 6 2 5x, and y 5 6 1 11x. 13. Any equation of the form y 5 23x 1 b with three different values for b; some examples are y 5 23x 1 5, y 5 23x, y 5 23x 212. 14. a. y 5 4x 1 9 b. y 5 223x 1 7
30,000 24,000
Dollars
b.
15
7. a. 4; (1, 4) b. 4; (21, 12)
–6
22 0 2
6 9 12 Years of education
6. 3
Algebra Aerobics 2.7
23 0 3
3
c. y 5 210x 2 7 d. y 5 2x 1 2.1
15. a. m 5 62 5 3 1 y 5 3x 2 1 b. m 5 219 1 y 5 219x 1 53
18,000 12,000
c. m 5 26 26 5 1 1 y 5 x 2 5
S = 12,000 + 3000x
6,000
0
1 2 3 4 Years of employment
5
16. a. y 5 234x 2 3 b. y 5 7x 2 5 c. y 5 214x 1 18
d. y 5 12x e. y 5 3x 1 5 f. y 5 25x 1 6
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CH. 2 Exercises Solutions for Section 2.7
CH. 2 Exercises Solutions for Section 2.7
17. Note the difference in scales on all graphs. a. Graph of y 5 2x 1 5 c. Graph of y 5 2 A34B x 2 1
c. y 5 2x 1 2
d. y 5 2
y 4
7 10
30
y
y
y
5
(4, 2)
(–5, 2) (0, 2)
y = –x + 2
y=2
20 10
–10
–10
–5
–5
5
x 10
5
x
x 10
(2, 0)
–5
5
x
–3
–5
0
–5
5
–10
7. a. y 5 23x 2 2; the slope is 2/3 and the y-intercept is 22. b. y 5 232x 1 3; the slope is 23/2 and the y-intercept is 3.
–10
–20
b. Graph of y 5 2 A12B x 1 6
d. Graph of y 1 3x 1 9 40
20
c. y 5 223x 1 12; the slope is 22/3 and the y-intercept is 12. d. y 5 32 x; the slope is 3/2 and the y-intercept is 0.
y
e. y 5 23 x; the slope is 3/2 and the y-intercept is 0. So the equation is equivalent to the one in part (d). f. y 5 34x 1 14; the slope is 3/4 and the y-intercept is 1/4.
30
y
20 10 10 –10
–5
5
x 10
9. a.
–20 –10
–5
x 10
5
x
ƒ(x) 5 0.10x 1 10
2100 0 100
–10
0 10 20
–30
Graph of f(x):
Exercises for Section 2.7
y
1. a. Slope 5 5, so equation is of the form y 5 5x 1b. The line passes through (22, 3), 1 3 5 5(22) 1 b 1 b 5 13. So y 5 5x 1 13 b. Slope 5 234 , so equation is of the form y 5 234 x 1 b. The line passes through (22, 3), 1 3 5 234 (22) 1 b 1 b 5 1.5. So y 5 20.75x 11.5 c. Slope 5 0, so equation is of the form y 5 b. The line passes through (22, 3), 1 3 5 b. So the equation is the horizontal line y 5 3 3. a. m 5 (5.1 2 7.6)/(4 2 2) 5 22.5/2 5 21.25 and y 2 7.6 5 21.25(x 2 2) or y 5 21.25x 1 10.1 b. m 5 (16 2 12)/(7 2 5) 5 4/2 5 2 and W 2 12 5 2(A 2 5) or W 5 2A 1 2 5. a. y 5 21.5x 1 6
20
f(x) = 0.10x + 10 10
–100
b.
x
x 100
0
h(x) 5 50x 1 100
20.5 0 0.5
75 100 125
Graph of h(x):
b. y 5 20.5x 1 2
y 140
y
y 14
4 h(x) = 50x + 100 (–2, 3) y = –1.5x + 6
y = –0.5x + 2 (2, 1)
(2, 3)
(4, 0)
–5 –2
5
x
5
–5 –1
x –0.5
0
x 0.5
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CH. 2 Exercises Solutions for Section 2.7
CH. 2 Algebra Aerobics Solutions for Section 2.8
13. C(n) 5 2.50 1 0.10n, where C(n) is the cost of cashing n checks that month. 15. a. Annual increase 5 (32,000 2 26,000)/4 5 $1500 b. S(n) 5 26,000 1 1500n, where S(n) is measured in dollars and n in years from the start of employment. c. Here 0 # n # 20 since the contract is for 20 years. 17. a. P 5 285 2 15t b. It will take approximately 16.33 years to make the water safe for swimming since 40 5 285 2 15t implies that 15t 5 245 or t < 16.33 years. 19. a. V(t) < 70,000 1 26,000t, where t 5 years since 1977. b. V(33) 5 $928,000. 21. The equation of the line illustrated is y 5 (8/3)x 2 4. The graph in the text can be altered to appear much steeper if the graph is stretched vertically—as in the accompanying graph—where the tick marks on the x-axis are 10 units apart.
b. The relationship is linear. The slope between any two points is $3000 per year of education. c. The linear equation is: M 5 3000E 2 4810, where M 5 mean earning measured in dollars and E 5 years of education. 25. The entries in the table argue that the relationship is linear. The average rate of change in salinity per degree Celsius is a constant: 20.054. Since the freezing point, P, for 0 salinity is 0 degrees Celsius, we have that P 5 20.054S, where S is the salinity measured in ppt and P, the freezing point, is measured in degrees Celsius.
Section 2.8 Algebra Aerobics 2.8a 1. If (0, 0) is on each line, then b 5 0 in y 5 mx 1 b. a. y 5 2x. y y = –x
10
5 y 4 –10
–5
0
5
x 10
y = (8/3)x – 4 –5 x –30
30 –10
b. y 5 0.5x. y
–4 10
If, however, we reversed the situation and vertically compressed the graph (as in the one below, where the tick marks on the y-axis are now 10 units apart), the graph would seem less steep than the one in the text.
5
–10
–5
0
y = 0.5x
5
x 10
y –5
30
–10 x –4
2. a. The variables x and y are directly proportional; the equation is y 5 23x. b. The variables x and y are not directly proportional; the equation is y 5 3x 1 5.
4
y = (8/3)x – 4 –30
23. a. The three data points are plotted in the accompanying graph.
Mean earnings (dollars)
$50,000
(16, $43190)
$40,000
3. E 5 euros; D 5 U.S. dollars a. E 5 0.79D b. E 5 0.79D 1 2.50. c. Only (a) because it is of the form y 5 mx for some constant m. 4. d 5 60t. This represents direct proportionality. If the value of t doubles, the value for d also doubles. If the value for t triples, then so does the value for d.
$30,000 (12, $31190) $20,000 (8, $19190) $10,000 $0 0
5
10 Years of education
15
20
5. Since C and N are directly proportional, C 5 kN for some constant k. Since 50 5 k ? 2, then k 5 $25, the cost per ticket. So the cost for 10 tickets is C 5 25 ? 10 5 $250.
Ch. 2
11. C(n) 5 150 1 120n, where C(n) is the cost of n credits, and 150 is measured in dollars and 120 in dollars per credit.
601
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CH. 2 Algebra Aerobics Solutions for Section 2.8
6. a. 4 5 k(12) 1 k 5 13 1 y 5 13x b. 300 5 k(50) 1 k 5 6 1 d 5 6t 7. a. d 5 kC
b. T 5 kI
8. a. a 5 kb; 10 5 k(15)
1 perpendicular to the given line is m 5 223/4 5 43 . If the line passes through the point (0, 3) then x 5 0, y 5 3, the vertical intercept b 53. So the equation is y 5 43x 1 3.
c. t 5 kc 1 23
CH. 2 Algebra Aerobics Solutions for Section 2.8
5 k;a 5 23 (6) 5 4
b. a 5 kb; 4 5 23b 1 b 5 6
12. 4x 2 y 5 6 1 y 5 4x 2 6. The slope of the given line is 4; therefore the slope of any line perpendicular to it is 21/4. If the line passes through (2, 23) 1 x 5 2, y 5 23. So y 5 2 A14B x 1 b, and (23) 5 21 4 (2) 1 b 1 5 1 23 5 212 1 b 1 b 5 25 2 1 y 5 2 A4 B x 2 2
Algebra Aerobics 2.8b 1. a. y 5 25
b. y 5 23
c. y 5 5
2. a. x 5 3
b. x 5 5
c. x 5 23
3. a. y 5 27
b. x 5 24.3
4. Slope is 21, y-intercept is 0, so m 5 21, b 5 0 1 y 5 2x. 5. m 5 360, C 5 4, W 5 1000 in W 5 360C 1 b. To solve for b, use the point (4, 1000), so 1000 5 (360)(4) 1 b 1 1000 5 1440 1 b 1 b 5 2440. So, the equation is: W 5 360C 2 440. 6. Let m 5 slope of given line; M 5 slope of perpendicular line. So M 5 2m1 . 1 a. m 5 23 1 M 5 223 5 13 b. m 5 1 1 M 5 211 5 21 1 < 20.32 c. m 5 3.1 1 M 5 23.1
13. a. vertical line b. neither Am 5 223B c. horizontal line 14. Slope of given line 5 223, so slope of a perpendicular line 5 32. So: a. y 5 32 x 1 5 b. Subsituting (26, 1), we have 1 5 32 (26) 1 b 1 b 5 10 1 y 5 32 x 1 10 15. Slope 5 2 for the given line and for lines parallel to it. So: a. y 2x 9 b. Subsituting (4, 3), we have 3 5 2(4) 1 b 1 b 5 25 1 y 5 2x 2 5
Algebra Aerobics 2.8c
1 5 53 b. m 5 235 1 M 5 223/5
1. a. Graph of f(x)
7. a. Slope of y 5 2x 24 is 2, so line perpendicular to it has slope 21/2 or 20.5. Since it passes through (3, 25), x 5 3 when y 5 25. So, y 5 mx 1 b is: 25 5 (20.5)(3) 1 b. Solve it for b. 25 5 21.5 1 b 1 b 5 23.5. So, equation is: y 5 20.5x 2 3.5. b. Any line parallel to (but distinct from) y 5 20.5x 2 3.5 will be perpendicular to y 5 2x 2 4, but will not pass through (3, 25). They have same slope m, (20.5), but different values of b, in y 5 mx 1 b. Two examples are y 5 20.5x and y 5 20.5x 2 7.5. c. The three lines are parallel.
2
1
1
2
3
–1
–2
b. Graph of g(x)
8. Ax 1 By 5 C 1 By 5 C 2 Ax 1 y 5 (C 2 Ax)/B 1 A A y5C B 2 B x, so m 5 2B
4 3
223
2x 1 3y 5 5 1 A 5 2, B 5 3 1 m 5 3 3x 2 4y 5 12 1 A 5 3, B 5 24 1 m 5 224 5 34 2 2x 2 y 5 4 1 A 5 2, B 5 21 1 m 5 221 52 1 x 5 25 1 A 5 1, B 5 0 1 m 5 20, which is undefined; this line is vertical 1 5 13 e. x 2 3y 5 5 1 A 5 1, B 5 23 1 m 5 223 0 f. y 5 4 1 A 5 0, B 5 1 1 m 5 21 5 0 (this line is horizontal)
9. a. b. c. d.
x 4
0
2 1 –4 –3 –2 –1 –1
1
2
3
x 4
–2 –3 –4
10. 2x 1 3y 5 5 1 3y 5 22x 1 5 1 y 5 223x 1 53 1 m 5 223 . Parallel lines have slopes that are equal, so m 5 223. If the line passes through the point (0, 4) then x 5 0, y 5 4, the vertical intercept b 5 4 1 y 5 223x 1 4.
2. Q(t) 5 e
t 2 2for 25 , t # 0 2 2 tfor 0 , t # 5
C(r) 5 e
r 1 1for 23 , r # 1 5 for 1, r # 4
11. 3x 1 4y 5 27 1 4y 5 27 2 3x 1 y 5 274 2 A34B x 1 m 5 234 . Because perpendicular lines have slopes that are negative reciprocals of each other, the slope of any line
3. a. 2
b. 6
c. 2
d. 22
d. 215
4. a. g(23) 5 u 26 u 5 6; g(0) 5 u 23 u 5 3; g(3) 5 u 0 u 5 0; g(6) 5 u 3 u 5 3
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CH. 2 Exercises Solutions for Section 2.8
3. a. The slope m 5 56.92 2 42.69 5 14.23 and since y is directly proportional to x, then y 5 14.23x is the equation. Thus, if x 5 5, then 5 ? 14.23 5 71.15 is the missing value. b. The coefficient of x is the cost of a single CD.
9
6
3 x –6
–3
0
3
6
Graph of g(x) = |x –3| for –6 ≤ x ≤ 6
c. The graph of g is the graph of f shifted three units to the right. d. g(x) 5 e
x 2 3if x $ 3 3 2 xif x , 3
5. a. 3 # t # 7, which means that the values of t lie between (and include) 3 and 7. b. 69 , Q , 81, which means that the values of Q lie between (but exclude) 69 and 81.
Total distance
b.
Distance between you and shop
6. |T 2558| # 208, or equivalently 358 # T # 758. 7. a.
5. a. The independent variable is the price P; the dependent variable is the sales tax T. The equation is T 5 0.065P. b. Independent variable is amount of sunlight S received; dependent variable is the height of the tree H. The equation is H 5 kS, where k is a constant. c. Time t in years since 1985 is the independent variable, and salary S in dollars is the dependent variable. The equation is S 5 25,000 1 1300t. 7. d 5 5t; yes, d is directly proportional to t; it is more likely to be the person jogging since the rate is only 5 mph. 9. a. b. c. d.
m 5 0; y 5 3 m 5 0; y 5 27 slope is undefined; x 5 23 slope is undefined, x 5 2
11. The horizontal line N 5 2300, where N is measured in millions of books. (Student answers may vary.)
Time
Time
8. a. 4.25% in week 4; 5.25% in week 52; the longest period in which the rate remained the same was week 26 through week 52. b. Typical rate increase 5 0.25%. The increases are not at regular intervals. c. Curbing inflation.
13. a. horizontal: y 5 24; vertical: x 5 1; line with slope 2: 24 5 2 ? 1 1 b 1 b 5 26 1 y 5 2x 2 6. b. horizontal: y 5 0; vertical: x 5 2; line with slope 2: 0 5 2 ? 21 b 1 b 5 24 1 y 5 2x 2 4. 15. a. The average rate of change is 10 lb per month. b. w(t) 5 175, where t 5 number of months after end of spring training and 175 is measured in pounds. The graph of this function is a horizontal line. 200
d. If R(w) 5 the federal funds rate in week w of 2006, then we can write R(w) as:
w(t) = 175 Pounds
4.25%for 1 # w , 5 4.50%for 5 # w , 13 R(w) 5 e 4.75%for 13 # w , 19 5.00%for 19 # w , 26 5.25%for 26 # w # 52 e.
100
t
0 0
3 Months
Federal Funds Interest Rates for 2006
17. a. 7 5 b 2 3 1 b 5 10 and therefore y 5 10 2 x. b. 7 5 b 1 3 1 b 5 4 and therefore y 5 4 1 x.
6 Interest rate (%)
603
5.5 5
19. The lines described by: a. x 5 0
4.5
b. y 5 0 c. y 5 x.
4 3.5 3 0
10
20
30 40 Week
50
60
Exercises for Section 2.8 1. a. 2 5 m ? 10 means m 5 0.2 b. 0.1 5 m ? 0.2 means m 5 0.5 c. 1 5 m ? 1/4 means m 5 4
21. a. Intercepts of one line are (0, 4) and (1, 0), and the intercepts of the other line are (0, 4) and (21, 0). Thus the slope of the first line is 24 and the slope of the second line is 4, and thus the lines are not perpendicular. b. Intercepts of the first line are (0, 2) and (1, 0) and the intercepts of the second line are (0, 2) and (24, 0). So the slopes are now 22 and 1/2, and thus the lines are perpendicular.
Ch. 2
b.
CH. 2 Exercises Solutions for Section 2.8
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CH. 2 Exercises Solutions for Section 2.8
CH. 2 Algebra Aerobics Solutions for Section 2.9
23. For Graph A: both slopes are positive; same y-intercept. For Graph B: one slope is positive, one negative; same y-intercept. For Graph C: the lines are parallel; different y-intercepts. For Graph D: one slope is positive, one negative; different y-intercepts. y 5 (2A/B)x 1 (C/B), B 2 0 The slope is 2A/B, B 2 0 The slope is 2A/B, B 2 0 The slope is B/A, A 2 0
27. a. Graph of f(x)
1 t 5 4 hours. A will have traveled 4 hr ? 60 miles/hr 5 240 miles. B will have traveled 4 hr ? 40 miles/hr 5 160 miles. (Note: Together they will have traveled 240 1 160 5 400 miles.) d. One hour before they meet (3 hours into the trip), DAB (3) 5 400 2 100 ? 3 5 1100 miles, which means that they are 100 miles apart and traveling toward each other. One hour after they meet (at 5 hours), DAB (5) 5 400 2 100 ? 5 5 2100 miles, which means that they are 100 miles apart and traveling away from each other. 400 2 100tfor 0 # t # 4 e. D(t) 5 u 400 2 100 t u 5 e 100 t 2 400for t . 4 Absolute Value of Distance between A and B
4 3 Distance (miles)
400
1 x 1
2
3
4
–2
200 100 0
–3
0
–4
4 3 2 1
8
–4
22 2 xfor x , 22 2 1 xfor x $ 22
2008
–3
2007
4
2006
3
2005
2
Price (cents)
1
–2
44 42 40 38 36 34 32 30
2001
Price of U.S. First-Class Stamp x
Year
b. g(x) 5 u 2 1 x u c. The graph of g is the same as the graph of f shifted horizontally two units to the left.
c. The domain would be 2001 # x , 2008, and the top (dotted) line segment would be added to the graph.
Section 2.9
31. a. dA (t) 5 60 t,dB (t) 5 40 t b. DAB (t) 5 400 2 dA (t) 2 dB (t) 5 400 2 60 t 2 40 t 5 400 2 100t Distance between A and B
Algebra Aerobics 2.9 1. a. In 1960 there were about 7.5 million college graduates, in 2005 about 27.5 million. b.
400
Number of College Graduates 30
200 t
0 0
2
4
6
8
–200 –400 Time (hours)
c. They will meet when the distance between them DAB (t) 5 0 miles 1 400 2 100 t 5 0 1 400 5 100t
Number (millions)
29. a. g(x) 5 e
4 6 Time (hours)
34for 2001 # x , 2002 33. a. S(x) 5 • 37for 2002 # x , 2006 39for 2006 # x , 2007 b. Note that the top dotted line segment is from part (c).
b. Graph of g(x)
–4 –3 –2 –1 –1
2
2004
–4 –3 –2 –1 –1
300
2003
25. a. b. c. d.
Distance (miles)
Ch. 2
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2002
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25 20 15 10 5 0 1960 1970 1980 1990 2000 2010 Year
c. Two estimated points on the line are (1960, 7.5) and (2005, 27.5). (27.5 2 7.5) Slope 5 (2005 2 1960) 5 20 45 < 0.44 million/yr.
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CH. 2 Exercises Solutions for Section 2.9
e. y 5 7.5 1 0.44x f. The number of graduates was about 7.5 million in 1960 and has been steadily increasing since—by about 0.44 million, or 440,000, persons each year. 2. a. 100%
Percentage of M.D. Degrees Awarded to Women
80%
50
60% 40
20% 2004
2003
Year
20 10
b. Letting x 5 years from 1996, two estimated points on the line are (0, 17) and (7, 80). The slope 5 (80217)/(720) 5 63/7 5 9 percentage points per year. This tells you that the adults with access to the Internet are increasing by 9 percentage points per year. (Note: You add 9% each year to the previous year’s % value, you are not calculating a percentage.) c. (0, 17) d. If we let I(x) 5 percent of adult population with access to the Internet x years from 1996, then I(x) 5 17 1 9x. e. In 1998, we have x 5 2. So I(2) 5 17 1 (9 ? 2) 5 35%. In 2002, we have x 5 6. So I(6) 5 17 1 (9 ? 6) 5 71%. f. As the percentage of adults with Internet access comes close to the maximum of 100%, we would expect the rate of change to slow down. In 2003 the percent is already close to 80%, so we should not expect the linear model to continue to be appropriate.
Exercises for Section 2.9 1. Equation a matches with table C. Equation b matches with table A. Equation c matches with table B.
1970
1975
1980
1985
1990
1995
2000
2005
Year
b. To extrapolate to 2010, it helps to generate the equation of the line. If we let x 5 years since 1970, the slope remains the same at 1.13 percentage points per year. If x 5 0 (at year 1970), the vertical intercept is at 10. So the equation is y 5 10 1 1.13x where x 5 years since 1970, and y 5 percentage of M.D. degrees awarded to women. If y 5 100%, we would have 100 5 10 1 1.13x 1 x 5 90/1.13 < 80 years since 1970 or in 2050. So if we extrapolated the model to 2050, it would predict that in that year 100% of all medical degrees would be awarded to women. c. It is highly unlikely that this would happen. The most likely scenario is that the graph (and correspondingly the percentage of female M.D. degrees awarded) will taper off to some maximum percentage value, say at 50%. The accompanying graph fits that description. (Student answers may differ somewhat.) Projected Medical Degrees Awarded to Women
3. a. Exactly linear; y 5 1.5x 2 3.5 b. An estimated best-fit line has intercepts at (0, 6) and (2.5, 0), so the slope 5 (6 – 0)/(0 – 2.5) 5 –2.4. The equation is then y 5 –2.4x 1 6. (Note that the data point (0, 6.5) is not on the best-fit line.) Your answer may be somewhat different. y 10
y = –2.4x + 6
–1
30
4
x
–4
c. Exactly linear; y 5 3x 1 35 5. a. Exactly linear; y 5 0.07 quadrillion Btus of solar units b. Approximately linear; farm output 5 220 billion dollars
Degrees awarded to women (%)
2002
2001
2000
1999
1998
1997
0%
Percentage
40%
1996
Percentage with access
Percentage of U.S. Adult with Internet Access
7. a. The accompanying graph shows a hand-drawn bestfit line. We can estimate two points on the line as (1970, 10) and (2000, 44). So the slope of the line is (44 2 10)/(2000 2 1970) 5 34/30 < 1.13. So the average rate of change in the percentage of female M.D. degrees between 1970 and 2002 was about 1.13 percentage points per year.
50
0 1970
2000 Year
2030
9. The graph on the next page gives an eyeballed bestapproximation line that goes through two estimated coordinates of (0, 150) and (80, 40). The y-intercept is 150. 2110 The slope is 408022150 0 5 80 5 21.375. Thus the equation of the graph can be estimated as y 5150 2 1.375x, where y stands for infant mortality rate (deaths per 1000 live births) and x stands for literacy rate (%).This means that on average for each 1% increase in literacy rate, the number of infant deaths drops by 1.375 per thousand.
Ch. 2
d. (0, 7.5) and (45, 27.5)
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CH. 2 Check Your Understanding
Infant Mortality Rate vs. Literacy Rate Infant mortality rate (deaths per 1000 live births)
160 140 X 120 100 80 60 40 20 0 0
X 20
40 60 Literacy rate (%)
b. 2010 corresponds to t 5 50 years and H(50) 5 5 1 0.225 ? 50 5 16.25% of GDP. c. One possible reason is that the GDP has grown very large and thus the percentage for health care has not grown as much as the health care costs themselves. 15. a. Below are the graph of the data and a hand-drawn best-fit line.
80
100
Tax vs. MPG on New Model Cars 7000
11. a. The hand-drawn best-fit model for the years 1990 to 2003 is in the graph below.
6000 Tax in $
5000
Percentage of All U.S. Households with Cell Phones from 1990 to 2003
Percentage
70 60
2002
2000
1998
1996
1994
1992
10 0 1990
14.5
16.5
18.5
20.5
22.5
MPG
40 30 20
Year
An eyeball estimate of two points on the best-fit line that is drawn are (1990, 5) and (2003, 72). The slope of the line between those two points is approximately 72 2 5 67 2003 2 1990 5 13 < 5.15 percentage points per year. This value indicates that on average the percent of U.S. households with cell phones grew approximately 5.15 percentage points each year between 1990 and 2003. Answers may vary slightly. b. The percent of households having cell phones appears to be leveling off. Perhaps the saturation point has been reached. c. Letting x 5 years since 1990, an estimated piecewise bestfit linear function for the combined time periods would be: 5 1 5.15xfor 0 # x , 13 P(x) 5 e 70 for 13 # x # 15 where P(x) 5 % of all households with cell phones. 13. a. Below is the graph of the data along with an estimated best-fit line. The two points (0, 5) and (40, 14) lie on the 9 2 5 line 1 slope 5 14 40 2 0 5 40 5 0.225 The equation of the line is H(t) 5 5 1 0.225t where t 5 years since 1960 and H(t) 5 health care costs as % of GDP. Health Care Costs as Percentage of GDP
b. The approximate coordinates of two points on this line are (17.5, 3000) and (22.5, 0) 1 the slope of this line is 0 2 3000 23000 5 2600 dollars per mpg. The 22.5 2 17.5 5 5 equation of this line is T(x) 5 2600x 1 13,500, where x is in mpg and T(x) 5 tax (in dollars). The vertical intercept is very large because its value is what one would get if mpg takes the value of 0. (This value, of course, represents an impossible situation.) c. The average rate of change for the given line is 2600 dollars per mpg. As the mpg rating increases, the tax paid goes down by $600 per mpg.
Ch. 2: Check Your Understanding 1. False
9. True
17. True
25. False
2. True
10. True
18. False
26. False
3. False
11. False
19. False
27. False
4. False
12. True
20. True
28. False
5. True
13. True
21. True
29. True
6. True
14. False
22. False
30. True
7. True
15. True
23. False
8. False
16. True
24. False
31. Possible answer: y 5 22x 1 5, y dependent variable, x independent variable. 32. Possible answer: D 5 0, D dependent variable, p independent variable. 33. Possible answer: 2x 2 3y 5 6, y dependent variable, x independent variable.
18 16 14 12
34. Possible answer: 3x 1 5y 5 215, y dependent variable, x independent variable.
10 8 6 4
35. Possible answer: T 5 37l, T dependent variable in minutes, l independent variable in laps.
2 0
3000 1000 0 12.5
Slope ≈ 5.15
50
4000 2000
80
Percentage of GDP
Ch. 2
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0
10
20
30
Years since 1960
40
t 50
36. Possible answer: V 5 19.25 1 0.25q, V dependent variable in dollars, q independent variable in number of quarters from now.
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39. Possible answer: For m 5 21/4 (or 20.25), 21/m 5 4, d 5 20.25t 1 3 and d 5 4t 2 1, d dependent variable, t independent variable. 40. Possible answer: (1990, $100), (1993, $133), (1995, $150), (2000, $175), (2003, $185). 41. True
48. False
42. False
49. False
43. False
50. False
44. False
51. False (it can be 0)
45. True
52. True
46. True
53. False (not a function)
47. True
Ch. 2 Review: Putting It All Together 1. a. $217,300 2 $147,300 5 $70,000 b. $70,000/$147,300 < 0.475, or about a 47.5% increase c. ($217,300 2 $147,300)/(2006 2 2000) 5 $70,000/6 < $11,670/year 3. a. The function is positive between points B and E excluding the endpoints; negative between A and B (excluding point B) and between F and G (excluding point F); zero at point B and between points E and F. b. The slope is positive between A and C; negative between D and E and between F and G; zero between C and D and between E and F.
11. a. Average rate of change between weeks 1 and 31 5 (3.04 2 2.22)/(31 2 1) 5 0.82/30 < 0.03, which means that the price of gasoline was increasing by roughly 3 cents per week. Average rate of change between weeks 31 and 43 5 (2.22 2 3.04)/(43 2 31) 5 20.82/12 < 20.07, which means that the price of gas was decreasing by about 7 cents per week. Average rate of change between weeks 1 and 43 5 (2.22 2 2.22)/(43 2 1) 5 0/42 5 0, which means, of course, that if we look only at weeks 1 and 43, we would have the impression that gas prices had not changed at all. b. During the summer of 2006 gasoline prices soared to an all-time high. In January the price per gallon was relatively low. By August the price per gallon had increased on average by 3 cents each week to reach a cost of $3.04 per gallon. In total the cost had increased by $0.82 per gallon or by about 37%. By October gas prices fell back to the January rate, decreasing at a rate of 7 cents per week from the August high. 13. Possible answers are: a. Q 5 3t 2 2 or Q 5 3t b. Q 5 22t 1 3 or Q 5 22t c. Q 5 3 or Q 5 22 d. Q 5 22t or Q 5 3t e. Q 5 3t 2 2 or Q 5 3t f. Q 5 22t or Q 5 22t 1 3 15. a. (3600 people/day) ? (365 days/year) 5 1,314,000 people/ year or, equivalently, an increase of 1.314 million people per year living in the coastal regions of the United States. b. The coastal population P(x) (in millions) can be modeled by P(x) 5 153 1 1.314x, where x is the number of years since 2003. 170
7. a. 34% of 296.4 5 0.34 ? 296.4 < 100.78 million Americans (about one-third) were without a cell phone in 2005. b. Average rate of change 5 (207.9 2 109.4) million/(2005 – 2000) 5 (98.5 million)/(5 years) 5 19.7 million cell phone subscriptions per year. c. C(t) 5 109.4 1 19.7 t, where t 5 years since 2000. d. In 2010 we would have t 5 10, and C(10) 5 109.4 1 (19.7 ? 10) 5 306.4 million cell phone subscriptions. The U.S. population did exceed 300 million in fall 2006, but this extrapolation seems somewhat excessive. 9. a. b. c. d.
i and ii i and ii iii both ii (which is decreasing) and iv (which is increasing).
Coastal population (millions)
5. a. They are both linear functions of U.S. sizes since the average rates of change are constant for both. b. The British size equals the U.S. size plus 6. c. If U.S. sizes are denoted by U, and French sizes by F, then F 5 U 1 34.
165 160 155 150 x 0 1 2 3 4 5 6 7 8 9 10 Years since 2003
d. In 2008 x 5 5, so P(5) 5 153 1 1.314(5) < 160 million people living in coastal regions. 17. a.
70 Paper recycled (%)
38. Possible answer: y 5 2x 1 4, y 5 5x 1 4, y 5 4 2 2x, y 5 4, y dependent variable, x independent variable.
607
65 60 55
(7, 55) (O, 51.5)
50 0
5
10
15
t 20
The slope 5 (55 2 51.5)/7 5 0.5 percentage points per year.
Ch. 2
37. Possible answer: C 5 2T 1 3 and C 5 2(T 2 2) 1 3 (or 2T 2 1), C dependent variable, T independent variable.
CH. 2 Review: Putting It All Together
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Exercises Solutions for EE on Education and Earnings
b. R(t) 5 51.5 1 0.5t, where R(t) is the percentage of paper recycled in t years from 2005. c. R(0) 5 51.5%; R(5) 5 54%; R(20) 5 61.5% would mean that in the year 2025, 61.5% of all paper is recycled.
27. a.
19. Graph A: y 5 5; Graph B: x 5 22: Graph C: y 5 2x 11; Graph D; y 5 2x 1 4; Graph E: y 5 2(1/2)x 1 6 21. a.
23 5
x g(x)
22 4
21 3
b.
0 2
1 3
2 4
3 5
5 4 g(x)
3 2 1
x –3
–2
–1
0
1
2
3
c. The graph of g(x) 5 u x u 1 2 is the graph of the absolute value function f (x) 5 u x u raised up two units. 23. Student answers will vary. 25. a. Over the last 250 years in Sweden the probability of a young child dying has steadily decreased. The child mortality rate has declined from about 40% in 1750 to less than 1% in 2000. The death rate is very similar for female and male children, though the male rates are consistently somewhat higher. b. 250 Years of Mortality in Sweden for Children between 0 and 5 Years 50% Female Male 40% Child mortality rate
EE
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30%
20%
(1800, 33%)
(1900, 13%)
10%
0% 1750
1800
1850
1900
1950
2000
Point
x-Coordinate of Point
y-Coordinate of Point
Average Rate of Change between Two Adjacent Points
A B C D E
0 1 2 3 4
4 1 0 1 4
n.a. 23 21 1 3
b. The function is increasing over the interval (2, 4). The average rates of change (or slopes) are increasing on that interval, which also suggests that the function is increasing. c. The function is decreasing over the interval (0, 2). The absolute value of the average rate of change (or slope) is decreasing over that interval, which also suggests that the function is decreasing. d. The function is concave up throughout, independent of whether the function is increasing or decreasing. The steepness of the curve (as approximated by the absolute values of the average rates of change) first decreases and then increases, also suggesting that the curve is concave up.
Exercises for EE on Education and Earnings 1. a. 0.65, 0.68, 0.07, 0.70 b. k 20.07 k , k 0.65 k , k 20.68 k , and k 0.70 k 3. a. The slope 5 6139 dollars per year of education past grade 8; vertical intercept 5 22105 dollars; cc 5 0.72 b. On average, for each increase of a year of education past grade 8 the median personal earnings increase by $6139. c. $6139; $61,390 5. a. The rate of change is 10,733 dollars per year of education past grade 8. b. Three points are (4, 34242), (8, 77334), and (12, 10426). A 77,334 2 34,242 43,092 5 4 5 10,773 sample computation: 8 2 4 dollars per year of education above grade 8. c. For each extra year of education past grade 8 the median personal total income of white males on average rises $10,733. d. From the FAM 1000 data:
Year
Two estimated points on the line are (1800, 33%) and 13 2 33 (1900, 13%). So the slope of the line is 1900 2 1800 5 220/100 5 20.2%/year. This means that on average the probability of a child between the ages of 0 to 5 dying was decreasing by two-tenths of a percentage point each year, or equivalently, there were 2 fewer children dying per thousand. c. If we let t 5 years since 1800, then (0, 33%) becomes the vertical intercept. The slope remains the same. So the linear model is P(t) 5 33 2 0.2t, where P(t) gives the female child mortality rate (as a percentage) at t years after 1800. d. After 1950 the mortality rates are very low and still declining, almost approaching 0% per year. (In fact, Sweden currently has one of the lowest child mortality rates in the world.)
Median Total Income for White Males ($) 225,000 200,000 175,000 150,000 125,000 cc = 0.87 100,000 75,000 50,000 25,000 0
2 4 6 8 10 12 14 16 Education (years past grade 8)
Median total income for white males 5 28850 1 10,773 ? yrs of educ. past grade 8
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females is: Percentage 5 20.4x 1 35.3, where x 5 years since 1965 for both. The cc for males is 20.972 and for females it is 20.973, both quite good. Percentage of Male and Female Smokers over 18
EE
7. a. 0.516 is the slope of the regression line, and it indicates that, on average, each increase of an inch in the mean height of the fathers means there is an increase of 0.516 inch in the mean height of the sons. b. If F 5 64, then S 5 33.73 1 0.516 ? 64 5 66.75 in. If F 5 73, then S 5 33.73 1 0.516 ? 73 5 71.40 in. c. If F 5 S, then F 5 33.73 1 0.516F or 0.484F 5 33.73 or S 5 F 5 69.69 in. d. For each of the data points the S value represents the mean height of all the sons whose father has the given mean height of F. There are 17 mean heights listed (from 57 to 75 inches) for the 1000 fathers.
60.0 50.0
Percentage
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Males = –0.7x + 48.7 cc = –0.972
40.0 30.0 20.0
Females = –0.4x + 35.3 cc = –0.973
10.0
9. a, b. Here is a graph that gives the Excel-generated regression lines, their equations, and the cc values for public and private 4-year colleges over the time period given in the table. Student estimates of the regression lines for each will vary.
0
10
20
30
40
Years since 1965 Male smokers Female smokers
Mean Tuition and Fee Costs at Public and Private 4-year College
Linear (Male smokers) Linear (Female smokers)
$30,000
e. Student answers will vary. The more notable factors are the downward trend overall for both males and females and the fact that the downward trend is more rapid among males and that the percentage of smokers is always higher among males than among females. Students, in citing these factors, should also be citing actual figures.
$25,000 $20,000 $15,000
Private = 896x + 6154 cc = 0.996
$10,000
Public = 191x + 1152 cc = 0.977
$5,000 0
5
10
15
20
13. a. The regression line and equation and the cc are given in the accompanying diagram.
Year
11. a.
Percentage of Smokers among Total Population 18 and Older 45.0 40.0
Percentage
35.0 30.0
Boston Marathon Winning Times for Women Winning times in minutes
c. In 2010 the cost at a public college will be 191 ? 25 1 1152 5 $5927 and in private colleges the cost will be 896 ? 25 1 6154 5 $28,554.
200 180 160 140 120 100 80 60 40 20
y = –0.9x + 166.4 cc = –0.752
0
25.0 y = –0.6x + 41.6 cc = –0.986
20.0 15.0 10.0 5.0 0.0 0
10
20
30
40
Years since 1965
20.9 2 42.4 222 i. 2005 2 1965 5 40 < 20.55 percentage point per year for all smokers from 1965 to 2005. 20.9 2 25.5 24.6 ii. 2005 2 1990 5 15 < 20.31 percentage point per year for all smokers from 1990 to 2005. b. Student estimates of the regression line will vary. The Excel-generated regression line is shown in the diagram in part (a), and the equation is Percentage 5 20.6x 1 41.6, where x 5 years since 1965. The average rate of change is 20.6 percentage point per year. c. Student estimates of the regression line will vary. The line drawn in the diagram is the Excel-generated regression line. The cc is quite good: 20.986. d. The regression line equation for males is: Percentage 5 20.7x 1 48.7, and the regression line equation for
10 20 Years since 1972
30
b. In 2010 x 5 38 and time predicted 5 20.9 ? 38 1 166.4 < 132.2 minutes. This does not seem to be a reasonable winning time. It is 10 minutes less than the recent winning times. c. Using an eyeball estimate, one gets a horizontal line that goes through approximately y 5 144. Yes, it is more realistic than the estimate in part (b). d. Student answers will differ, but all should mention that the data seem to fall along a horizontal line in the constricted time span, indicating a leveling at times just above 140 minutes. 15. a. The plot of the data and the regression line are given in the accompanying diagram. The formula relating boiling temperatures in 8F to altitude is F 5 211.80 2 0.0018H, or when suitably rounded off, F 5 212 2 0.002H, where H is feet above sea level. The correlation coefficient is 20.9999. The answer to the second part depends on where the student lives. Other factors could be something put into the water, such as salt, or variations in the air pressure.
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Boiling Temperature of Water
CH. 3 Algebra Aerobics Solutions for Section 3.2
3. a. 12x 2 9y 5 18 is equivalent to 4x 5 6 1 3y because
Temperature in degrees F
Ch. 3
250
1 1 (12x 2 9y) 5 (18) 1 4x 2 3y 5 6 3 3 1 4x 5 6 1 3y.
200 150
So these equations represent the same line. b. There is an infinite number of solutions to the system of equations in part (a) since every solution to the equation (every point on the line) is a solution to the system.
100 50
0
5,000 10,000 15,000 20,000 25,000 30,000 Altitude in feet above sea level
b. On Mt. McKinley water boils at 212 2 0.002 ? 20320 5 171.36 8F; in Death Valley water boils at 212 2 0.002 ? (2285) 5 212.57 8F. c. 32 5 212 2 0.002H or 2180 5 20.002H or H 5 90,000 feet or approximately 17 miles. But this seems unreasonable since then the water would be outside earth’s atmosphere layer (which goes to 9 miles above the earth). 17. a. By computer or calculator one gets that the regression line equation as: y 5 3.5x 1 26.7, where x 5 years since 1945 and y is measured in millions of registrations. [The cc. is 0.998]. b. On average the number of registrations goes up approximately 3.5 million per year since 1945. c. 2004 is 59 years after 1945, and thus y 5 3.5 ? 59 1 26.7 5 233.2 million. It is about 4.9 million too high. d. 2010 is 65 years from 1945. Thus y 5 3.5 ? 65 1 26.7 5 254.2 million. This seems to be too high since the entire U.S. population in 2007 was just over 300 million. 19. a. Of itself high correlation does not mean that there is causation involved. More studies would have to be done and in fact were done. The research leaves no doubt that cigarette smoking is indeed a cause of lung cancer. The high correlation coefficient was nevertheless an important factor.
4. Graph A: (0, 2.5) Graph B: (4, 24)
Exercises for Section 3.1 1. a. 4(5) 2 3(210) 5 50 and 2(5) 1 2(210) 5 210, and thus (5, 210) does not solve the given system. b. The coordinates must satisfy both equations. 3. A solution to a system of equations is a number (or set of numbers) that satisfies all of the equations in the system. 5. a. The coordinates are approximately (1995, $380,000) b. To the left of the intersection point the population of Pittsburgh is larger than that of Las Vegas, and to the right the population of Las Vegas is greater than that of Pittsburgh. 7. a.
b.
y 3
y 10 x+y=9 (5, 4)
(–1, 1)
x + 4y = 3 x
–3
3 x + 2y = 1
–3
x 10
–10 2x – 3y = –2
–10
9. y 5 2x 2 2 and y 5 2x – 8 is the system, and the solution is (2, 24). Check: 22 2 2 5 24 and 2 ? 2 2 8 5 24, and thus the claimed solution works.
Section 3.2 Algebra Aerobics 3.2a
CHAPTER 3 Section 3.1 Algebra Aerobics 3.1 1. Gas is the cheapest system from approximately 17.5 years of operation to approximately 32.5 years of operation. Solar becomes the cheapest system after approximately 32.5 years of operation. 2. a. (3, 21) is a solution since: 4(3) 1 3(21) 5 9 and 5(3) 1 2(21) 5 13. It is a solution of both equations. b. (1, 4) is not a solution since: 5(1) 1 2(4) 5 13 but 4(1) 1 3(4) 5 16, not 9. So it is not a solution of 4x 1 3 5 9.
1. a. y 5 7 2 2x b. y 5 6 25 3x 5 65 2 35 x c. x 5 2y 2 1 2. a. no solution, because the lines have the same slope but different y-intercepts, so they are parallel. b. one solution, because the lines have different slopes. 3. a. Set y 5 y 1 x 1 4 5 22x 1 7 1 3x 5 3 1 x 5 1; y 5 22(1) 1 7 5 5. Check: y 5 (1) 1 4 1 y 5 5 1 solution (x, y) is (1, 5). b. Set y 5 y 1 21700 1 2100x 5 4700 1 1300x 1 800x 5 6400 1 x 5 8; y 5 4700 1 1300(8) 5 15,100 1 solution (x, y) is (8, 15100).
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b. 3x 1 y 5 6 1 y 5 6 2 3x; 6x 1 2y 5 5 1 y 5 5 22 6x or y 5 52 2 3x 1 no solution since the lines have the same slope of 23 and different y-intercepts of 6 and 52 .
4. a. Substitute y 5 x 1 3 into 5y 2 2x 5 21 1 5(x 1 3) 2 2x 5 21 1 5x 1 15 2 2x 5 21 1 3x 5 6; x 5 2; so, y 5 (2) 1 3 5 5 1 y 5 5. Solution (x, y) is (2, 5).
c. 2x 1 3y 5 1 1 y 5 13 2 23x; 4x 1 6y 5 2 1 y 5 13 2 23x 1 equivalent equations and an infinite number of solutions, since the lines have same slopes and same y-intercepts.
d. Solve: r 2 2s 5 5 for r, and substitute the resulting expression for r into 3r 2 10s 5 13. r 5 2s 1 5 1 3(2s 1 5) 2 10s 5 13 1 6s 1 15 2 10s 5 13 1 24s 522 1 s 5 12; so r 5 2 A12B 1 5 5 6 1 r 5 6. Solution (r, s) is A6, 12B .
Algebra Aerobics 3.2b 1. a. By the elimination method, add equations: 2y 2 5x 5 21 and 3y 1 5x 5 11 1 5y 5 10 1 y 5 2; so 3(2) 1 5x 5 11 1 5x 5 5 1 x 5 1. Solution (x, y) is (1, 2). b. Multiply the equation (3x 1 2y 5 16) by 3 and the equation (2x 2 3y 5 211) by 2 1 9x 1 6y 5 48 and 4x 2 6y 5 222. By the elimination method, add these equations 1 13x 5 26 1 x 5 2, so 3(2) 1 2y 5 16 1 2y 5 10 1 y 5 5. Solution (x, y) is (2, 5). c. By substitution of t 5 3r 2 4 into 4t 1 6 5 7r 1 4(3r 2 4) 1 6 5 7r 1 12r 2 16 1 6 5 7r 1 210 1 5r 5 0 1 5r 5 10 1 r 5 2, so t 5 3(2) 2 4 5 2 1 t 5 2. Solution (r, t) is (2, 2). d. Substitute z 5 2000 1 0.4(x 2 10,000) into z 5 800 1 0.2x 1 2000 1 0.4(x 2 10,000) 5 800 1 0.2x 1 2000 1 0.4x 2 4000 5 800 1 0.2x 1 0.2x 5 2800 1 x 5 14,000; so, z 5 800 1 0.2(14,000) 5 3600 1 z 5 3600. Solution (x, z) is (14000, 3600). 2. a. By substitution: 2x 1 4 5 2x 1 4 1 3x 5 0 1 x 5 0, so y 5 2(0) 1 4 5 4 1 y 5 4. So solution (x, y) is (0, 4). b. By substitution of (y 526x 1 4) into (5y 1 30x 5 20) 1 5(26x 1 4) 1 30x 5 20 1 230x 1 20 1 30x 5 20 1 20 5 20. So both equations must be equivalent. There are infinitely many solutions since both equations describe the same line. c. 2y 5 700x 1 3500 1 y 5 350x 1 1750. The slopes of the lines of both the equations are 350, but the y-intercepts are different (1500 and 1750), so the lines are parallel. There is no solution. 3. In order for a system of equations to have no solutions, they must produce parallel lines with the same slope, but different y-intercepts. One example is: y 5 5x 1 10; y 5 5x 1 3. 4. a. 2x 1 5y 5 7 1 y 5 22x51 7; 3x 2 8y 5 21 1
y 5 3x 81 1 1 one solution since the lines have unequal slopes of 225 and 38.
y
5. a. 6 A2x 1 3B 5 6(3) 1 3x 1 2y 5 18. Substitute y 5 x 1 4 1 3x 1 2(x 1 4) 5 18 1 5x 1 8 5 18 1 5x 5 10 1 x 5 2; so, y 5 (2) 1 4 5 6 1 y 5 6. Solution (x, y) is (2, 6). b. 260(0.5x 1 0.7y) 5 260(10) 1 230x 2 42y 5 2600. Add to 30x 1 50y 5 1000 1 8y 5 400 1 y 550, so 30x 1 50(50) 5 1000 1 30x 1 2500 5 1000 1 30x 5 21500 1 x 5 250, so the solution (x, y) is (250, 50). 6. a. 4 (39) 1 3q 5 240 1 3q 5 84 1 q 5 28 gals b. 4p 1 3(20) 5 240 1 4p 5 180 1 p 5 $45 per gal c, d. Solve for p: 4p 5 240 2 3q 1 p 5 60 2 3/4q 1 p 5 60 2 0.75q. p
Supply
70
q
p
0 20 40
60 45 30
Price (dollars)
c. Substitute x 5 2y 2 5 into 4y 2 3x 5 9 1 4y 2 3(2y 2 5) 5 9 1 4y 2 6y 1 15 5 9 1 22y 5 26 1 y 5 3; so x 5 2(3) 2 5 5 1 1 x 5 1. Solution (x, y) is (1, 3).
d. 3x 1 y 5 8 1 y 5 8 2 3x; 3x 1 2y 5 8 1 y 5 4 2 32 x 1 one solution since the lines have unequal slopes of 23 and 232.
p = 0.85q
50 30
Demand p = 60 – 0.75q q
10 –10
0 10 20 30 40 50 60 70 80 Quantity (gallons)
e. Solve by substitution of p 5 0.85q into 4p 1 3q 5 240 1 4(0.85q) 1 3q 5 240 1 q 5 37.5; p 5 0.85(37.5) 5 31.9. So the equilibrium point is ,(38, $32), which means that when the price is around $32, the demand will be around 38 gallons. f. There is a surplus of supply because where the line p 5 39 crosses the supply curve, it is above the demand curve, so the supply is greater than the demand.
Price (dollars)
b. Substitute z 5 3w 1 1 into 9w 1 4z 5 11 1 9w 1 4(3w 1 1) 511 1 9w 1 12w 1 4 5 11 1 21w 5 7 1 w 5 1/3; so z 5 3 A13B 1 1 5 2 1 z 5 2. Solution (w, z) is A13, 2B
80 70 60 50 40 30 20 10 0
Supply p = $39
Demand
10 20 30 40 50 60 70 80 Quantity (gallons)
7. For this system to have an infinite number of solutions, the two linear equations should represent the same line. The slope of the line of the first equation is 2, and the y-intercept is 4. Solving the second equation for y, we get y 5 2??/2x 1 4. Thus, ??, the coefficient for which we are solving, must be 24, so the slope of that line is also 2.
Ch. 3
c. Set F 5 F 1 C 5 32 1 95 C 1 5C 5 32(5) 1 9C 1 24C 5 160 1 C 5 240 F 5 C 1 F5 240 1 solution (C, F) is (240, 240).
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CH. 3 Exercises Solutions for Section 3.2
CH. 3 Exercises Solutions for Section 3.2
Exercises for Section 3.2
S 70,000
After approximately 9 months. Approximately $8.30 per hour. WA(m) 5 7.00 1 0.15m and WB (m) 5 7.45 1 0.10m The exact place where they meet is m 5 9, and WA(9) 5 WB(9) 5 $8.35 per hour. e. The exact common hourly wage is $8.35 per hour. f. Before 9 months, the monthly wage rates at company B are higher; after 9 months the monthly wage rates at company A are higher. If one had more information about the number of hours worked at each place, one could judge the companies on accumulated wages instead of hourly rates. But no such information is given.
3. a. The graph of the linear system is given with the intersection point marked.
40,000 (26.3, 33950)
C solar = 30,000 + 150n Cost in $
Ch. 3
1. a. b. c. d.
C elec = 5000 + 1100n
n 0
30 Years
b. 1100 and 150 are the slopes of the heating cost lines; 1100 represents the rate of change of the total cost in dollars for electric heating per year since installation; 150 is the rate of change in the total cost for solar heat in the same units. c. 5000 is the initial cost of installing the electric heat in dollars; 30,000 is the initial cost in dollars of installing solar heating. It cost a lot more initially to install solar heating than to install electrical heating. d. The point of intersection is approximately where n 5 26 and C 5 34,000. e. n < 26.32, C < 33,947.37 is a more precise answer; the values have been rounded off to two decimal places; they were obtained by setting the equations equal to each other. f. Assuming simultaneous installation of both heating systems, the total cost of solar heat was higher than the total cost of electric heat up to year 26 (plus nearly 4 months); after that the total cost of electric heat will be greater than that of solar heat. 5. a. Setting the two equations equal gives 20,000 1 2500n 5 25,000 1 2000n or 500n 5 5000 or n 5 10. Plugging in that n value gives S 5 20,000 1 2500 ? 10 5 45,000 b. The graphs of the two linear equations are given in the diagram below. From inspecting the graphs it seems that the intersection occurs when n 5 10 and S 5 45,000.
S = 25,000 + 2000n (10, 45,000)
S = 20,000 + 2500n
0
n 20
10
7. a. The method that is easiest is often a judgment by the person solving the problem. i. Either is easy iv. Substitution ii. Elimination v. Either is easy iii. Substitution vi. Substitution b. i. Setting the y values equal to each other gives 6 5 24. Thus there is no solution. ii. Letting y 5 2x – 5 in the second equation gives 5x 1 2(2x 2 5) 5 8 or 9x 2 10 5 8 or 9x 5 18 or x 5 2. Thus y 5 2 ? 2 – 5 5 21 and therefore the solution is (2, 21). Check: 2(2) 2 (21) 5 5; and 5 ? 2 1 2 ? (21) 5 8. iii. Substituting x 5 7y 2 30 into the first equation gives 3 ? (7y 2 30) 1 2y 5 2 or 21y 2 90 1 2y 5 2 or 23y 5 92 or y 5 4 and then x 5 28 2 30 5 22. Thus the solution is (22, 4). Check: 3 ? (22) 1 2 ? 4 5 26 1 8 5 2 and 7 ? 4 2 30 5 22. iv. Substituting y 5 2x 2 3 into second equation gives: 4(2x 2 3) 2 8x 5 212 or 8x 2 12 2 8x 5 212 or 0 5 0; thus the two equations have the same line as their graph. Thus all points on the line y 5 2x 2 3 are solutions. v. Elimination yields 0 5 26 and thus there is no solution. vi. Substituting y 5 3 into the second equation gives x 1 2 ? 3 5 11 or x 5 5. Thus the solution is (5, 3). Check: 3 ? 3 5 9 and 5 1 2 ? 3 5 11. 9. a. Subtracting the first equation from the second yields 4x 5212 or x 5 23. Putting this value into the first equation gives 23 1 3y 5 6 and thus 3y 5 9 and y 5 3. Putting this value of x into the second equation gives 5(23) 1 3(3) 5 26 and thus the solution is (23, 3). b. The graphs of the two equations and the coordinates of the intersection point are shown in the accompanying figure. y 5 (–3, 3) x + 3y = 6 x
–5
5 5x + 3y = –6
–5
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CH. 3 Exercises Solutions for Section 3.2
CH. 3 Exercises Solutions for Section 3.2
11. Let x 5 amount to be invested at 4% and let y 5 amount to be invested at 8%. Then the system of equations to be solved is x 1 y 5 2000 and 0.04x 1 0.08y 5 100. Substituting y 5 2000 2 x into the second equation gives, after simplification, x 5 $1500 and thus y 5 $500. [Check: 1500 1 500 5 2000 and 0.04 ? 1500 1 0.08 ? 500 5 100.]
15. a. The two equations in m and b are: 22 5 2m 1 b and 13 5 23m 1 b and the solution is m 5 23 and b 5 4. [Check: 2(23) 1 4 5 22 and 23(23) 1 4 5 13.] b. The two equations in m and b are: 38 5 10m 1 b and 24.5 5 1.5m 1 b. The solution is m 5 5 and b 5 212. [Check: 38 5 5 ? 10 – 12 and 24.5 5 5 ? 1.5 2 12.] 17. a. 12.5 is the production cost per shirt in dollars. b. 15.5 is the selling price of a shirt in dollars. c. 15.5x 5 12.5x 1 360 or 3x 5 360 or x 5 120 and y 5 15.5 ? 120 5 $1860. d. When x 5 120, then C 5 12.5 ? 120 1 360 5 $1860 and R 5 15.5 ? 120 5 $1860.
d or 720 2 20d 5 12d or 2 d 1 1 3 2 12 d 5 20 d or 36 12 5 20 720 5 32d or d 5 22.5 ft and H 5 22.5/20 5 1.125 ft. Thus the point of meeting is where d 5 22.5 ft from the platform and H 5 1.125 ft above ground level. 25. a. b. c. d. e.
Eliminate z; 11x 1 7y 5 68 (4) Eliminate z; 9x 1 7y 5 62 (5) x 5 3 and y 5 5 satisfy (4) and (5) Thus z 5 2 ? 3 1 3 ? 5 2 11 5 10 Thus the solution is x 5 3, y 5 5, z 5 10 and the check is below: (1) 2 ? 3 1 3 ? 5 2 10 5 11 (2) 5 ? 3 2 2 ? 5 1 3 ? 10 5 35 (3) 1 ? 3 2 5 ? 5 1 4 ? 10 5 18
27. Answers will vary. a. The system y 5 x 1 5 and y 5 x 1 6 has no solution. b. The system y 5 x 1 5 and y 5 2x 1 5 has exactly one solution. c. Algebraically: setting x 1 5 5 2x 1 5 gives 2x 5 0 or x 5 0 and thus y 5 5; alternatively, adding the two equations together gives 2y 5 10 or y 5 5 and thus x 5 0. The graphs of the two lines intersecting at the point claimed is in the accompanying diagram. The answers agree. y 10 y=x+5
19. a. The graphs of the supply and demand equations are in the accompanying diagram. b. The equilibrium point is shown in the diagram. It is the spot where supply meets the demand; i.e., if the company charges $410 for a bike it will sell exactly 4000 of them and have none left over.
(0, 5)
y = –x + 5 x
–5
p
0
5
700
29. The system of equations has no solution if the graphs of the two equations are parallel and distinct lines. This occurs when m1 5 m2 (parallel means same slope) and b1 2 b2 (different vertical intercepts).
Selling price in $
Equilibrium point (4, 410)
Supply
Demand
q 0
Bicycles made/sold in 1000
9
31. a. The equations of this pair of lines are: y 2 5 5 24(x 2 2) and y 2 5 5 3.5(x 2 2), or in simplified form: y 5 24x 1 13 and y 5 3.5x 2 2. b. The graphs of the two equations are given in the accompanying diagram. [Check: 24 ? 2 1 13 5 5 and 3.5 ? 2 2 2 5 5.]
21. Two equations are equivalent if their graphs are the same, i.e., they have the same sets of solutions. An example is the system 2x 1 y 5 1 and 4x 1 2y 5 2. 23. If we make the origin the spot on the diagram where the height (in feet) or H-axis meets the ground and let d be the distance (in feet) from the origin, then the equation of 1 the ramp is H 5 3 2 12 d. The equation to describe the rising 1 ground is H 5 20 d . Setting them equal to each other gives
y 40
y = –4x + 13 (2, 5) x –5
5 y = 3.5x – 2 –20
Ch. 3
13. a. Letting y 5 x 2 4/3 from the second equation and substituting this value in the first equation, we get x 2 4/3 x 5 1. Multiplying both sides by 6 gives 3 1 2 2x 1 3(x 2 4/3) 5 6 or 5x 2 4 5 6 or x 5 2 and thus y 5 2 2 4/3 5 2/3. [Check: 2 2 2/3 5 4/3 and 2/3 1 1/3 5 1.] b. Substituting y 5 x/2 from the second equation into the first equation we get x/4 1 x/2 5 9 or (3/4)x 5 9 or x 5 12. Then y 5 12/2 5 6. [Check: 12/4 1 6 5 9.]
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CH. 3 Exercises Solutions for Section 3.2
CH. 3 Algebra Aerobics Solutions for Section 3.3
b. Lower birth rate: y Old birth rate
Old equilibrium point
New equilibrium point
x 0
35. a. For a given price the new supply curve shows more items being made. b. The requested graph is given in the diagram. In going from the old equilibrium point to the new one, the price goes down and the quantity made/sold goes up at the equilibrium point.
Death rate
New birth rate Rates
33. a. yB 5 30, yA 5 0.625x. b. The common point in space that both planes will eventually occupy is where x 5 48 and yB 5 30. (It is the intersection point of the graphs of y 5 30 and yA 5 0.625x. These are equations for constant altitude of the flight paths of the two planes.) c. B, in going from (230, 30) to (48, 30), travels a distance of 78 miles, and this takes B 13 minutes to do (since it is traveling at 6 miles/minute). A, in traveling from (80, 50) to (48, 30), covers "(30 2 50) 2 1 (48 2 80) 2 5 "400 1 1024 5 "1424 < 37.7 miles, and this will take approximately 18.9 minutes (since plane A is traveling at 2 miles per minute). Thus plane A will arrive at this point nearly 6 minutes after plane B. It is a safe situation.
Population density
If the birth rate decreases (and the death rate stays the same) the equilibrium moves to the left and down. This means the equilibrium point will occur at a lesser population density and lower birth rate.
Section 3.3 Algebra Aerobics 3.3 1. a.
Price
Demand curve
d.
y 5
Old supply curve
y
4
New supply curve
Old equilibrium point New equilibrium point
x 5
–2 –2
x –8 Quantity
b.
y
8
e.
4
y 100
–200
37. a. Higher birth rate:
–100
100
x 200
200
x 400
x 4
–2
–100
–4
c.
New birth rate
Old equilibrium point Rates
Ch. 3
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Death rate
–200
f.
1000
y 400
New equilibrium point
Old birth rate
200
x –40
40
–400
–200 –200
–400
0
Population density
If the birth rate increases (and the death rate stays the same) the equilibrium point moves to the right and up. This means that the equilibrium point will occur at a greater population density and a higher birth rate.
2. a. (2, 3) is a solution because: 3 . 2(2) 2 3 1 3 . 1 and 3 # 3(2) 1 8 1 3 # 14 are true. b. (24, 7) is not a solution because: 7 # 3(24) 1 8 1 7 # 24 is not true. c. (0, 8) is a solution because: 8 . 2(0) 2 3 1 8 . 23 and (8) # 3(0) 1 8 1 8 # 8 are true. d. (24,26) is a solution because: 26 . 2(24) 2 3 1 26 . 211 and 26 # 3(24) 1 8 1 26 # 24 are true.
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CH. 3 Exercises Solutions for Section 3.3
CH. 3 Exercises Solutions for Section 3.3
e. (20, 28) is not a solution because: 28 . 2(20) 2 3 1 28 . 37 is not true. f. (1, 21) is not a solution because: 21 . 2(1) 2 3 1 21 . 21 is not true. 3. A. y # 2 2 x
B. y . 1 1 2x
C. y $ 23
b. y $ 232 x 1 3
3. a. y , 23 x 1 2
5. a. Yes, (0, 0) satisfies the inequality. y
D. x . 4
6 2 1 x+ 3 3
x –6
6 y<
Dulcimer Books Cost and Revenue
2 1 x+ 3 3
R
1400 1200 1000 800 600 400 200 0
–6 C
b. No, (0, 0) does not satisfy the inequality.
0
y
25 50 75 100 125 150 175 200 Number of books
5
c. $400 because that is the cost for selling 0 books (vertical intercept of the cost equation). d. Assuming fixed costs at $400, C1 $ 3x 1 400, R1 # 7x. 5. a. C (n) 5 $50,000 1 235n; R(n) 5 270n. b. C (n) 5 R(n) at breakeven point 1 50,000 1 235n 5 270n 1 n < 1429 tons. Selling about 1429 tons will yield a profit of $0 since cost 5 revenue at the breakeven point. c.
x≥3 x –5
5 x=3
–5
c. Yes, (0, 0) satisfies the inequality.
R(n) $600,000
C(n)
y
Dollars
8
Breakeven point
$200,000
y > 3x – 7
y = 3x – 7 x
–8 0
8
n 5,000 10,000 15,000 200,00 Tons of fertilizer –8
Exercises for Section 3.3 1. a.
c.
y 8
d. No, (0, 0) does not satisfy the inequality.
y 6 y=4
y 5 y>x+5 x –8
y < 2x + 2
x
8
–6
6
y=x+5
y<4
y = 2x + 2
x –5 –8
b.
–6
d.
y 8
y
–5
6
y = –3x – 3
y ≥ (2/3)x – 4 x
x –8
8
–6
6
y ≥ –3x – 3
y = (2/3)x – 4
–8
5
–6
7. One looks to see if (0, 0) satisfies the inequality; it does here and thus the region is that half of the plane that contains (0, 0). In this case the shaded region is above the line. 9. a. goes with g. b. goes with i.
c. goes with j. d. goes with f.
e. goes with h.
Ch. 3
y=
4. a. Approximately (100, $700). For sales of 100 books, the cost is equal to the revenue, which is $700. b. The region between the two graphs to the left of the breakeven point.
Dollars
615
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CH. 3 Exercises Solutions for Section 3.3
CH. 3 Exercises Solutions for Section 3.3
11. a. l1 has the equation y 5 1 1 0.25x and l2 has the equation y 5 3 2 1.5x. b. 3 2 1.5x # y # 1 1 0.2x
But its graph is lower from W 5 100 to W 5 200. It is a safe rule. 17.
13.
W 180
T 900
Ch. 3
Weight in pounds
Overweight zone
Unsafe times
Healthy zone W = Whi
Underweight zone T = 20S W = Wlo 100
Safe times
0
H 15
S 45
0 2
a. The two formulae and the three zones are graphed in the diagram. b. 100 1 3.5H # W # 118.2 1 4.2H lb for 0 # H # 15 in above 5 ft. c. Wlo(2) 5 107 lb and Whi(2) 5 126.6 lb. Thus the shorter woman is overweight. For the taller woman: Wlo(5) 5 117.5 and Whi(5) 5 139.2. Thus the taller woman is in the healthy range. d. Whi(4) 5 135 and (165 2 135)/1.5 5 30/1.5 5 20. Thus it would take 20 weeks for this woman to reach the top of the healthy range.
a. T 5 20S b. The graph is in the diagram. A suitable domain is 2 # S # 45. c. T . 20S denotes unsafe times. d. The shading and labels are found in the diagram. e. The equation would be T 5 40S and its slope would be steeper.
15.
Height above 5 ft. in inches
N 14
19. Not legally safe region
2,000,000
(200, 12.65)
Money in $
R = 266.67T N = 6.4 + 0.0625(W – 100) (100, 6.4)
C = 235,487 + 206.68T (3925.4, $1,046,796)
Legally safe region
44,000 1000 0 100
W 200
a. N . 6.4 1 0.0625(W 2 100), where W is measured in pounds and N measures the number of ounces of beer that gets one to the legal limit for safe driving. b. The sketch of the shaded areas is found in the diagram for 100 # W # 200. c. If W 5 100 lb, then N 5 6.4 oz. Thus one may legally drink 6.4 oz or less; for W 5 150 lb we have N 5 9.525 oz and for W 5 200 lb we have N 5 12.65 oz. d. N 5 0.0625W 1 0.15 e. The given rule of thumb translates into the formula N 5 6 1 0.05(W 2 100) 5 0.050W 1 1. Thus it starts out higher and grows more slowly than the legal one.
T 8000 Tons
a. C 5 235,487 1 206.68T gives the cost in dollars when T is measured in tons of fertilizer produced. R 5 266.67T gives the revenue in dollars from selling T tons of fertilizer. b. The graph is found in the accompanying diagram and the breakeven point is marked on the graph. It is where T < 3925.4 tons and M < $1,046,800 dollars c. The inequality R 2 C . 0 describes the profit region, and this occurs when T . 3925.4. It is shaded in the accompanying graph. 21. For Graph A: x $ 0, y $ 0, and y , 21.5x 1 3 For Graph B: x 1 1 # y , 2x 1 2.
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CH. 3 Algebra Aerobics Solutions for Section 3.4
23.
CH. 3 Exercises Solutions for Section 3.4
C
b. g(i) 5 e
5. a. ƒ(30,000) 5 0.0595(30,000); g(30,000) 5 0.055(30,000) 5 $1785 5 $1650 Flat tax is $135 higher for $30,000 income.
(5000, 15,000)
(15,000, 15,000)
C = 15,000 T + C = 30,000
Feasible region T = 5000
(30,000, 0) T 30,000
(5000, 0) 0 Amount invested in Treasury bonds
a. 0 # T 1 C # 30,000, 0 # C # 15,000 and 5000 # T # 30,000 b. The feasible region is the shaded area of the graph. c. Intersection points and interpretations: (5000, 0) is where $5000 is invested in T bonds; (30000, 0) is where all $30,000 is in T bonds; (5000, 15000) is where $5000 is in T bonds and $15,000 is in C bonds; and (15000, 15000) is where $15,000 is in each kind.
c. ƒ(120,000) 5 0.0595(120,000) 5 $7,140 g(120.000) 5 6263 1 0.098(120,000 2 90,000) 5 6263 1 2940 5 $9,203 Graduated tax is $2063 higher for $120,000 income.
6. y 5 e
100 1 1.5x for 0 # x , 200 400for 200 # x # 500
1. ƒ(210) 5 2 ? (210) 1 1 5 219 ƒ(22) 5 2(22) 1 1 5 23 ƒ(0) 5 2 ? 0 1 1 5 1 ƒ(2) 5 3 ? 2 5 6 and ƒ(4) 5 3 ? 4 5 12
Algebra Aerobics 3.4 c.
y
y 80
3
x –4
40
4
3. a. goes with Graph B. x
–3 0
b.
b. ƒ(60,000) 5 0.0595(60,000) 5 $3570 g(60,000) 5 2761 1 0.088(60,000 2 50,200) 5 2761 1 0.088(9800) 5 2761 1 862.40 5 $3623.40 Graduated tax is $53.40 higher for $60,000 income.
Exercises for Section 3.4
Section 3.4 1. a.
0.06i for 0 # i # $30,000 1800 1 0.09(i 2 30,000) for i . $30,000
25
5. a. y 5 b
y 10
b. y 5 b
1 x
b. goes with Graph A.
for 0 # x # 1 for x . 1
12x 1.5x 2 1.5
for 0 # x # 1 for x . 1
7. a. Graph of y 5 h(x) 0
x 8
2. Graph A: ƒ(x) 5 e Graph B: ƒ(x) 5 e
x 1 3 for x # 3 22x 1 12for x . 3 22 for x # 3 2x 2 8for x . 3
y 40
(0, 20)
3. a. P(25) 5 3, P(0) 5 3, P(2) 5 23, P(10) 5 219 b. W(25) 5 29, W(0) 5 24, W(2) 5 6, W(10) 5 14. 4. a. g(i) 5 e
0.05i for 0 # i # $50,000 2500 1 0.08(i 2 50,000) for i . $50,000
(5, 10)
0
(10, 10) x 20
Ch. 3
Amount invested in corporate bonds
30,000
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CH. 3 Exercises Solutions for Section 3.4
CH. 3 Check Your Understanding
b. Graph of y 5 k(x) 10,000 (50,000, 8400)
Ch. 3
y = k(x)
(20,000, 3000)
(0, 0) x 70,000
0
9. Answers will vary from state to state. Check student answers against the local tax form itself.
15. The graphs for parts (a) and (b) are shown in the accompanying diagram. a. For 0 # T # 20: Dbeginner 5 (3.5/60)T or 0.0583T since there is 1/60 of an hour in a minute; note that T is measured in minutes and Dbeginner is measured in miles. b. For 0 # T # 10: Dadvanced 5 (3.75/60)T or 0.0625T and for 10 , T # 20 we have Dadvanced 5 0.625 1 (5.25/60) (T 2 10) 5 0.0875T 2 0.25.
11. a. 45 mph is the speed limit. b. Speed (mph) Fine ($) 40 45 50 55 60 65 70 75 80
40 if 0 # x # 450 40 1 0.45(x 2 450) if x . 450 L(x) 5 60 1 0.05x if x $ 0 where x measures minutes used for long distance and C(x) and L(x) are measured in dollars. The two cost functions are graphed in the diagram. They meet at x 5 556.25 minutes and C(x) 5 L(x) < $87.81. Thus the two plans cost the same at the point where one uses 556.25 minutes for long distance. c. It would be more advantageous to use the cell phone for 0 # x , 556.25 minutes d. It would be more advantageous to use the local company plan if x . 556.25 minutes.
a. C(x) 5 b
y
0 0 75 100 150 200 300 400 500
d 1.6
c. Check the range category in which the clocked speed is found and then apply the formula for that range category. d. In each category this number represents how much more a person is fined for each increase of 1 mph in speed. e. F(30) 5 0; F(57) 5 100 1 10 ? 2 5 $120; and F(67) 5 200 1 20 ? 2 5 $240. 500 f. (80, 500)
Advanced Beginner
Fine = f (speed)
Fine in $
t 20
0
(65, 200) (55, 100)
T
Dadvanced
Dbeginner
0 5 10 15 20
0.0000 0.3125 0.6250 1.0630 1.5000
0.0000 0.2915 0.5830 0.8745 1.1660
(45, 0) 0
Speed in mph
80
c. The graphs intersect only at T 5 0.
13. The graph is given here to help one see the answers:
Ch. 3: Check Your Understanding
100
Local phone cost
(556.25, 87.81)
Cost in $
y = L(x)
1. False
8. True
15. True
22. True
2. True
9. True
16. True
23. True
3. True
10. True
17. True
24. True
4. False
11. False
18. False
25. False
5. False
12. True
19. False
26. True
6. False
13. False
20. True
27. False
7. False
14. False
21. False
28. True
Cell phone cost y = C(x)
0
Minutes used
600
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CH. 3 Review: Putting It All Together
CH. 3 Review: Putting It All Together
2x 1 y 5 7 26x 2 3y 5 221
32. Possible answer: b
y . 2x 1 1 y , 2x 2 5
33. Possible answer: b
c5r11 c 5 2r 2 1
34. Possible answer: b
C 5 25q 1 2500 R 5 50q
35. Possible answer: p 5 100 2 3q 36. b
x.0 y,0
37. False
38. False
39. True
40. True
Ch. 3 Review: Putting It All Together 1. a. Maximum: approx. 1650 MMT in 2004; minimum: approx. 1420 MMT in 2002. b. The three intersection points show when production equaled consumption. c. In 2002 there was a deficit of about 100 MMT. d. Original title for the graph: Grain Consumption Outstrips Production Again. (Answers will vary.)
9. a. Estimates: production 5 2900 thousand barrels/day and consumption 5 2300 thousand barrels/day. The net difference is 600 thousand barrels per day. In 1990 China was producing more oil than it was consuming. It may have exported or stored this difference. b. 1993. The amount of oil consumed is the same as the amount produced. c. Estimates for 2006: production 5 3800 thousand barrels/day and consumption 5 7400 thousand barrels/day. The net difference is approximately 3600 thousand barrels per day. In the year 2006 China consumed almost double the amount of oil it produced. China needed to use its oil reserves or import this difference. 11. a. i. x 5 25 and y 5 5/7; ii. a 5 3 and b 5 0.5 b. Answers will vary. A system of two equations whose graphs are two distinct parallel lines will not have a solution. 13. a. s 1 r # 60 minutes: 8s 1 10r $ 560 calories; s $ 0 and r $ 0. b.
r 60 50 Minutes running
2x 1 3y 5 6 30. Possible answer: b 4x 1 6y 5 10
3. a. Recall that if two lines are perpendicular to each other (and neither is horizontal) then their slopes are negative reciprocals of each other. y 5 24x 1 26 and y 5 x/4 1 9
(20, 40) Intersection point
40 30 20 10 0
b.
0
y 30
y = x/4 + 9
(4, 10)
y = –4x + 26 –15
60
s 70
15. a. A(m) 5 e
y
5.
20 30 40 50 Minutes swimming
c. There are many answers, for example: s 5 10, r 5 50 minutes is in the solution set and s 5 10, r 5 40 minutes is not in the solution set. d. r 1 s # 70 minutes; 10r 1 8s $ 560 calories; s $ 0 and r $ 0. The intersection point of the boundary lines changes and the shaded area representing the solution set increases.
x 15
–15
10
8
39.99 for 0 # m # 450 39.99 1 0.45(m 2 450)for 450 , m # 2500
b. (1/2, 2)
(5, 2) x
–2
6
–8
(5, –7)
B(m) 5 e
59.99 for 0 # m # 900 59.99 1 0.40(m 2 900)for 900 , m # 2500
c. C(m) 5 e
79.99 for 0 # m # 1350 79.99 1 0.35(m 2 1350)for 1350 , m # 2500
Ch. 3
7. a. C 5 10,000 1 7x; R 5 12x, where x is the number of CDs b. C 5 R 1 10,000 1 7x 5 12x 1 x 5 2000 CDs. Breakeven point is (2000, $24000). c. If p is the new price per CD, then P(1600) 5 10,000 1 7(1600) 5 $21,200 1 p 5 $13.25 per CD. She would need to raise the price to $13.25 for each CD. d. If c 5 new fixed cost, then c 1 7(1600) 5 12(1600) 1 c 5 $8000. She would need to reduce fixed costs by $2000.
29. True
31. Possible answer: b
619
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CH. 3 Review: Putting It All Together
CH. 4 Algebra Aerobics Solutions for Section 4.1
d.
d.
Recommended Heart Rate for Exercise 250
Cost Plan B
Plan C
500
39.99 1 0.45(50) 5 $62.49
59.99
79.99
800
39.99 1 0.45(350) 5 $197.49
59.99
79.99
1000
39.99 1 0.45(550) 59.99 1 0.40(100) 5 $287.49 5 $99.99
Heart rate, beats/minute
200
Plan A
150
100 Hb Hi Ha Hmax
50
79.99
0 0
17. There have been wide swings in the real (adjusted for inflation) price of crude oil since the early 1860s. Dramatic increases occurred in the early 1860s and in the 1970s, with the maximum occurring in 1864 at over $100 a barrel. In the early 1860s the price of crude oil increased over tenfold, and in 1974 the price was over three times the price in 1972. Even with the sharp rise in oil prices from 2004 to 2006, using real dollars adjusted for inflation, the price of oil in 2006 was only about two-thirds the price of crude oil in 1860s. The nominal price or price actually paid for crude oil showed less variation than the price adjusted for inflation. It showed a huge surge during the Civil War, then remained fairly constant until the 1970s, after which the pattern more closely followed that of the price adjusted for inflation. 19. a. Cp 5 4.39N Cc 5 3.85N 1 4.00 4.99Nfor 0 , N , 10 Ci 5 e 4.79N 1 2.50for N $ 10 b. From the graphs of the three formulae given below, it can be seen that if one orders less than 7 bottles, then the Cp formula gives the best buy but if one orders 7 bottles or more, then the Cc formula gives the best buy. 30 Cp Ci Cc Cost in dollars
Ch. 4
Number of Minutes Used/Month
(7.4, 17.5) (3.5, 17.5)
0
21. a. Hb 5 132 2 0.60A b. Hi 5 154 2 0.70A c. Ha 5 187 2 0.85A
40 Age in years
60
80
Athletes are recommended to work in the zone on and between the top two lines, Ha and Hmax. e. Hb 5 120, Hi 5 140, Ha 5 170, Hmax 5 200 f. 65-year-old: I < 86%. She is just below her Hmax 5 200 ⫺ 65 5 135 beats per minute. 45-year-old: I < 77% 25-year-old: I < 69%
CHAPTER 4 Section 4.1 Algebra Aerobics 4.1 1. a. 1010: to express 10 billion as a power of 10, start with 1.0, then count the ten place values the decimal must be moved to the right, in order to produce 10 billion. b. 10214: the decimal point in 1.0 must be moved 14 place values to the left to produce 0.000 000 000 000 01. c. 105 d. 1025 2. a. b. c. d.
0.000 000 01 10,000,000,000,000 0.000 1 10,000,000
3. a. 1029 or 0.000 000 001 sec b. 103 or 1000 m c. 109 or 1,000,000,000 bytes (a byte is a term used to describe a unit of computer memory). 1m 1 5 7 ? 2 m 5 7 · 1022 or 0.07 m 100 cm 10 1m 1 5 9 ? 3 m 5 9 · 1023 or 0.009 m b. 9 mm ? 1000 mm 10 1000 m 5 5 · 103 or 5000 m c. 5 km ? 1 km
4. a. 7 cm ?
12 Number of bottles
20
5. 602,000,000,000,000,000,000,000 6. 3.84 · 108 m 7. 1 · 10–8 cm
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CH. 4 Exercises Solutions for Section 4.1
CH. 4 Algebra Aerobics Solutions for Section 4.2
8. 0.000 000 002 m 9. a. 2705,000,000 b. 20.000 040 3
d. One meaning for the slope is that for each increase of a billion years in the estimated age of Earth, time advances, on average, 11.5 years. However, it would make more sense to say that on average for every 11.5 years, estimates of Earth’s age increased by a billion years.
c. 5,320,000 d. 0.000 000 102 1
10. a. 24.3 · 107 b. 28.3 · 1026
621
c. 5.83 · 103 d. 2.41 · 1028
1970 Tilton
1 1 11. a. 5 5 5 1025 100,000 10 1 1 b. 5 9 5 1029 1,000,000,000 10
Patterson
3. a. 1 ? 10 m b. 4 ? 103 m
Clarke 1920 Barrell 12
1910
1900 1
7. a. 2.9 ? 1024 b. 6.54456 ? 102 c. 7.2 ? 105
d. 10211 e. 2.45 ? 1026 f. –1.98 ? 106
g. 24.9 ? 1023
9. a. 723,000 b. 0.000526
c. 0.001 d. 1,500,000
e. 0.000188 f. 67,800,000
d. False; 1.596 ? 109 e. True f. False; 6 ? 10212 second
b. 9
c. 1000
d. 21000
15. a. True
b. False
c. False
d. True
17. a. k x 2 1 k , 5 if x 5 5; k x 2 1 k . 5 if x 5 25 b. 2 k 3 2 x k , 10 if x 5 5; 2 k 3 2 x k . 10 if x 5 25. c. k x 2 1 k . 0 whether x 5 5 or –5 e. k 2x 2 1 k , 11 if x 5 5; k 2x 2 1 k 5 11 if x 5 25 f. k 2x k , 6 whether x 5 5 or –5 19. Note: Since the coordinates given below come from eyeball estimates, student answers may vary from those given here. a. Wolcott (0.1, 1900); Sollas (0.2, 1908); and Clarke (0.3, 1921). b. Barrell did it in 1918; the coordinates are approximately (1.3, 1918). c. Estimating two points on a hand-drawn line could result in coordinates such as (0, 1910) and (4, 1956). Then the slope, m, of the line is
5
3
4
5
6
Billions of years
Algebra Aerobics 4.2a 1. a. b. c. d. e. f. g. h. i.
105 ? 107 5 10517 5 1012 86 ? 814 5 86114 5 820 z5 ? z4 5 z514 5 z9 Cannot be simplified because bases, 5 and 6, are different. 73 1 73 5 73(1 1 1) 5 2 ? 73 5 ? 56 5 51 ? 56 5 57 34 1 7 ? 34 5 34(1 1 7) 5 8 ? 34 or 23 ? 34 23 1 24 5 23 1 23 ? 21 5 23(1 1 21) 5 3 ? 23 Cannot be simplified because bases, 2 and 5, are different.
1015 5 101527 5 108 107 86 b. 4 5 8624 5 82 8 35 c. 4 5 3524 5 31 or 3 3 d. Cannot be simplified because bases, 5 and 6, are different.
2. a.
d. k 2x k . 4 whether x 5 5 or –5
1956 2 1910 4 2 0
2
Section 4.2
13. a. 9
m5
Sollas Wolcott
5. gigabyte 5 109 bytes; terabyte 5 1012 bytes.
11. a. False; 7.56 ? 1023 b. True c. False; 4.9 ? 107 watts
Ellsworth
1930
c. 3 ? 10 m d. 6 ? 1029 m
21
Holmes
1940
d. 1023 e. 1013 f. 1028
Ch. 4
1950
Exercises for Section 4.1 1. a. 106 b. 1025 c. 109
Polkanov
1960
46 4
5 11.5
A plot of the various estimates and the graph with this slope and y-intercept are given at the top of the next column.
51 56 23 g. 1 2 6 h. 4 2 e.
3. a. b. c. d.
5 525
f.
34 5 33 31
? 34 23 34 321 ? 3422 5 22 ? 32 2 5 1 ? 2 5 2 ?3 2 3 2?3 21 3 5 4 5 4 ? 5 2124 ? 3 5 3 ? 223 2 ?1 2 1
105 ? 106 5 10516 5 1011 103 ? 1026 5 1031(26) 5 1023 10211 ? 1025 5 102111(25) 5 10216 109 ? 1024 5 1091(24) 5 105
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CH. 4 Algebra Aerobics Solutions for Section 4.2
Ch. 4
e. 106 ? 102(23) 5 10613 5 109 f. 1025 ? 102(24) 5 102514 5 1021 g. 1026 ? 1024 5 10261(24) 5 10210 4. a. b. c. d. e. f. g. h. i. 5. a. b. c. d. e. f.
6.
104(5) 5 1020 72(3) 5 76 x4(5) 5 x20 (2x) 4 5 24x4 or 16x4 (2a4 ) 3 5 23 (a4 ) 3 5 23a12 5 8a12 (22a) 3 5 (22) 3a3 5 28a3 (23x2 ) 3 5 (23) 3 (x2 ) 3 5 227x6 ((x3 ) 2 ) 4 5 (x3?2 ) 4 5 (x6 ) 4 5 x6?4 5 x24 (25y2 ) 3 5 (25) 3 (y2 ) 3 5 2125y6 (22) 3 ? x3 28x3 2x3 (22x) 3 5 3 5 3 3 3 5 (4y) 4 ?y 64y 8y3 2 (25) 5 (25)(25) 5 25 252 5 2(5)(5) 5 225 23(yz2 ) 4 5 23(y) 4 (z2 ) 4 5 23y4z8 (23yz2 ) 4 5 (23) 4 (y) 4 (z2 ) 4 5 81y4z8 (23yz2 ) 3 5 (23) 3 (y) 3 (z2 ) 3 5 227y3z6
(capacity of hard drive) 5 number of disks (capacity of disk) 4.0 ? 1010 < 0.54 ? 102 < 54 disks 7.37 ? 108
b. 33 1 53 5 27 1 125 5 152 c. 3 ? 52 5 3 ? 25 5 75 d. 23 ? 52 5 23 ? 25 5 275
Algebra Aerobics 4.2b 1. a. (0.000 297 6)(43,990,000) < (0.000 3)(40,000,000) 5 3 ? 1024 ? 4 ? 107 5 12 ? 103 5 12,000 453,897 ? 2,390,702 500,000 ? 2,000,000 < 0.004 38 0.004 (5 ? 105 ) (2 ? 106 ) 10 1011 5 5 ? 4 4 ? 1023 1023
5 2.5 ? 10 < 3 ? 10 14
14 27
000 3 3 ? 10 < 0.000 200,000 5 2 ? 105 5 1.5 ? 10212 < 2 ? 10212 d. 28,000,000 ? 7629 < 30,000,000 ? 8000 5 3 ? 107 ? 8 ? 103 5 24 ? 1010 5 2.4 ? 1011 < 2 ? 1011 e. 0.000 021 ? 391,000,000 < 0.000 02 ? 400,000,000 5 2 ? 1025 ? 4 ? 108 5 8 ? 103 5 8,000
c.
0.000 000 319 162,000
2. a. (3.0 ? 10 )(4.0 ? 10 ) 5 12 ? 10 5 1.2 ? 106 5 1,200,000 3
b. c.
d. (4.0 ? 102 ) 3 (2.0 ? 103 ) 2 5 (43 ? 106 )(4 ? 106 ) 5 44 ? 1012 5 256 ? 1012 5 2.56 ? 1014 5 256,000,000,000,000 3. If we use 3.14 to approximate p: a. Surface area of Jupiter 5 4pr2 < 4p(7.14 ? 104 km) 2 5 4p(7.14) 2 (104 ) 2 km2 < 4(3.14)(50.98)108 km2 < 640 ? 108 km2 < 6.4 ? 102 ? 108 km2 < 6.4 ? 1010 km2 5 43pr3 < (1.3) ? (3.14)(7.14 ? 104 km) 3 < (4.08)(7.14) 3 (104 ) 3 km3 < (4.08)(364)1012 km3 < 1486 ? 1012 km3 < 1.486 ? 103 ? 1012 km3 < 1.486 ? 1015 km3
b. Volume of Jupiter
4. If only 3/7 of the farmable land is used, the people/sq. mi. of used farmland is: 6.6 ? 109 people (3/7)12 ? 106 sq.mi.
6.6 ? 109 people 5 Q73R 12 ? 106 sq.mi.
< 1.283 ? 103 or 1283 people/sq.mi. For fractions . 0, if the denominator is decreased, the value of that fraction is increased. So, one expects this ratio to be larger than the ratio of people to farmable land.
Exercises for Section 4.2
7. a. (3 1 5) 3 5 83 5 512
b.
CH. 4 Exercises Solutions for Section 4.2
2
(5.0 ? 102 ) 2 ? 104 5 25 5 2.5 ? 103 25 ? 102 2.0 ? 105 20 ? 104 5 5.0 ? 103 5 4 5.0 ? 103
5
1.0 ? 102 5 100 ? 101 5 40
107 1.1 ? 104 2 ? 103 x15 x50 47 1 52—this expression cannot be simplified without multiplying out the values and adding them together. i. 321 or 1/3 g. z5 h. 1 or as is j. 411
1. a. b. c. d. e. f.
3. a. b. c. d.
16a4 22a4 2x15 28a3b6
e. 32x20 f. 18x6 g. 2500a20 h. as is—nothing is simpler
5. a. 2 A58B 5 225 64 2
3 3 33x9 27x9 b. Q3x 5y2 R 5 53y6 5 125y6 5 4 10,000x20 104x20 x20 c. Q210x 2b2 R 5 42 b8 5 16b8 5 625 b8 5 3 9 d. A2x x2 B 5 2x
7. a. b. c. d.
(2 ? 106) ? (4 ? 103) 5 8 ? 109 (1.4 ? 106) 4 (7 ? 103) 5 0.2 ? 103 5 2 ? 102 (5 ? 1010) ? (6 ? 1013) 5 30 ? 1023 5 3 ? 1024 (2.5 ? 1012) 4 (5 ? 105) 5 0.5 ? 107 5 5 ? 106
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CH. 4 Exercises Solutions for Section 4.2
9. a. x13y4 b. 5x4y 11. a. b. c. d.
c. 28x9y9 d. 16x10y8
CH. 4 Algebra Aerobics Solutions for Section 4.3
per person than the United Kingdom, but the United Kingdom produces 4 times more nuclear energy than the United States per square mile.
e. 81x8y20 9 4 f. 25 x
109/106 5 103 5 1000 1000/10 5 102 5 100 1000/0.001 5 106 5 1,000,000 1026/1029 5 103 5 1000
15. (1.5 ? 104 beverages/sec)(8.64 ? 104 sec/day) 5 1.296 ? 109 beverages/day, or over a billion Coca-Cola beverages were consumed each day worldwide in 2005. 17. a. If a is positive, then 2a is negative. If n is even, then (2a)n is positive; but if n is odd, then (2a)n is negative. If a is negative, then 2a is positive, then (2a)n is positive whether n is even or odd. b. This is answered in part (a). 2 3 2 2 3 2 4n6 2 4 19. a. Qmmnn R 5 (mn2 ) 2 5 m2n4 and Qmmnn R 5 m m2n2 5 m n 2 3 4 b. Q2aabb2 R 5 (2ab) 4 5 16a4b4 and 8 b12 2 3 4 4 4 Q2aabb2 R 5 16a a4 b8 5 16a b
3 4 16a12 24a12 21. Q2a 5b2 R 5 54b8 5 625b8
23. Two cases are distinguished: If n 5 0, then (ab)0 5 1 and a0 ? b0 5 1 ? 1 5 1 If n . 0, then an 5 a ? c ? a (n factors) and bn 5 b ? c ? b (n factors) and thus an ? bn 5 (a ? c ? a) ? (b ? c ? b) 5 (ab) ? c ? (ab) (n factors), after rearrangement, and this product is what is meant by (ab)n when n . 0. 25. a. Generated in the United Kingdom: (81 terawatt-hours)/(60.6 ? 106 persons) < 1.34 ? 1026 terawatt-hours/person. (81 terawatt-hours)/(94,525 miles2) < 8.57 ? 1024 terawatt-hours/mile2. b. Generated in the United States: (780 terawatt-hours)/(3.0 ? 108 persons) < 2.60 ? 1026 terawatt-hours/person. (780 terawatt-hours)/(3,675,031 miles2) < 2.12 ? 1024 terawatt-hours/mile2. c. The United Kingdom produces about four times more terawatt-hours per square mile than the United States, because UK/US (8.57 ? 1024 terawatt-hours/mile2)/ (2.12 ? 1024 terawatt-hours/mile2) < 4.04. d. Answers will vary. The United States produces (uses) (2.60 ? 1026 terawatt-hours/person)/(1.34 ? 1026 terawatt-hours/person) < 1.9 times more nuclear energy
Section 4.3 Algebra Aerobics 4.3 1. a. 10527 5 1022 5 101 2 b. 1162(24) 5 11614 5 1110 c. 3252(24) 5 32514 5 321 5 1/3 d. Cannot be simplified: different bases, 5 and 6. e. 7323 5 70 5 1 f. a221(23) 5 a25 5 a15 g. 34 ? 33 5 37 h. (22 ? 3) ? (26) ? (24 ? 3) 5 22 ? 26 ? 24 ? 3 ? 3 5 212 ? 32 2. Time for a TV signal to travel across the United States 5 (time to travel 1 kilometer) ? (number of kilometers) 5 (3.3 ? 1026) sec/km ? (4.3 ? 103) km 5 (3.3 ? 4.3) ? (1026 ? 103) sec 5 14 ? 1023 sec 5 1.4 ? 10 ? 1023 sec 5 1.4 ? 1022 sec or 0.014 sec So it would take less than two-hundredths of a second for the signal to cross the United States. 3. a. x22 (x5 1 x26 ) 5 x22 (x5 ) 1 x22 (x26 ) 5 x3 1 x28 5 x3 1 x18 b. 2a2 (b2 2 3ab 1 5a2 ) 5 b2 (2a2 ) 2 3ab(2a2 ) 1 5a2 (2a2 ) 5 2a2b2 1 3a112b 2 5a212 5 2a2b2 1 3a3b 2 5a4 4. a. 10(4)(25) 5 10220 5 10120 b. 7(–2)(–3) 5 76 c. (2a13 )2 5 4a1 6 22 x2 d. A8x B 5 A8x B 2 5 64 2 e. 221x2 5 x2 f. 2x2
24
g. Q2y3 2R
2y2 4
24y8
16y8
5 Q 3 R 5 34 5 81
h. 3 ? (2y2 ) 4 5 3 ? 24y8 5 48y8 t23 (1)
5. a. t212 5 t232(212) 5 t9 v23w7 b. v26 5 v232(26)w72(210) 5 v3w17 w210 728x21y2
c. 725x1 y3 5 7(28)2(25)x(21)21y223 d.
5 723x22y21 5 73x12y 1 2 22 6 5 a 51?a15b2bc26c 5 5221a121b(22)22c62(26)
a(5b21c3 ) 2 5ab2c26
12
5 5b24c12 5 5cb4
Ch. 4
13. a. Japan’s population density 5 (1.275 ? 108 people)/ (1.525 ? 105 miles2 ) < 836 people/mile2. b. The U.S. population density 5 (3.0 ? 108 people)/ (3.62 ? 106 miles2 ) < 83 people/mile2. c. Japan’s population density is (836)/(83) 5 10 times larger, or one order of magnitude larger than that of the United States.
623
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CH. 4 Exercises Solutions for Section 4.3
CH. 4 Algebra Aerobics Solutions for Section 4.4
4 qt
Ch. 4
Exercises for Section 4.3 26
1. a. 101 b. 1025 5 101 5
c. 10 5 d. 105
3. a. 221x4 5 4x1 4 b. xy12 c. x31y2
d. (x 1 y)11 7 e. ab c6
5. a. 4.6 ? 1010 b. 4.07 ? 103 c. 1.525 ? 1011
d. 5.1669 ? 1028 e. 2.3833 ? 10158 f. 2.601 ? 10221
7. a. b. c. d. e. f.
ml f. 1 gal ? 1 gal ? 946 1 qt 5 3784 ml
1 106
5280 ft mile 1 hour g. 11 hour ? 3600 sec ? 1 mile < 1.47 ft/sec
4. 1 km 5 1000 m, so the conversion factor from meters to 1 km kilometers is 10 3 m . Hence, 1 km 3.84 ? 108 m ? 10 5 3.84 ? 105 km 3 m
5. 1 km 5 1000 m, so the conversion factor from kilometers to 3m meters is 10 1 km . Hence 3m 8 3 11 7.8 ? 108 km ? 10 1 km 5 7.8 ? 10 10 5 7.8 ? 10 m
1 ? 1021 ? 1025 5 1 ? 1026 5 ? 1025/(5 ? 104) 5 1 ? 1029 3/(6 ? 1023) 5 0.5 ? 103 5 5 ? 102 8 ? 103/(8 ? 1024) 5 1 ? 107 6.4 ? 1023/(8 ? 103) 5 0.8 ? 1026 5 8 ? 1027 5 ? 106 ? 4 ? 104 5 20 ? 1010 5 2 ? 1011
9. a. 26 b. 26
6. 1 km 5 0.62 mi, so the conversion factor to convert from km mi to mi is 0.62 1 km . Hence mi 12 9.46 ? 1012 km ? 0.62 1 km 5 5.87 ? 10 mi
which is close to 5.88 ? 1012 mi. 7. 1 m 5 100 cm, so the conversion factor for converting from 1m cm to m is 100 cm .
c. 4 d. 25
y3
Hence 1 Å 5 1028 cm ? 1012 mcm 5 10210 m.
5xy9
11. a. 223x9
c. 3
8. 1 km 5 0.62 mi, so the conversion factor for converting from mi km to miles is 0.62 1 km .
6 24 d. 3 xz6
b. 221x4
13. 200 times longer or 1.6 ? 1022 second y 2 x2
y
15. a. Qx12 2 1y R ? sxy 2 d 5 Q x 2y R ? sxy 2 d 5 sy 2 x 2 d ? QxR y 2 2 x 2y
5 x (This one is difficult because you need to find a common denominator.) b.
x 4y 6 2 5
5
x 4y 6 25
17. 5.23 ? 1023, 0.00523 and 5.23/1000 p 19. x1p 5 A1x B for any p. Here we have p 5 2n; thus
1 x
2n
2n 5 Q1x R 5 sx 21 d 2n 5 x s21d?s2nd 5 x n
21. (9 ? 109)/(35 ? 106) < 257 kernels per pound.
Section 4.4 Algebra Aerobics 4.4 1 qt
oz 1. 2 l ? 0.946 l ? 32 1 qt < 67.65 oz 1 in 2. 120 cm ? 2.54 cm < 47.24 in. cm 3. a. 12 in. ? 2.54 1 in. 5 30.48 cm m b. 100 yd ? 0.914 1 yd 5 91.4 m 1 lb c. 20 kg ? 0.4536 kg < 44.09 lb 1 year $40,000 < $19.23 per hour d. 1 year ? 52 weeks ? 1 workweek 40 hr 60 min 60 sec e. 241 hours day ? 1 hour ? 1 min 5 86,400 sec/day
mi Hence, 218 km 5 218 km ? 0.62 1 km 5 135.16 mi or ,135 mi.
9. If a dollar bill is 6 in long, then two dollar bills/12 in 5 2 dollars/ft. The number of dollars needed to reach from Earth to the sun is: ft 93,000,000 mi ? 5,280mi ? 2 dollars ft 713 < 9.3 ? 5.3 ? 2 ? 10 dollars
< 98.6 ? 1010 < 9.9 ? 1011 or 990,000,000,000 dollar bills, almost a trillion dollars. 2,560 mi
4096 km 4096 km km 1 hr 10. 4.2 hrs ? 1.6 1 mi 5 4.2 hrs 5 4.2 hrs ? 60 min km 5 4096 252 min (or approximately 16.25 km/min) mm 11. 5 m ? 1016 m ? 1000 5 5 ? 1023 mm 1 1 spore is 1m m 5 ? 1023 mm in diameter. To compare the 1 mm diameter of a pencil to the diameter of a spore, we divide: 1 mm 5 200. The diameter of a pencil is approximately 5 ? 1023 mm
200 times larger than the diameter of a spore. ? 1012 km 13 12. a. 4.3 light-year ? 9.46 1 light-year < 4.07 ? 10 km; ? 1012 miles 13 4.3 light-year ? 5.88 1 light-year < 2.53 ? 10 miles ? 1012 km 1000 m b. 108 light-years ? 9.46 1 light-year ? 1 km
5 9.46 ? 1023 m ? 1012 miles 5.28 ? 103 ft c. 1.6 ? 103 light-year ? 5.881 light-year ? 1 mile 18 19 < 49.67 ? 10 5 4.97 ? 10 ft
8 ? 10210 m 5 5 ? 1021 ? 10210 5 5.0 ? 10211 m 13. a. 0.5 A 8 1A
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CH. 4 Exercises Solutions for Section 4.4
CH. 4 Algebra Aerobics Solutions for Section 4.5
625
ft 14. 364 Smoot ? 1 5.6 Smoot 5 2038.4 ft
21. a. Volume 5 p ? 92 ? 4 cu. ft 5 324p cu. ft < 1018 cu. ft, and 1018 cu. ft ? 1728 (cu. in.)/(cu. ft) 5 1,759,104 cu. in., and since there are 231 gal per cubic inch, there are 1,759,104/231 < 7615 gal. b. 7615 ? 24 gal/(2500 gal/hr) < 3.05 hr c. 7615 ? 24 gal/(10,000 gal/lb) < 0.761 lb
Exercises for Section 4.4
23. a. Bush’s BMI < 72 in./39.37 in./m2 < 1.832m2 < 26.33 kg/m2.
8 ? 10210 m 5 1025 m b. 105 A 8 1A
8 ? 10210 m 5 10215 m c. 1025 A 8 1A
m b. (3 ft) ? 0.305 1 ft < 0.92 m
c. (5 lb) ?
0.4536 kg 1 lb
< 2.27 kg
cm d. (12 in.) ? 2.54 1 in. 5 30.48 cm m e. (60 ft) ? 0.305 1 ft 5 18.3 m liters f. s4 qtd ? 0.946 < 3.78 liters 1 qt
3. a. Student estimates will vary. There are approximately 30.5 cm per foot. b. From conversation table, 1 ft 5 0.305 m or approximately 30% of a meter. 5. Converting the units to decimeters we get: (100 km) ? (250 m) ? (25 m) 5 106 ? (2.5 ? 103) ? (2.5 ? 102) cubic decimeters 5 6.25 ? 1011 cubic decimeters 5 (6.25 ? 1011 liters) ? (103 droplets/liter) 5 6.25 ? 1014 droplets 7. 1 m < 3.28 ft and thus 9.8 m/sec < 9.8 ? 3.28 5 32.144 ft/sec 9. a. 186,000 miles per second or one hundred and eighty-six thousand miles per second. ? 105 mi 1609 m 60 sec 60 min 24 hr 365 days ? mi ? min ? hr ? day ? yr b. 1.86 sec < 9.438 ? 1015 m/yr
11. 500 seconds; 500 seconds or 8 minutes and 20 seconds from now. 13. U.S. barrel of oil 5 42 gal 5 42 ? 4 qt 5 168 qt < 168 ? 0.946 liter 5 158.928 liters. Thus the British barrel of oil is larger than the U.S. barrel of oil by approximately 4.727 liters. 15. a. 1 acre 5 43,560 ft2 and thus is "43,560 < 208.71 ft on each side. b. 150x 5 1.5 ? 43,560 or x 5 435.6 ft.; perimeter 5 2 ? (150 1 435.6) 5 1171.2 feet or 390.4 yards. c. 1 hectometer 5 100 meters; 1 square hectometer 5 10,000 square meters. 1 meter < 3.28 ft. Thus 1 sq. meter < 10.7584 square ft. Thus 1 acre 5 43,560 sq. ft 5 43,560/10.7584 < 4048.929 square meters < 0.4049 square hectometer. d. 1 hectare 5 100 acres 5 4,356,000 ft2/[(5280)2 ft2/mi2] 5 0.15625 sq. mi. Thus 1 acre 5 0.0015625 sq. mile. Thus in a square mile there are 1/0.0015625 5 640 acres. This is 6.4 hectares. 17. Light travels 186,000 ? 5280 ft/sec and thus it travels 186000 ? 5280/109 < 0.98208 ft per nanosecond. 19.
15,000 bev second
1 sec 60 min 24 hr 365 day ? 60 1 min ? 1 hr ? 1 day ? 1 year ? 6.45 ? 109 persons
< 73.34 beverages/year/person
88.18 kg
< 26.33 kg/m2. He is slightly overweight. b. BMI in pounds and inches is BMI 5 (pounds/2.2)/ (inches/39.37)2 5 [(39.7)2/2.2](lb/in2), or approximately 704.5 lb/in2. Thus for Bush we get 704.5 ? 194/722 < 26.36, which is about the same. c. Answers will vary, but note that 0.45 < 1/2.2 and 0.254 < 1/39.37. d. A kilogram, more precisely, is approximately 2.2046 lb, and 39.372/2.2046 < 703.07. Thus Kigner is correct. 25. a. 10235 m 5 10235 ? 1023 5 10238 km b. 10235 m < 10235 ? 0.00062 5 6.2 ? 10239 mile c. x 5 10235/3 ? 108 < 3.33 ? 10244 sec
Section 4.5 Algebra Aerobics 4.5a 1. a. "81 5 9 b. "144 5 12 c. "36 5 6 d. 2 "49 5 27 e. not a real number 2. a. "9x 5 3x1/2 b.
x2
325 5
"x2 "25
5 5x
c. "36x2 5 "36 "x2 5 6x 9y2
5
d.
325x
e.
3x
f.
3169 5
4
49 2
4a
"9y2
3y
"25x4
5 5x2
5 "492 5 7x "x
"4a "169
1/2 "a 5 2 13 5 2a13
3. a. S 5 "30 ? 60 5 "1800 < 42 mph b. S 5 "30 ? 200 5 "6000 < 77 mph 4. a. "25 , "29 , "36 1 5 , "29 , 6; 5 and 6 b. "81 , "92 , "100 1 9 , "92 , 10; 9 and 10 c. "100 , "117 , "121 1 10 , "117 , 11; 10 and 11. d. "64 , "79 , "81 1 8 , "79 , 9; 8 and 9. e. "36 , "39 , "49 1 6 , "39 , 7; 6 and 7.
Ch. 4
km 1. a. s50 milesd ? 1.609 1 mile 5 80.45 km
(194 lb) /(2.2 lb/kg)
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CH. 4 Algebra Aerobics Solutions for Section 4.5
CH. 4 Exercises Solutions for Section 4.5
b. 51/2 ? 51/4 5 51/211/4 5 53/4
3 5. a. " 27 5 3
c. 31/2 ? 91/3 5 31/2 ? (32 ) 1/3 5 31/2 ? 32/3 5 37/6
4 b. "16 5 2
c.
1 3 "8
d. x1/4 ? x1/3 5 x7/12
5 12
e. x3/4 ? x1/2 5 x5/4
5 d. "32 5 2
f. x1/3 ? y2/3 ? x1/2 ? y1/2 5 x5/6 ? y7/6
e. 2711/3 5 31 5 13 "27 f. 2511/2 5 1 5 15
1/2
2. a. 221/3 5 21/221/3 5 21/6
Ch. 4
"25
g.
8 21/3 A27 B
5 5
h.
1
1 b. 221/4 5 2121/4 5 23/4
1/3 A27 8B
27 8
3 3
3
"27 3 "8
5
5
1/4 1 c. 551/3 5 51/421/3 5 521/12 or 51/12
3 2
1/2 d. xx3/4 5 x2 orx11/4 1 4
1
316 5 4
x1/3 ? y2/3
6. a. (227)1/3 5 23 since (23)3 5 227
3. a. c 5 17.1(0.25) 3/8 cm
b. There is no real number solution to the fourth root of a negative number, since a negative or positive number raised to the fourth power is always positive.
c < 17.1(0.59) cm 1 c < 10.2 cm b. c 5 17.1(25) 3/8 cm c < 17.1(3.34) cm 1 c < 57.2 cm
c. (21000)1/3 5 210 since (210)3 5 21000. d. 2 "16 5 22, since 2 5 16 1 2 "2 5 22. 4
4
4
4
4. a. "20x2 5 "22 ? 5 ? x2 5 2x "5
3 e. "28 5 22, since (22) 3 5 28.
b. "75a3 5 "52 ? 3 ? a2 ? a 5 5a "3a
f. "2500 5 50, since 502 5 2500.
3 3 3 c. "16x3 y4 5 "23 ? 2 ? x3 ? y3 ? y 5 2xy "2y
3V 5 r3 1 7. a. V 5 43pr3 1 34V 5 pr3 1 4p
r5
3 3
y1/6
e. x1/2 ? y1/2 5 x21/6 ? y1/6orx1/6
3 ? 2 feet3 4p
5
3 feet3 2p
3 3
< 0.78 feet
4
d.
< "0.478 feet 3
3
"32x4y6 4
#81x y
8 5
5
We can express that with a more meaningful figure if we convert it into inches. Since 1 ft 5 12 in., we can use a conversion factor of 1 5 (12 in.)/(1 foot). The radius of the balloon is:
4
"24 ? 2 ? x4 ? y4 ? y2
5
4
#34 ? (x2 ) 4 ? y4 ? y
4
2xy ? "2y2 4
3x2y ? "y
2 2 4 2y 2 4 5 3x 3 y 5 3x "2y
5. a. "4a2 b6 5 (22 a2 b6 ) 1/2 5 22/2a2/2b6/2 5 2ab3 4 b. "16x4 y6 5 (24 x4 y6 ) 1/4 5 24/4 x4/4 y6/4
5 2xy3/2 (0.78 ft) A121 ftin.B 5 9.36 in. or < 9.4 in.
3 c. "8.0 ? 1029 5 (23 ? 1029 ) 1/3 5 23/3 ? 1029/3 5 2 ? 1023 5 0.002
8. a. "25 5 5
d. "8a24 5 (23a2 4 ) 1/2 5 23/2a22
b. 2 "49 5 27 c. 25
Exercises for Section 4.5
d. "45 2 3 "125 5 "9 ? 5 2 3 "25 ? 5 9. a. "36 5 (62 ) 1/2 5 62/2 5 6
3. a.
3 b. "27x6 5 (33 x6 ) 1/3 5 33/3 x6/3 5 3x2
c. "81a b 4
4
12
5 (3 a b ) 4
4
V 1r 5 10. a. r2 5 ph
3ph
3V b. r2 5 ph 1r 5
3ph
12 1/4
3V
S
36
1. a. 2
?2
b
12/4
52
1/211/3
c.
c
52
49x
By
5 3ab
2
5. a. 5 "5a
b.
7. a. 6 ? 103 9. a. 0.1
1 2xy 3
d.
5 7 "x 3
x 4y 2 6 100z B
y
5
x 2y 10z 3
d. 8x2y2 "y
b. 2 ? 103
c. 5 ? 105
d. 1.0 ? 1022
b. 1/5
c. 4/3
d. 0.1
c. "2 < 1.414 d. not defined
b. "3 1 2 , 5 2
6
c. 2xy "2x
13. a. "3 1 "7 . "3 1 7 5/6
e. 210 f. 210
3
11. a. 4 b. 24
Algebra Aerobics 4.5b 1/3
a
4/4
2 5 ab3
b. "36x y 5 6x "y
V
d. a 5 "c2 2 b2
1/2
53
4/4
a 2b 4 6 B c
4
3 c. S 5 " V
e. x 5
c. 1/10 5 0.1 d. 21/10 5 20.1
1. a. 10 b. 210
5 3 "5 2 15 "5 5 212 "5
2
e. 2 "2 < 2.8284 f. 1 c. "52 2 42 . 2
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CH. 4 Algebra Aerobics Solutions for Section 4.6
15. a. 9
b. 81
1 c. 125
1 d. 27
17. V 5 (4/3)pr3 and thus if V 5 4, then r3 5 3/p. If one uses 3 as a crude estimate of p, then r is approximately 1 foot. (A more precise estimate, from using a calculator, is 0.985 ft.)
T 5 2p 32.67 32 < 0.578p sec < 1.816 sec. b. Solving for L, we get: L, 5 8T2/(p2). So if T is 2 seconds, then L 5 8(2)2/(p2) 5 32/p2 < 3.24 feet. b. 6345 lb
Section 4.6 Algebra Aerobics 4.6 6.9 1 10 5 101, the Armenian earthquake had tremors 105.9
about one order of magnitude larger than those in Los Angeles. magnitude Chile earthquake
2. Since magnitude 2003 Little Rock earthquake 9.5 1 10 5 103, the maximum tremor size of the Little 106.5
Rock earthquake of 2003 was 1000 times smaller (or three orders of magnitude smaller) than the maximum tremor size of the Chile earthquake. your salary
3. Since my salary of $100,000 5 101 1 your salary is $1,000,000; Henry’s salary my salary $100,000
5 1022 1
Henry’s salary my salary $100,000
1 1 Henry’s salary is $1000. 5 10 2 21 5 10 5 102129 5 1012, 109
the radius of the Milky Way is 12 orders of magnitude larger than the radius of the sun, or equivalently, the radius of the sun is 12 orders of magnitude smaller than the radius of the Milky Way. radius of a proton
c. 7
d. 6
3. a. 52.61 is one order of magnitude larger than 5.261. b. 5.261 is three orders of magnitude smaller than 5261. c. 526.1 is four orders of magnitude smaller than 5.261 ? 106. 5. a. (50 ? 103) ? (3 ? 108) ? (365 ? 24 ? 602) < 4.7304 ? 1020 5 number of meters in 50,000 light-years. Hence, the Milky Way is (4.7302 ? 1020)/(0.5 ? 1024) 5 9.4608 ? 1024 times larger than the first life form. Thus the Milky Way is nearly 25 orders of magnitude larger than the first living organism on Earth. b. (100 ? 106)/(100 ? 103) 5 103; thus Pleiades is three orders of magnitude older than Homo sapiens. 7. The raindrop is 1024 times heavier and this is an order of magnitude of 24.
magnitude 1988 Armenia earthquake magnitude 1987 LA earthquake
radius of the Milky Way radius of the sun
b. 5
215
2152(211) b. Since radius of the hydrogen atom 5 10 5 1024, 10211 510 the radius of the proton is four orders of magnitude smaller than the radius of the hydrogen atom, or equivalently, the radius of the hydrogen atom is four orders of magnitude larger than the radius of a proton.
5. a. $4 million since $400,000 ? 101 5 $4 million b. $65,000 since $650,000 ? 1021 5 $65,000 6. 1 km 5 1000 m 5 103 meters, so three orders of magnitude; m 1000 mm 1 km ? 1000 5 106 mm, so six orders of magnitude 1m 1 km ?
9. a. 2.0 ? 1025 in. b. 2.0 ? 1025/39 < 5.128 ? 1027 m c. The name is a bit off — by two orders of magnitude (looking at meters). d. The tweezers would have to be made able to grasp things two orders of magnitude smaller than they can grasp now. i. Radius of the moon 5 1,758,288.293 meters, or about 1.76 ? 106 meters. ii. Radius of Earth 5 6,400,000 meters, or about 6.4 ? 106 meters. iii. Radius of the sun 5 695,414,634.100 meters, or about 6.95 ? 108 meters. b. i. The surface area of a sphere is 4pr 2, and thus the ratio of the surface area of Earth to that of the moon is the same as the ratio of the squares of their radii, or [6.4 ? 106/(1.76 ? 106)]2 < 13.22. Thus the surface area of Earth is one order of magnitude bigger. ii. The volume of a sphere is (4/3)pr3, and thus the ratio of the volume of the sun to that of moon is the same as the ratio of the cubes of their radii, or [6.95 ? 108/(1.76 ? 106)]3 < 6.16 ? 107. Thus the volume of the sun is seven orders of magnitude bigger than the volume of the moon.
11. a.
13. Scale (a) is additive because the distances are equally spaced, and scale (b) is multiplicative or logarithmic because the distances are spaced like the logarithms of numbers. 15. a. Being 5 orders of magnitude larger than the first atoms means that it is 1025 m and thus would appear at 25 on the log scale. b. Being 20 orders of magnitude smaller than the radius of the sun means that it is 10211 m and thus would appear at 211 on the log scale.
Ch. 4
21. a. If L < 2.67 ft, then
4. a. Since
7. a. 1,000,000,000 m 5 109 m 1 plot at 109 m b. 0.000 000 7 m 5 7 ? 1027 m < 10 ? 1027 m 5 1026 m 1 plot at 1026 m c. Plot at 107 ? 1022 5 105 m
1. a. 12
b. area 5 16 "3 cm2 < 27.7 cm2
1. Since
627
Exercises for Section 4.6
19. a. height 5 4 "3 cm < 6.9 cm
23. a. 3673 lb
CH. 4 Exercises Solutions for Section 4.6
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CH. 4 Algebra Aerobics Solutions for Section 4.7
17. a. b. c. d.
1 watt 5 100 watts 10 kilowatts 5 104 watts 100 billion kilowatts 5 102 ? 109 ? 103 5 1014 watts 1000 terawatts 5 1015 watts
a
100
Ch. 4
10:22 AM
b
101
102
103
104
105
106
107
108
109
1010
1011
1012
1013
CH. 4 Exercises Solutions for Section 4.7
then combine powers 3,000,000 < 106.48 and rewrite as a logarithm log 3,000,000 < 6.48 A calculator gives log 3,000,000 < 6.477121255.
c
d
1014
1015
Section 4.7 Algebra Aerobics 4.7a 1. a. b. c. d. e. f. g.
Since 10,000,000 5 107, log 10,000,000 5 7. Since 0.000 000 1 5 1027, log 0.000 000 1 5 27. Since 10,000 5 104, log 10,000 5 4. Since 0.0001 5 1024, log 0.0001 5 24. Since 1000 5 103, log 1000 5 3. Since 0.001 5 1023, log 0.001 5 23. Since 1 5 100, log 1 5 0. c. 10 5 101 d. 0.01 5 1022
2. a. 100,000 5 105 b. 0.000 000 01 5 1028 3. a. b. c. d. e.
log N 5 3 1 N 5 103 5 1000 log N 5 21 1 N 5 1021 5 0.1 log N 5 6 1 N 5 106 5 1,000,000 log N 5 0 1 N 5 100 5 1 log N 5 22 1 N 5 1022 5 0.01
4. a. b. c. d. e. f.
log 1000 5 c, c 5 3 log 0.001 5 c, c 5 23 log 100,000 5 c, c 5 5 log 0.000 01 5 c, c 5 25 log 1,000,000 5 c, c 5 6 log 0.000 001 5 c, c 5 26
5. a. b. c. d.
x 2 3 5 2, so x 5 5 2x 2 1 5 4, so x 5 5/2 101 5 x 2 2, so x 5 12 1021 5 5x, so x 5 0.1 5 5 0.02
b. Write 0.006 in scientific notation 0.006 5 6 ? 1023 and substitute 100.78 for 6 0.006 < 100.78 ? 1023 then combine powers 0.006 < 100.7823 < 1022.22 and rewrite as a logarithm log 0.006 5 22.22 A calculator gives log 0.006 < 22.22184875. 3. a. b. c. d.
4. a. You want to estimate a number x such that 104.125 5 x. Since 104 5 10,000, an estimate for x is 12,000. With a calculator, x < 13,335. b. You want to estimate a number x such that 105.125 5 x. Since 105 5 100,000, an estimate for x is 120,000. With a calculator, x < 133,352. c. You want to estimate a number x such that 102.125 5 x. Since 102 5 100, an estimate for x is 120. With a calculator, x < 133. 5. a. You want to estimate a value x such that log 250 5 x. 102 , 250 , 103 1 2 , x , 3. With a calculator x < 2.398. b. You want to estimate a value x such that log 250,000 5 x. 105 , 250,000 , 106 1 5 , x , 6. With a calculator, x < 5.398. c. You want to estimate a value x such that log 0.075 5 x. 1022 , 0.075 , 1021 1 22 , x , 21. With a calculator, x < 21.125. d. You want to estimate a value x such that log 0.000 075 5 x. 1025 , 0.000 075 , 1024 1 25 , x , 24. With a calculator, x < 24.125. 6. a. log 57 < 1.756 1 57 < 101.756 b. log 182 < 2.26 1 182 < 102.26 c. log 25,000 < 4.398 1 25,000 < 104.398 d. log 7.2 ? 109 < 9.857 1 7.2 ? 109 < 109.857
Algebra Aerobics 4.7b
1.756
10
2.26
10
10
1. a. log 3 is the number c such that 10c 5 3. With a calculator you can find that: 100.4 < 2.512 and 100.5 < 3.162 1 0.4 , log 3 , 0.5. A calculator gives log 3 < 0.477. b. log 6 is the number c such that 10c 5 6. With a calculator you can find that: 100.7 < 5.012 and 100.8 < 6.310 1 0.7 , log 6 , 0.8. A calculator gives log 6 < 0.778. c. log 6.37 is the number c such that 10c 5 6.37. Since 100.8 < 6.310 and 100.9 < 7.943, then 0.8 , log 6.37 , 0.9. A calculator gives log 6.37 < 0.804. 2. a. Write 3,000,000 in scientific notation and substitute 100.48 for 3
0.000 000 7 5 10log 0.0000007 < 1026.1549 780,000,000 5 10log 780,000,000 < 108.892 0.0042 5 10log 0.0042 < 1022.3768 5,400,000,000 5 10log(5,400,000,000) 5 109.732
3,000,000 5 3 ? 106 3,000,000 < 100.48 ? 106
1
x 10 log x
1
102 2 1.756
3
10
3
4
2.26
4.398
4
10
9.857
10 10
5
10
5
6
6
7
8
9
10
10
10
7
8
9
10 9.857
4.398
Exercises for Section 4.7 1. a. log(10,000) 5 4 b. log(0.01) 5 22
c. log(1) 5 0 d. log(0.00001) 5 25
3. Since log(375) < 2.574, we have that 375 < 102.574. 5. a. log10(100) 5 2 b. log10(10,000,000) 5 7
10
10
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CH. 4 Exercises Solutions for Section 4.7
c. d. e. f.
CH. 4 Review: Putting It All Together
log10(0.001) 5 23 101 5 10 104 5 10,000 1024 5 0.0001 x 5 101.255 < 17.9887 < 18 x 5 103.51 < 3235.94 < 3236 x 5 104.23 < 16,982.44 < 16,982 107.65 < 44,668,359.22 < 44,668,359
9. a. b. c. d.
x 5 log(12,500) < 4.097 x 5 log(3,526,000) < 6.547 x 5 log(597) < 2.776 x 5 log(756,821) < 5.879
Ch. 4: Check Your Understanding
11. a. b. c. d.
x 5 100.82 < 6.607 x 5 log(0.012) < 21.921 x 5 100.33 < 2.138 x 5 log(0.25) < 20.602
13. a. b. c. d.
x – 5 5 3 or x 5 8 2x 1 10 5 100 or 2x 5 90 or x 5 45 3x – 1 5 24 or 3x 5 23 or x 5 21 500 – 25x 5 1000 or –25x 5 500 or x 5 220
15. a. 1 , log(11) , 2 since 1 , 11 , 100 and log(11) < 1.041. b. 4 , log(12,000) , 5 since 10,000 , 12,000 , 100,000 and log(12,000) < 4.079. c. 21 , log(0.125) , 0 since 1021 5 0.1 , 0.125 , 1 5 100 and log(0.125) < 20.903. 17. a. b. c. d.
Multiplying by 1023 Multiplying by "10 Multiplying by 102 Multiplying by 1010
pH 5 2log(1027) 5 7 pH 5 2log(1.4 ? 1023) 5 3 2 log(1.4) < 2.85 11.5 5 2log([H1]); thus [H1] 5 10211.5 < 3.16 ? 10212 A higher pH means a lower hydrogen ion concentration, and one can see this because in plotting pH values one uses the numbers on the top of the given scale. e. Pure water is neutral, orange juice is acidic, and ammonia is basic. In plotting, one uses the top numbers to find the right spots. Thus water would be placed at the 7 mark, orange juice 85% of the way between the 2 and 3 marks, and ammonia halfway between the 11 and 12 marks.
19. a. b. c. d.
21.
23. We measure all in seconds, the smallest time unit. a. 1 heartbeat < 100 sec b. 10 minutes 5 600 sec < 102.8 sec c. 7 days 5 7 ? 24 ? 3600 sec < 105.8 sec d. 1 year 5 365 ? 24 ? 3600 sec < 107.5 sec e. 38,000 years 5 38,000 ? 365 ? 24 ? 3600 < 1012.1 sec f. 2.2 ? 106 ? 365 ? 24 ? 3600 sec < 1013.8 sec
1. True
8. True
15. True
22. True
2. False
9. False
16. True
23. True
3. False
10. False
17. False
24. False
4. False
11. False
18. False
25. False
5. True
12. False
19. False
26. True
6. True
13. False
20. False
7. False
14. True
21. False
27. Possible answer: population of city A 5 583,240 and population of city B 5 3615. 28. Possible answer: x 5 150,000,000. 29. Possible answer: x 5 0.45. 30. Possible answer: b such that b 5 14. 31. b 5 1. 32. Possible answer: b 5 23. 33. False
36. True
39. False
34. False
37. False
40. False
35. True
38. False
41. False
10
(102.85, 101.40)
1. a. 4200
b. 240
c. 216
e. 27
3. When x is an odd integer the statement is true. ft 60 sec 60 min 1 mile 5. 320sec ? 1 min ? 1 hr ? 5280 ft < 4.55 miles/hour
7. Tiger Woods makes $111.9 ? 106/yr. 5 $111.9 ? 108/yr. The worker makes 40 $13,312 ? wkhr ? 52wk 5 $1.3312 ? 104 /yr. yr 5 yr
$1.119 ? 10 The worker must work $1.3312 < 0.84 ? 104 years or ? 104 about 8,400 years to make Tiger Wood’s annual pay. 8
100 102
1 d. 10
Tiger Wood makes 4 orders of magnitude or about 10,000 times more than the minimum-wage worker.
101
101
42. True
Ch. 4 Review: Putting It All Together
$6.40 hr
2
100
Ch. 4
7. a. b. c. d.
629
103
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CH. 4 Review: Putting It All Together
CH. 5 Algebra Aerobics Solutions for Section 5.1
Item
Value
Ch. 5
Mass-energy of electron The kinetic energy of a flying mosquito An average person swinging a baseball bat Energy received from the sun at the Earth’s orbit by one square meter in one second Energy released by one gram of TNT Energy released by metabolism of one gram of fat Approximate annual power usage of a standard clothes dryer
4184 38,000 320,000,000
9. Convert to same unit of measure—for example, km2. Country Russia
Area
Scientific Notation
17,075,200 km2
1.70752 ? 107 km2
2
Chile
(290,125 mi )
km 2 A1.609 1 mi B
7.51099 ? 105 km2
2
< 751,099 km
km 2 6 2 (3,830,840 mi2) A1.609 1 mi B 9.917589 ? 10 km
Canada
< 9,917,589 km2 1.184825 ? 106 km2
Norway
323,895 km
3.23895 ? 105 km2
Monaco
km 2 0.5 mi2 A1.609 1 mi B
1.29 ? 100 km2
2
< 1.29 km
a. Russia is the largest in area. Monaco is the smallest in area. b. Russia, Canada, South Africa, Chile, Norway, Monaco c. Russia is seven orders of magnitude larger than Monaco. 11. False. An increase in one order of magnitude is the same as multiplying by 10. A 100% increase would only double the original amount. 13. a.
?
60 sec 1 min
?
60 min 1 hr
?
24 hr 1 day
5 12,342.9
babies day
<
sec 60 min 24 hr ? 60 1 min ? 1 hr ? 1 day 5 2787.1 2787 immigrants/day
immigrants day
60 sec 60 min 24 hr deaths c. 113death sec ? 1 min ? 1 hr ? 1 day 5 6646.15 day < 6646 deaths/day
15. Volume of a sphere 5 43pr3, so the volume of Earth 5 ? 10 m) 5 6
3
A43p(6.3) 3B
? 10 m 18
3
< 1047.4 ? 1018 m3 < 1.0474 ? 1021 m3 5.97 ? 1024 kg mass 5 1.0474.1021 m3 < 5.7 ? 103 kg/m3 Density 5 volume
17. The patient’s weight in kilograms is: 130 lb ?
0.4536 kg 1 lb
< 59 kg.
5 mg
daily dosage ⫽ kg ? 59 kg < 295 mg Since each tablet is 100 mg, the patient should take 295 mg 100 mg/tablet
21. a. See chart at top of page for solution. ? 108 J b. 3.2 5 0.4 ? 107 5 4 ? 106, or 4 million times more 8 ? 101 J energy. 3.8 ? 10 J < 2.372 ? 1011, or approximately 200 billion c. 1.602 ? 1027 J times more energy.
W 5 180 lb ?
0.4536 kg 1 lb
< 81.6 kgand
cm H 5 6 ft ? 30.5 1 ft 5 183 cm
BSA 5 71.84 ? 81.60.425 ? 1830.725 < 20,376 cm2 Converting cm2 to m2: 2
1m 20,376 cm2 ? (100 5 2.04 m2 cm) 2 mg
1 immigrant 31 sec
4 3 p(6.3
19. a. log 1 ⫽ 0 b. log 1,000,000,000 5 log 109 ⫽ 9 c. log 0.000 001 5 log 1026 5 26
daily dosage ⫽ 15 m2 ? 2.04 m2 ⫽ 30.6 mg
12,343 babies/day b.
4.184 ? 103 J 3.8 ? 104 J 3.2 ? 108 J
J J J
23. (Requires scientific calculator.) Converting height and weight to kilograms and centimeters, respectively:
2
1 baby 7 sec
5.1 ? 10214 J 1.602 ? 1027 8 ? 101 J 1.360 ? 103 J
4
1,184,825 km2
South Africa
Value in Scientific Notation
0.000 000 000 000 051 J 0.000 000 160 2 J 80 J 1360 J
5 2.95, or 3 tablets each day.
<
CHAPTER 5 Section 5.1 Algebra Aerobics 5.1 1. a. Initial value 5 350 Growth factor 5 5 b. Initial value 5 25,000 Growth factor 5 1.5
c. Initial value 5 7000 Growth factor 5 4 d. Initial value 5 5000 Growth factor 5 1.025
2. a. P 5 3000 ? 3 t b. P 5 (4 ? 107)(1.3)t
c. P 5 75 ? 4t d. P 5 $30,000(1.12)t
3. a. P 5 28 ? 1.0650 5 28 million or 28,000,000 people; the initial population b. P 5 28 ? 1.06510 < 52.6 million or 52,600,000 people; P 5 28 ? 1.06520 < 98.7 million or 98,700,000; P 5 28 ? 1.06530 < 185.2 million or 185,200,000 c. t < 11 years for P to double or reach 56,000,000.
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CH. 5 Exercises Solutions for Section 5.1
4.
CH. 5 Algebra Aerobics Solutions for Section 5.2
t
0
1
10
15
20
P
80
84.80
143.27
191.72
256.57
631
For every unit increase in t, N increases by 3, so the graph is linear with a slope of 3. The vertical intercept is 10. N 25
a. 80 b. 84.8 c. The amount doubles (reaches 160) between t 5 10 and t 5 15 d. t < 12 time periods.
N = 10 + 3t
Exercises for Section 5.1 e. 8000; 2.718 f. 4 ? 105; 2.5
Initial Value C
Growth Factor a
Exponential Function y 5 Ca x
1600 6.2 ? 105 1400
1.05 2.07 3.25
y 5 1600 ? 1.05x y 5 6.2 ? 105 ? (2.07)x y 5 1400 ? (3.25)x
5. a. f(x) 5 6 ? 1.2x
4
N 5 10 ? 3t The rate of change here is not constant as in the previous problem. For example, the average rate of change between 0 and 1 is (30 2 10)/1 5 20 and between 1 and 2 is (90 2 30)/1 5 60. But the vertical intercept for the graphs of both functions is the same, 10. N 800
b. f (x) 5 10 ? 2.5x
600 N = 10 • 3t
400
7. a. 1.17
b. 63 cells per ml.
d
C
0 1 2 3 4 5 6 7 8 9 10
63.0 73.7 86.2 100.9 118.1 138.1 161.6 189.1 221.2 258.8 302.8
2. a.
b.
d. C doubles somewhere between d 5 4 and d 5 5 days. 9. Expenditures 5 2016 ? (1.076)5 < 2907.7 billion dollars. 11. a. G(t) 5 5 ? 103 ? 1.185t cells/ml, where t is measured in hours. b. G(8) 5 5 ? 103 ? 1.1858 5 19,440.92 cells/ml 13. a. 10 fish b. It doubled in 5 months.
c. 80 fish d. approx. 35 months
Section 5.2 Algebra Aerobics 5.2 1.
200 0
c.
t
N 5 10 1 3t
N 5 10 ? 3t
0 1 2 3 4
10 13 16 19 22
10 30 90 270 810
N 5 10 1 3t
t
t 1
2
3
4
t
ƒ(t) 5 200 1 20t
t
g(t) 5 200(1.20)t
0 1 2 3 4 5
200 220 240 260 280 300
0 1 2 3 4 5
200 240 288 345.6 414.72 497.66 g(t)
500 400 f(t)
300 200 100
t 0
1
2
3
4
5
c. The function ƒ is linear and the function g is exponential. While both functions have the same initial value, g is growing faster than ƒ. 3. ƒ is linear because the rate of change is constant; that is, the slope is 6. g is linear because the rate of change is constant; that is, the slope is 0. h is exponential because the ratio of any two consecutive terms is constant; that is, there is a constant growth factor of 1.5 or 150%. p is linear because the rate of change is constant; that is, the slope is 90. r is exponential because there is a constant growth factor of 1.04 or 104%
Ch. 5
3.
0
c. 6 ? 108; 5 d. 25; 1.18
1. a. 275; 3 b. 15,000; 1.04
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CH. 5 Exercises Solutions for Section 5.2
CH. 5 Exercises Solutions for Section 5.2
500 4. a. Linear: m 5 6201 2 2 0 5 120; (0, 500) 1 b 5 500 1 y 5 500 1 120t Exponential: growth factor 5 620 500 5 1.24; (0, 500) 1 initial amount 5 500 1 y 5 500(1.24)t 2 3 b. Linear: m 5 3.2 1 2 0 5 0.2; (0, 3) 1 b 5 3 1 y 5 3 1 0.2t Exponential: growth factor 5 3.2 3 < 1.067; (0, 3) 1 initial amount 5 3 1 y 5 3(1.067)t t 5. a. growth factor 5 3750 1500 5 2.50 1 P 5 1500 ? 2.50 82,300
b. growth factor 5 80,000 5 1.029 1 A 5 80,000 ? 1.029t
11.
x
6 1 1.5x
6 ? 1.5x
1.5 ? 6x
0 1 2 3 4 5
6.0 7.5 9.0 10.5 12.0 13.5
6 9 13.5 20.25 30.375 45.5625
1.5 9 54 324 1,944 11,664
The graphs are found in the accompanying diagrams.
6. a. 2 b.
t (years)
Q
0 2 4 6 8
10 20 40 80 160
y 14
y = 6 + 1.5x
x 0
c. Let a 5 growth factor. The 2-yr. growth factor 5 2 1 a2 5 2 1 a 5 "2 < 1.41. The annual growth factor < 1.41. d. Q 5 10 ? 1.41t
5
y 45
Exercises for Section 5.2 1. a. linear; 5 b. exponential; 3
y = 6 • 1.5 x
c. neither; 3 d. exponential; 6
e. exponential; 7 f. linear; 0
3. a. P(t) 5 50 1 5t b. Q(t) 5 50 ? 1.05t c. The graph of the two functions is in the accompanying diagram. d. Graphing software gives that the two are equal at (0, 50) and at (26.6, 183). Thus the populations were both 50 people at the start and were both approximately 183 persons after 26.6 years. (Student eyeball estimates may differ.) 200
x 0
5
y 4500
y = 1.5 • 6 x
x
(26.6, 183) 0 P(t) = 50 + 5t
Population
Ch. 5
t c. growth factor 5 32.7 30 5 1.09 1 Q 5 30 ? 1.09
5
13. a. For both populations we are looking for an exponential growth factor, i.e., for the value of a, where A(t) 5 C ? at and where t measures years from 1986. For the entire Atlantic coast: we have C 5 5800 and A(16) 5 14,313 and thus 14,313 5 5800 ? a16 or a 5
Q(t) = 50 • 1.05t (0, 50)
14,313 1/16
0
30 Years
5. linear: y 5 3x 1 6; exponential: y 5 6 ? 1.5
x
7. a. linear: y 5 10x 1 10; exponential: y 5 10 ? A "3B x 3 b. linear: y 5 100x 1 100; exponential y 5 100 ? A "4B x
9. a. exponential; f (x) 5 7 ? 2.5x b. linear: g(x) 5 0.2x 1 0.5
Q 5800 R < 1.058. Thus the growth factor is about 1.058. For Massachusetts: we have C 5 585 and A(16) 5 1/16 < 1.106. 2939 and thus 2939 5 585 ? a16 or a 5 Q2939 585 R
Thus 1.106 is the approximate growth factor. The growth factor in Massachusetts is somewhat larger than that of the entire Atlantic coast. b. For the Atlantic coast, the average rate of change is the slope of the line between (0, 5800) and (16, 14313). This 14,313 2 5800 is: 5 8513 16 16 < 532 swans/year.
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CH. 5 Algebra Aerobics Solutions for Section 5.3
CH. 5 Exercises Solutions for Section 5.3
For Massachusetts, the average rate of change is the slope of the line between (0, 585) and (16, 2939). This is 2939 2 585 5 2354 16 16 < 147 swans/year. The average rate of change for the Atlantic coast is more than three times as large as the Massachusetts rate. c. The linear model for Massachusetts since 1986 is A(t) 5 147t 1 585; the exponential model since 1986 for Massachusetts is A(t) 5 585 ? 1.106t. d. Now 2010 is 24 years from 1986; thus S(24) 5 147 ? 24 1 585 5 4113 swans and A(24) 5 585 ? 1.10624 < 6566 swans. So the predictions are quite different for the two models. Years after 2004
Plan A
Plan B
0 1 2 3 4
$7.00 $7.35 $7.72 $8.10 $8.51
$7.00 $7.50 $8.00 $8.50 $9.00
2. a. y 5 2300(1/3)t b. y 5 s3 ? 109 ds0.35d t 3. y 5 12(5)2x 5 12(521)x 5 12(1/5)x
Initial Value, C
Decay Factor, a
Exponential Function y 5 Cax
500 1.72 ? 106 1600
0.95 0.75 0.25
y 5 500 ? 0.95x y 5 1.72 ? 106 ? (0.75)x y 5 1600 ? (0.25)x
Food
b. matches graph A. Formula
Amount in 2000
B(t) 5 72.1 ? 0.994 C(t) 5 32.7 ? 1.024t P(t) 5 52.1 ? 0.996t F(t) 5 12.4 ? 1.01t t
Beef Chicken Pork Fish
63.9 lb 52.5 lb 48.1 lb 15.1 lb
b. Growth: chicken and fish; decay: beef and pork. c. Answers will vary with students. It is expected that the growth rates and per-capita consumption figures will be cited. 13. a. f (t) 5 24t 1 20 b. g(t) 5 20 ? 0.25t y c. 20
Algebra Aerobics 5.3 c. decay d. growth
7.
11. a.
Section 5.3 1. a. growth b. decay
5. a. This is not exponential since the base is the variable. b. This is exponential; the decay factor is 0.5; the vertical intercept is 100. c. This is exponential; the decay factor is 0.999; the vertical intercept is 1000.
9. a. matches graph B.
b. Plan A: FA(t) 5 7 ? 1.05t and plan B: FB(t) 5 7 1 0.50t c. FA(25) 5 7 ? (1.05)21 < $19.50 and FB(25) 5 7 1 0.50 ? 21 5 $17.50 d. Student answers will vary, but one should note that plan B is more expensive until year 15; from then on plan A would be more expensive. Going for plan A for the short term seems like a better option.
c. h(x) 5 219 ? 0.1x
e. decay f. growth
f (t) = 20 – 4t
c. y 5 s375ds0.1d t g(t) = 20 • 0.25 t
t 0
so it represents decay.
5
x
1 4. a. y 5 23 ? 2.42x 1 y 5 23s2.421 d x 1 y 5 23 A2.4 B 1 x y 5 23(0.42) ; decay x
1 B 1 b. ƒ(x) 5 8000 ? (0.521)x 1 ƒsxd 5 8000 A0.5 x ƒsxd 5 8000 ? s2d ; growth
c. P 5 52,000 ? 1.0752t 1 P 5 52,000(1.07521)x 1 1 x P 5 52,000 A1.075 B 1 P 5 52,000(0.93) x; decay
15. The accompanying diagram contains the graphs of f (x) 5 25 ? 52x 5 25(0.2)x and g(x) 5 25 ? 0.5x. It is clear that the graph of f has the faster rate of decline. Note that the decay factor for f is 0.20 and the decay factor for g is 0.50. y (0, 25)
25
5. a. g(x) decreases more rapidly. b. For g(x), 70% of the previous amount remains each time; that is, there is 30% less each time period. For ƒ(x), 90% of the previous amount remains each time; that is, there is 10% less each time period.
g(x) = 25 • 0.5x
Exercises for Section 5.3 1. a. 0.43
b. 0.95
f (x) = 25 • 5–x
c. 1/3
x 0
5
Ch. 5
15. a.
3. a. f (t) 5 10,000 ? 0.4t b. g(T) 5 2.7 ? 1013 ? 0.27T
633
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CH. 5 Algebra Aerobics Solutions for Section 5.4
17. a. P(t) 5 100 ? bt; 99.2 5 100 ? b or b 5 0.992 and therefore P(t) 5 100 ? 0.992t b. P(50) 5 100 ? 0.99250 < 66.9 grams; P(500) 5 100 ? 0.992500 < 1.8 grams
Section 5.4 Algebra Aerobics 5.4 1. a.
y 3000 2500 2000
y = 1000(1.8)x y = 1000(1.5)x
Ch. 5
1500 1000
y = 1000(1.1)x
500 –10
–5
0
x 10
5
b. Yes, they intersect at (0, 1000). c. In the first quadrant, y 5 1000(1.8)x is on top, y 5 1000(1.5)x is in the middle, and y 5 1000(1.1)x is on the bottom. d. In the second quadrant, y 5 1000(1.1)x is on top, y 5 1000(1.5)x is in the middle, and y 5 1000(1.8)x is on the bottom. 2.
y
400
Q = 250(0.6)t Q = 250(0.3)t
300
Q = 250(0.2)t
500
CH. 5 Exercises Solutions for Section 5.4
c. The function ƒ approaches zero more rapidly as x S 2` . d. The graphs intersect at approximately (3, 35); ƒ is greater than g after the point of intersection.
Exercises for Section 5.4 1. Let y1 5 2x, y2 5 5x, and y3 5 10x. a. C 5 1 for each case; a 5 2 for y1, a 5 5 for y2 , and a 5 10 for y3. b. Each represents growth; y1’s value doubles, y2’s value is multiplied by 5, and y3’s value is multiplied by 10. c. All three graphs intersect at (0, 1). d. In the first quadrant the graph of y3 is on top, the graph of y2 is in the middle, and the graph of y1 is on the bottom. e. All have the graph of y 5 0 (or the x-axis) as their horizontal asymptote. f. Small table for each: x
y1
y2
y3
0 1 2
1 2 4
1 5 25
1 10 100
g. The graphs of the three functions are given in the accompanying diagram. They indeed confirm the answers given to the questions asked in parts (a) through (f). y
200
100
100 y3 = 10 x
t –1
–0.5
0
0.5
1
b. Yes, they intersect at (0, 250). c. In the first quadrant: Q 5 250(0.6)t is on top, Q 5 250(0.3)t is in the middle, and Q 5 250(0.2)t is on the bottom. d. In the second quadrant: Q 5 250(0.2)t is on top, Q 5 250(0.3)t is in the middle, and Q 5 250(0.6)t is on the bottom. 3. a.
y2 = 5x
y1 = 2 x
Q
t
300 250 200 150 100
P = 150 • 3 –6
–4
–2
x –5 P = 50 • 3t
50 0
t –50 2
b. No, the curves do not intersect. c. P 5 150 ? 3t is always above P 5 50 ? 3t. 4. a. The horizontal axis is the horizontal asymptote. b. The horizontal axis is the horizontal asymptote. c. No horizontal asymptotes. The graph of this function is a line. 5. Comparing the growth factors: 1.23 . 1.092 . 1.06, so the function h has the most rapid growth. Comparing the decay factors: 0.89 , 0.956, so the function g has the most rapid decay. 6. a. The function g has the larger initial value since it crosses the vertical axis above the function ƒ. b. The function ƒ has the larger growth factor since it is steeper, growing at a faster rate.
0
5
3. Let y1 5 3x, y2 5 (1/3)x, and y3 5 3 ? (1/3)x. a. C 5 1 for y1 and y2 and C 5 3 for y3. b. y1 represents growth; y2 and y3 represent decay. c. y1 and y3 intersect at approx. (0.5, 1.7); y1 and y2 intersect at (0, 1). d. For x . 0.5, the graph of y1 is on top, the graph of y3 is in the middle, and the graph of y2 is on the bottom. For x , 0, the graph of y3 is on top, that of y2 is in the middle, and that of y1 is on the bottom. e. All the graphs have the x-axis as their horizontal asymptote. f. A small table for each: x
y1
y2
y3
0 1 2
1 3 9
1 1/3 1/9
3 1 1/3
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CH. 5 Exercises Solutions for Section 5.4
CH. 5 Exercises Solutions for Section 5.4
g. The graphs of these three functions are given in the accompanying diagram, and they confirm the answers given to the questions in parts (a) through (f).
635
2680 , x , 2700 and another over 20.5 , x , 0. From these one can see that f(x) . g(x) if 20.196 , x , 2690.51; otherwise g(x) . f (x).
y
y
8
1400000
(2690.51, 1345354.02)
y1 = 3 x
f (x) = 100 + 500x
g(x) = 2 • 1.005x
y2 = (1/3) x
x –5
0
5 y
5. The function P goes with Graph C, the function Q goes with Graph A, the function R goes with Graph B, and the function S goes with Graph D. 7. a. The smaller the decay factor, the faster the descent; thus h has the smallest, then g, and finally f. b. The point (0, 5) and no other point. c. It will approach 0 more slowly than the other two functions. d. h is on top and will stay on top for x , 0. 9. The accompanying diagram contains the graphs of f (x) 5 30 1 5x and g(x) 5 3 ? 1.6x with the points of intersection marked. a. f (0) 5 30; g(0) 5 3; f (0) . g(0) b. f (6) 5 60; g(6) < 50.33; f (6) . g(6) c. f (7) 5 65; g(7) < 80.53; f (7) , g(7) d. f (25) 5 5; g(25) < 0.286; f (25) . g(25) e. f (26) 5 0; g(26) < 0.179; f (26) , g(26) f. f and g go to infinity; f (x) , g(x) for all x . 6.5, approximately. g. f goes to 2` and g(x) goes to 0; thus f(x) , g(x) for all x , 26, approximately. 200
4
g (x) = 2 • 1.005 x (–0.196, 1.998)
x –0.5
0
f(x) = 100 + 500x
–4
13. The graphs of the three functions are given in the accompanying diagram. The graphs of F and H are mirror images of each other and are equally steep when the absolute values of average rates of change are considered. H(x) 5 100(1/1.2)x < 100(0.83)x and G(x) 5 100 ? 0.8x, so G(x) has a smaller decay factor and thus decays faster than H(x) and is steeper than H(x). Thus, the graph of G is steeper than the other two, again when considering the absolute values of the average rates of change. y 200 G(x) = 100 • 0.8 x
y
H(x) = 100 • 1.2(–x)
F(x) = 100 • 1.2x
(6.45, 62.26) x
f (x) = 30 + 5x (–5.96, 0.182) –10
–20
g (x) = 3 • 1.6x x 10
11. a. As x S 1`, g(x) will dominate over f (x), i.e., g(x) . f (x) b. There is no one coordinate window that will display the graphs of f and g to help one see the answer to his question. Thus two displays are given: one over
–5
0
5
15. The graphs are given in the accompanying diagram. a. It is clear that the graphs of f and h are mirror images of each other with respect to the y-axis. b. The graphs of f and g are mirror images of each other with respect to the x-axis. c. The graphs of g and h are mirror images of each other with respect to the origin.
Ch. 5
x 2700
1300000 2680
y3 = 3 • (1/3) x
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CH. 5 Algebra Aerobics Solutions for Section 5.5
d. The graphs of these functions are mirror images of each other with respect to the x-axis. e. These functions are mirror images of each other with respect to the y-axis.
25
y
–x
f (x) = 3 • 1.5 x
h(x) = 3 • 1.5
x 5
Ch. 5
–5
g (x) = –3 • 1.5 x –25
Section 5.5 Algebra Aerobics 5.5 1. a. b. c. d.
22% 106.7% or 1.067 1 2 0.972 5 0.028 or 2.8% 1 2 0.123 5 0.877
CH. 5 Exercises Solutions for Section 5.5
5. Initial Value a. b. c. d. e. f. g.
Growth Growth Growth or Decay or Decay or Decay? Factor Rate
600 1200 6000 1.5 ? 106 1.5 ? 106 7 60
growth growth decay decay growth growth decay
2.06 3.00 0.25 0.75 1.25 4.35 0.35
7. a. A(T) 5 12,000 ? 1.12T and B(T) 5 12,000 ? 0.88T, where T 5 decades since 1990. b. If t 5 years since 1990, then t 5 10T or T 5 t/10, and thus a(t) 5 12,000 ? [1.12]t/10 5 12,000 ? [1.120.1]t 5 12,000 ? 1.0114t people and b(t) 5 12,000 ? [0.88]t/10 5 12,000 ? [0.880.1]t 5 12000 ? 0.9873t people where t is the number of years from 1990. c. A(2) < 15,053; a(20) < 15,053; B(2) < 9293 and b(20) < 9293. A(2) and B(2) represent the population after two decades, whereas a(20) and b(2) represent the population after 20 years. Note that A(2) should equal a(20) and B(2) should equal b(20). Any small differences are because of round-off error.
11. A(x) 5 225 ? 1.015x; growth rate of 1.5% 13. a. P(t) 5 150 ? 3t
c. P(t) 5 150 ? 0.93t
b. P(t) 5 150 212t
A 5 4(1.03)t A 5 10(0.98)t A 5 1000(1.005) t A 5 30(0.96) t A 5 50,000(1.0705) t A5 200(0.51) t
3. a. b. c. d. e.
Initial Value
Growth Growth or or Decay Decay? Factor
4 growth 10 decay 1000 growth 30 decay $50,000 growth 200 g decay
growth factor 5 1.06 5 106% decay factor 5 0.89 5 89% growth factor 5 1.23 5 123% decay factor 5 0.956 5 95.6% growth factor 5 1.0925109.2%
1.03 0.98 1.005 0.96 1.0705 0.51
Growth or Decay Rate 0.03 or 3% 0.02 or 2% 0.005 or 0.5% 0.04 or 4% 0.0705 or 7.05% 0.49 or 49%
growth rate 5 6% decay rate 511% growth rate 5 23% decay rate 5 4.4% growth rate 5 9.2%
4. a. $6.00 1.65 < 3.64, so 3.64 (the growth factor) 5 1 1 growth rate. The growth rate 5 2.64 or a 264% increase in 30 years. b.
0.167 0.299
< 0.56, so 0.56 (the decay factor) 5 1 2 decay rate. The decay rate 5 0.44 or a 44% decrease in one season.
c. 23 17 < 1.35, so 1.35 5 1 1 growth rate. The growth rate 5 0.35 or a 35% increase in one month.
Exercises for Section 5.5 1. a. Q 5 1000 ? 3T b. Q 5 1000 ? 1.3T 3. a. 5% b. 18%
c. Q 5 1000 ? 1.03T d. Q 5 1000 ? 1.003T c. 55% d. 34.5%
N(t) 5 600 ? 2.06t N(t) 5 1200 ? 3t N(t) 5 6000 ? 0.25t N(t) 5 1.5 ? 106 ? 0.75t N(t) 5 1.5 ? 106 ? 1.25t N(t) 5 7 ? 4.35t N(t) 5 60 ? 0.35t
106% 200% 75% 25% 25% 335% 65%
9. A(x) 5 500 ? 0.85x; decay rate of 15%
2. Exponential Function
Exponential Function
e. 0.4% f. 27.5%
d. P(t) 5 150 1 t
15. In 2030 the population size will be approximately 200 ? 1.42 5 392 million, and in 2060 it will be approximately 200 ? 1.43 5 548.8 million. Using a graphing calculator to estimate the intersection point of the graphs of y1 5 200 ? 1.4t and y2 5 1000 (where y1 and y2 are measured in millions and t is measured in 30-year periods) gives t < 4.78. Thus, according to this model, sometime in the first part of 2113, the U.S. population will reach approximately 1 billion. 17. a. A(n) 5 A0 ? 0.75n, where n measures the number of years from the original dumping of the pollutant and A0 represents the original amount of pollutant. b. We are solving 0.1 ? A0 5 A0 ? 0.75n for n and we first divide out by A0. Then we graph y 5 0.01 and y 5 0.75n and estimate their intersection. This gives n < 16 years. 19. a. 1.00712 < 1.087 and thus the inflation rate is about 8.7% per year. b. (1 1 r)12 5 1.05 means that r 5 (1.05)1/12 21 < 0.0041. Thus the rate is about 0.4% per month. 21. a. They are equivalent, since 1.20.1 5 1.0184 to four decimal places. b.
x
0
5
10
15
20
25
f (x) 15,000.0 16431.7 18,000 19,718.0 21,600.0 23,661.6 g(x) 15,000.0 16431.7 18,000.1 19,718.2 21,600.3 23,662.0
c. If x is the number of years, then ƒ(x) 5 15,000(1.2)x/10 represents a 20% growth factor over a decade, and g(x) 5 15,000(1.0184x) represents a 1.84% annual growth factor.
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CH. 5 Algebra Aerobics Solutions for Section 5.6
23. a. A graph with the data plotted and the best-fit exponential is given in the diagram. Alcohol Content over Time of an Initially Legally Drunk Person 0.12 0.10 y = 0.1 • 0.67 t BAC g/dl
0.08 0.06 0.04 0.02
t 1
3 2 Time in hours
4
b. If one looks at the exponential graph it is easy to see that it fits the data quite well. Moreover, the ratios of successive g/dl values are nearly a constant 0.67. Another equation, easier to handle, is y 5 0.1 ? 0.67t, where t is measured in hours and y in g/dl. c. In this model, the g/dl decreases each hour by about 33%. d. A reasonable domain could be 0 to 10 hours and the corresponding range would be 0.002 to 0.100 g/dl. e. 0.005 5 0.1 ? 0.67t implies that t is between 7 and 8 hours. This can be obtained by computing g/dl values for successive t values using the formula given in part (b). 25. a. If
the
model is linear, then the slope is 5 20.8 10 5 20.08 and the equation is MRL(y) 5 20.08y 1 7.2, where y is the number of years since 1996 and MRL is the number of deaths per 1000 live births. b. If the model is exponential, then the decay factor is 1/10 A6.4 < 0.988 and thus the equation is MRE(y) 5 7.2 B 7.2 ? 0.988y, and the units are the same as in part (a). c. The graphs requested are in the accompanying diagram. (6.4 2 7.2) (2006 2 1996)
Infant Mortality per 1000 Live Births 7.2
Mortaility rate
7 6.8 6.6 6.4 6.2
MRL = –0.08x + 7.2
MRE = 7.2(0.988)x
6 0
2
4
6
8
10 12 14 16
Years after 1996
d. In 2010, MRL(14) 5 6.08 and MRE(14) < 6.11; both are measured in deaths per 1000 live births.
Section 5.6 Algebra Aerobics 5.6a 1. a. Exponential growth; doubling period is 30 days since ƒs30d 5 300 ? 230/30 5 600; Initial 5 300; one day later: ƒ(1) 5 300 ? 21/30 < 307.01; one month later: ƒs30d 5 300 ? 230/30 5 600; one year later: ƒs360d 5 300 ? 2360/30 5 1,228,800
b. Exponential decay; half-life period is 2 days since gs2d 5 32s0.5d 2/2 5 16; Initial 5 32; one day later: g(1) 5 32 ? 0.51/2 < 22.6; one month later: g(30) 5 32 ? 0.530/2 5 0.00098; one year later: g(360) 5 32 ? 0.5360/2 5 trace amount (almost zero) c. Exponential decay; half-life period is one day since P 5 32,000 ? 0.51 5 16,000; Initial 5 32,000; one day later: P 5 32,000 ? 0.51 5 16,000; one month later: P 5 32,000 ? 0.530 < 0.000 03; one year later: P 5 32,000 ? 0.5360 5 trace amount (almost zero) d. Exponential growth; doubling period is 360 days since hs360d 5 40,000 ? 2360/360 5 80,000; Initial 5 40,000; one day later, hs1d 5 40,000 ? 21/360 < 40,077; one month later, hs30d 5 40,000 ? 230/360 < 42,379; one year later, hs360d 5 40,000 ? 2360/360 5 80,000 2. a. b. c. d.
70/2 5 35 yr 70/0.5 5 140 months 70/8.1 5 8.64 yr < 9 yr growth factor 5 1.065 1 growth rate 5 6.5% 70/6.5 5 10.77 < 11 yr
3. a. 70 10 5 7; R < 7% per year b. 70/5 5 14; R < 14% per minute c. 70/25 5 2.8; R < 2.8% per second 4. a. Since a 5 0.95, r 5 0.05, so this is decay of R 5 5%, and half-life is 70/5 5 14 months. b. Since a 5 0.75, r 5 0.25, so this is decay of R 5 25%, and half-life is 70/25 5 2.8 sec. c. Half-life is 70/35 5 2 yr.
Algebra Aerobics 5.6b 1. a. 70/3 < 23.3 yr b. 70/5 5 14 yr c. 70/7 5 10 yr 2. a. 70/5 5 14; 14% b. 70/10 5 7; 7% c. 70/7 5 10; 10% 3. a. P 5 $1000s1.04d n c. P 5 $1000s2.10d n
b. P 5 $1000s1.11d n
4. If inflation is 10%/month, then what cost 1 cruzeiro this month would cost 1.10 cruzeiros next month. We have 1 cruzeiro < 91% of 1.10 cruzeiros (since 1/1.1 < 0.91), so a month later 1 cruzeiro would only be worth 0.91 cruzeiros or 91% of its original value. Thus, the decay factor is 0.91. So the exponential decay function Q 5 100s0.91d n gives the purchasing power of 100 of today’s cruzeiros at n months in the future. When n 5 3 months, then Q, the value of 100 of today’s cruzeiros, will be 100(0.91)3 < 100(0.75) 5 75 cruzeiros. When n 5 6 months, then Q, the value of 100 of today’s cruzeiros, will be 100(0.91)6 < 100(0.57) 5 57 cruzeiros. When n 5 12 months or 1 yr, then Q, the value of 100 of today’s cruzeiros, will be 100(0.91)12 < 100(0.32) 5 32 cruzeiros. With a 10% monthly inflation rate, the value of a cruzeiro will shrink by more than two-thirds by the end of a year. 5. a. Initial investment is $10,000; growth factor 5 1.065; growth rate 5 6.5%; doubling time: using the rule of 70, 70/6.5 < 10.8 years
Ch. 5
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CH. 5 Exercises Solutions for Section 5.6
b. Initial investment is $25; growth factor 5 1.08; growth rate 5 8%; doubling time: using the rule of 70, 70/8 < 8.8 years c. Initial investment is $300,000; growth factor 5 1.11; growth rate 5 11%; doubling time: using the rule of 70, 70/11 < 6.4 years d. Initial investment is $200; growth factor 5 1.092; growth rate 5 9.2%; doubling time: using the rule of 70, 70/9.2 < 7.6 years
Doubling Time (approx.)
Function A 5 50000 ? 1.072 t
$50,000
1.072
7.2%
$53,600
9.7 years
A 5 100000 ? 1.067t
$100,000
1.067
6.7%
$106,700
10.4 years
A 5 49622 ? 1.058
$49,622
1.058
5.8%
$52,500
12
$3000
1.13
13%
$3390
t
A 5 3000 ? 1.13t
Growth Growth Factor Rate
Amount 1 Year Later
Initial Investment
years
5.4 years
Algebra Aerobics 5.6c 1. a. b. c. d. e. f.
x 5 5 since 2 5 32 x 5 8 since 28 5 256 x 5 10 since 210 5 1024 x 5 1 since 21 5 2 x 5 0 since 20 5 1 x 5 21 since 221 5 12 5
Population of US 1790–2000 450.0 400.0 y = 6.06 • 1.0204x
350.0 300.0 250.0 200.0 150.0 100.0
0
h. x 5 1/2 since 21/2 5 "2 b. N 5 3L
3.
7.
50.0
g. x 5 23 since 223 5 213 5 18
2. a.
5. a. 3 ? 2x/5 , 3(1.225)x, if x . 0; the inequality is reversed if x , 0. b. 50 ? (1/2)x/20 < 50 ? 0.9659x. c. 200 ? 2x/8 < 200 ? 1.0905x. d. 750 ? (1/2) x/165 . 750 ? 0.911x if x . 0, and the inequality is reversed if x , 0.
Level
# of People Called
Total Called
0 1 2 3 4 5 6 7 8
30 31 32 5 9 33 5 27 34 5 81 35 5 243 36 5 729 37 5 2187 38 5 6561
1 4 13 40 121 364 1093 3280 9841
200
250
9. a. The accompanying diagram contains the graphs of the best-fit linear and exponential models for the given data. The linear and exponential formulae are also given there. The linear model seems to fit the data better because of a slightly higher correlation coefficient. Internet Users in U.S. 1991–2007 300
L = 15.693x – 15.235 cc = 0.981
200
100
E = 4.8291(1.3196)x cc = 0.963 0
So it will take eight levels to reach 8000 people.
Exercises for Section 5.6 1. a. Has a fixed doubling time b. Has neither c. Has a fixed half-life
150 100 Years since 1790
50
a, b. The accompanying diagram contains the graph of both the data and the best-fit exponential function, via Excel. There y 5 population in millions and x 5 years since 1790. The annual growth factor is 1.0204; the annual growth rate is 2.04%; the estimated initial population is 6.06 million. One notes that the best-fit model is lower than the plotted data from 1850 to 1930. Its values surpass those of the data from 1940 on. It is a good model for the first 160 years but is not that good thereafter. c. The model predicts the population to be 516.2 million in 2010 and 698.8 million in 2025. These seem very high indeed.
Internet users (in millions)
Ch. 5
6.
3. b. f (x) 5 1000 ? 2t/7; 21/7 5 1.1041 per year; 10.41% per year c. f (x) 5 4 ? 2t/25; 21/25 5 1.0281 per minute; 2.81% per minute d. f (x) 5 5000 ? 2t/18; 2 1/18 5 1.0393 per month; 3.93% per month
Population in millions
638
10/11/07
d. Has a fixed doubling time e. Has neither f. Has a fixed half-life
2
4
6
8 10 12 14 16 18
Years from 1991
b. 2010 is 19 years after 1991. The linear model predicts L(19) 5 15.693 ? 19 2 15.235 < 283 million users, and the exponential model predicts E(19) 5 4.8291(1.3196)19 < 938 million users. c. Student answers will vary.
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CH. 5 Exercises Solutions for Section 5.6
11. The graph in the accompanying diagram goes through (0, 100) and (12.3, 50), where the first coordinate is measured in years and the second is measured in grams. Using the graphing utility gives: A(t) 5 100 ? 0.945t. Using algebra we get A(t) 5 100 ? (0.5)t/12.3, which gives the same formula as the best-fitting exponential formula. The graph of this function is given in the accompanying diagram. 100
y
Tritium (kg)
y = A(t) = 100 • 0.945t
(12.3, 50)
c. The initial value is 799.96 or approximately 800 for the world’s population in millions in 1800 (when t 5 0). (Note that it is approximately 180 million smaller than the actual size.) The growth factor is 1.00955, which gives a growth rate of 0.955% per year. The domain of the function P technically is any real number, but concretely, its values should not go much farther back than 1800 or much past 2005. The range in that domain is about 800 million to about 5600 million. d. 1.00955 represents a growth rate of 0.955% per year. e. i. Using a calculator gives the P(t) values in the accompanying table. Eyeball estimates from students should be close to these.
t 50
0
Year
t
P(t)
1750 1920 2025 2050
250 120 225 250
498.6 2487.9 6714.1 8504.5
Years
13. a. 0.55 5 1/32 5 0.03125 and thus 3.125% of the original dosage is left. b. i. Approximately 2 hours. ii. A(t) 5 100 ? 0.5t/2 where t 5 hours after taking the drug and A(t) is in mg. iii. 5 ? 2 5 10 hours and 100 ? (1/2)5 5 100/32 5 3.125 milligrams. Also A(10) 5 100 ? 0.510/2 5 3.125. iv. Student answers will vary. They should mention the half-life and present a graph to make the drug’s behavior clear to any prospective buyer. 15. a. b. c. d. 17. a.
R 5 70/5730 < 0.012% per year R 5 70/11,460 < 0.0061% per year R 5 70/5 5 14% per second. R 5 70/10 5 7% per second. Years from 1800
World Population (in millions)
0 50 100 150 170 180 190 200 205
980 1260 1650 2520 3700 4440 5270 6080 6480
19. a. 10 half-lives gives a 0.510 5 0.001 or 0.1% of the original or a 99.9% reduction. b. A(2) 5 0.25A0, A(3) 5 0.125A0, A(4) 5 0.0625A0. After n half-lives, the amount left is A(n) 5 0.5n ? A0. 21. a.
World Population 1800–2005 Population in millions
8000
y = 799.96 (1.00955)x cc = 0.972
4000 2000
0
t, Time (months) 0 3 6 9 12
b. P(t) 5 799.96 ? 1.00955t is the best-fit exponential function formula, where t is measured in years since 1800 and P(t) is measured in millions. Its graph is given in the accompanying diagram.
6000
ii. Using a calculator or the graph of the model one can see that 1 billion was reached during 1823 (or about 23.6 years after 1800); 4 billion was reached in the first half of 1970 (or about 170.2 years from 1800); and 8 billion will be reached during 2043 (about 243.5 years from 1800). f. As can be seen from the answers in (ii) it takes approximately 73 years for the population to double.
50
200 100 150 Years from 1800
250
M, mass (g) 10 20 40 80 160
In general M 5 10(21/3) t 5 10 ? 1.2599t after t months. b. Using a calculator or a graph of the model, when M 5 2000, then t < 23 months or nearly 2 years. c. 2000/10 5 200 and thus at 2000 grams, it is 20,000% of its original size. i. A(n) 5 5000 ? (1.035)n ii. B(n) 5 5000 ? (1.0675)n iii. C(n) 5 5000 ? (1.125)n b. A(40) 5 19,796.30 B(40) 5 68,184.45 C(40) 5 555,995.02.
23. a.
25. a. V(n) 5 100 ? 1.06n. Solving V(n) 5 200 for n graphically gives n < 12 years, which is the approximate doubling time. b. W(n) 5 200 ? 1.03n. Solving W(n) 5 400 for n graphically gives n < 23 years, which is the approximate doubling time.
Ch. 5
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CH. 5 Exercises Solutions for Section 5.7
b. From the equation N 5 100 ? 2t, we know that every three time periods, the quantity is multiplied by 23 or 8. c. Every four time periods, the quantity is multiplied by 24 or 16. d. The answers are consistent, since somewhere between three and four time periods the quantity should be multiplied by 10 (which is between 8 and 16).
c. The graphs of V(n) and W(n) are in the accompanying diagram. The two graphs intersect at approximately n 5 24 years, when V(n) 5 W(n) < $410. The values for V(n) are larger than those for W(n) after that. y
Dollars
(23.45, 400)
y = 400
400
y = W(n)
300
y = 200
200
(11.9, 200) y = V(n)
100
y 1,000,000,000
0
n 25
0 Year
27. Since 1.00/1.06 5 0.943, then the value of a dollar after t years of such inflation is given by V(t) 5 0.943t. Using a calculator or the “rule of 70,” when V(t) 5 0.50, gives t < 12 years. 29. a. and b. Below is a scatter plot of the data and the Excel generated exponential curve that is a best fit.
x
y 5 25(10)x
0 1 2 3 4 5
25 250 2,500 25,000 250,000 2,500,000
100,000,000 10,000,000 1,000,000 100,000 10,000 1,000 100 10 1
x 0
2
4
6
8
x Semi-log plot of y = 25(10)
Federal Debt y = 146.48 (1.074)x cc = 0.974
12000
3. a. If x 5 3.5, y is about 80,000. If x 5 7, y 5 250,000,000.
Exercises for Section 5.7
10000 8000
1. a. Graph of given table of values for y 5 500 ? 3x
6000 4000
Graph of y vs. x
2000 1.2E + 17 0
10 20 30 40 50 60 70 Years since 1950
The model would predict the debt of 2008 to be 146.48 (1.074)58 < $9205 billion. c. Answers will vary depending on when the student invokes the debt clock. 31. a. In theory, the number recruited is Mnew(n) 5 10n, where n measures the number of rounds and Mnew(n) measures the number of people participating in the nth round of recruiting. Note that this formula assumes that all who are recruited stay and that all recruits are distinct. b. MTotal(n) 5 1 1 10 1 c 1 10n. c. MTotal(5) 5 111,111, but only 11,110 of those stem from the originator. After 10 rounds the number recruited (not including the originator) would be 11,111,111,110, which is larger than the 2005 world population. d. Comments will vary, but all will probably note how fast the number of recruits needed grows and how the amounts expected are not quite what one would have thought from the advertisements. If the chain is initially successful, then you would get a large number of new recruits in a short period. However, as can be seen in part (c), it quickly becomes unrealistic for each new person on the chain to recruit 10 new people.
Section 5.7 Algebra Aerobics 5.7 1. a. Judging from the graph, the number of E. coli bacteria grows by a factor of 10 (for example, from 100 to 1000, or 100,000 to 1,000,000) in a little over three time periods.
1E + 17 y-values
0
8E + 16 6E + 16 4E + 16 2E + 16 0
5
10 15 20 25 30 35 x-values
b. Graph of table for log(500 ? 3x) vs. x Graph of log y vs. x
log(y) values
Ch. 5
2. The graph of y 5 25 (10) x will be a straight line on a semi-log plot.
18 16 14 12 10 8 6 4 2 0
5
10 15 20 25 30 35 x-values
c. y 5 10log(y) and thus each y can be written as 10 to the power listed in the third column. 3. a. A uses the linear scale. B uses a power of 10 scale on the vertical axis and C uses a logarithm scale on the vertical axis. b. The two graphs look the same because log(10 n) 5 n and the powers of 10 are spaced out like n. c. On Graphs A and B, when y 5 1000 is multiplied by a factor of 10 to get y 5 10,000, x has increased by about 5.5 units. d. On graphs B, y labels go up by factors of 10. e. On graph A, since the scales on both axes are linear. 5. a. The accompanying graph is of the white blood cell counts on a semi-log plot. The data from October 17 to
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CH. 5 Check Your Understanding
CH. 5 Review: Putting It All Together
October 30 seem to be exponential, since the data in that range seem to fall along a straight line in the plot. There does not seem to be a discernible exponential decay pattern in this graph of the data. Semi-log plot for 5(a).
25. Possible answer: P 5 2.2(1.005) t million people, t 5 years. 26. Possible answer: P 5 2.2(1.005) 4t million people, t 5 years. 27. Possible answer: M 5 1.4(0.977) t/10 billion dollars, t 5 years. 28. Possible answer: y 5 3125(0.2) t.
Log(Cell Count) vs. Date log(cell count) values
641
4.500 4.000 3.500 3.000 2.500 2.000 1.500 1.000 0.500 0.000 19-Sep 29-Sep 9-Oct 19-Oct 29-Oct 8-Nov Date
29. Possible answer: y1 5 300(0.88) t and y 2 5 300(0.94) t. 30. Possible answer: y 5 21.458x 1 5 and y 5 5(0.5) x. 31. Possible answer: V 5 1(0.97) t, t 5 years. 32. Possible answer: y 5 5 ? (1.051) t, t 5 years.
b. Below are two graphs of the E. coli counts; one has regular horizontal and vertical axes and the other is a semi-log plot. These data look very exponential from the third to the thirteenth time periods in the regular plot and fairly exponential in the semi-log plot (since that plot looks rather linear). For contrast, the regular linear plot is given as well.
34. True
37. True
40. False
35. False
38. True
41. False
36. True
39. False
42. True
43. False
Cell count per ml
Semi-log plot for 5(b). E. Coli Counts vs. Time 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E
+ + + + + +
Ch. 5 Review: Putting It All Together
10 08 06 04 02 00 0
5
10 15 20 20-minute time periods
25
1. a. P 5 150 ? 2t b. P 5 150 2 12t 3.
y 16
Regular plot for 5(b). Cell count per ml
c. P 5 150 (1.05)t d. P 5 150 1 12t
y = 4x
E. Coli Counts vs. Time 2.00E 1.50E 1.00E 5.00E 0.00E
+ + + + +
09 09 09 08 00
y = 3x y = 2x 0
5
10 15 20 25 20-minute time periods
30
7. a. This is a time-series, semi-log plot with time scale linear. b. It says that the growth was roughly exponential between 1993 and 2000. c. It says that the decline was roughly exponential between 2000 and 2002. d. Student answers will vary. Only Fidelity can give the real answer. But one might surmise that Fidelity wanted to show overall that it has done as well as the 500 Index Funds and to visually deemphasize its decline after 2000.
Ch. 5: Check Your Understanding 1. False
7. False
13. False
19. False
2. True
8. False
14. False
20. True
3. False
9. True
15. True
21. True
4. False
10. True
16. True
22. True
5. False
11. False
17. False
23. False
6. True
12. False
18. True
24. True
x –2
2
5. a. Males who just stopped smoking are about 22 times more likely (the relative risk) to get lung cancer than a lifelong nonsmoker. Females who stopped smoking 12 years ago are about three times more likely to get lung cancer than a lifelong nonsmoker. b. One reason could be that the longer it has been since someone quit smoking, the more time the lungs have had to heal. Another reason could be that the death rate of smokers is higher than that of nonsmokers. c. A relative risk of 1 means each group is equally likely to get lung cancer. It is highly unlikely for the relative risk of smokers vs. nonsmokers to go below 1 since that would mean smokers are less likely to get lung cancer than nonsmokers, which does not make sense. 7. Initial value 5 500 and growth rate 5 1.5 1 G(t) 5 500(1.5)t, where G(t) 5 number of bacteria and t 5 number of days. 9. a. If the world population in 1999 was 6 billion people, and it grew at a rate of 1.3% per year, then it is only in the first year that there is a net addition of 78 million people. The next year the increase would be 1.3% of 6,078,000,000, which is 79,014,000. The population would continue to have an increase that becomes larger
Ch. 5
33. Possible answer: R 5 200(0.966) t mg, t 5 years.
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CH. 5 Review: Putting It All Together
Ch. 5
and larger than 78 million each year. The increase is a fixed amount only in a linear model. b. Let P 5 world population (in millions) and t 5 number of years since 1999. Linear model: P 5 6000 ⫹ 78t Exponential model: 6000(1.013)t c. Linear model prediction for 2006: P 5 6000 ⫹ 78(2006 2 1999) 1 P 5 6546 million Exponential model prediction for 2006: 6000(1.013)7 1 P < 6568 million Answers may vary for best predictor depending on current population. The U.S Census Bureau website gives the 2006 world population as 6.567 billion, so our exponential model is a more accurate predictor for 2006. 11. If R 5 the annual growth rate, then using the “rule of 70” we have 70/25 5 2.8 1 R 5 2.8%. So the number of motor vehicles increases by almost 3% per year. 13.
Projected Year
2003
2004
2005
2006
Health care expenditures (billions)
$1740.6
$1877.6
$2016.0
$2169.5
Annual growth factor
n.a.
1877.6 1740.6
2016.0 1877.6
2169.5 2016
Annual percent growth rate
n.a
<1.079
<1.074
<1.076
7.9%
7.4%
7.6%
15. The ratio of consecutive values over 5-year intervals is approximately constant at 1.09, so an exponential model would be appropriate for the data in the table. If we let E(t) 5 energy consumption (in quadtrillion Btu), where t 5 number of 5-year intervals since 2015, then an exponential function to model the data would be: E(t) 5 563 (1.09)t. 17. a. 2002 to 2015: 13 year growth factor 5 (2 billion/605 million) 5 2,000 million/605 million < 3.306 annual growth factor 5 3.3061/13 < 1.096 annual growth rate < 0.096 or 9.6% b. 2015 to 2040: 25 year growth factor 5 (3 billion/2 billion) 5 1.5 annual growth factor 5 1.51/25 < 1.016 annual growth rate < 0.016 or 1.6% 19. Initial value 5 20 grams gives A(t) 5 20(1/2)t/8 grams, where t is in days. 21. The graph shows a semi-log plot. When an exponential function is plotted on a semi-log plot, its graph is a straight line. The data for “Internet hosts” is plotted as a straight line and thus could be modeled with an exponential function. The data for “Pages” and “Websites” is approximately linear on this semi-log plot, and thus an exponential model may be a close approximation.
23. a. Each linear dimension of the model is one-tenth that of the actual village, so the area of the model (which is twodimensional) would be (1/10) ? (1/10) 5 1/100 or onehundredth of Ao, the area of the actual village. b. Here each linear dimension is one-hundredth that of the actual church. The weight depends on volume (which is threedimensional) and would be (1/100) ? (1/100) ? (1/100) 5 1/1,000,000, or one-millionth of Wo, the weight of the actual church. c. An 5 Ao ? A101 2B n Wn 5 Wo ? A101 3B n 25. a. S(t) 5 0.5 ? 2t, the tumor size in cubic centimeters, where t 5 number of 2-month time periods A(t) 5 0.5 ? 2t/4, the tumor size in cubic centimeters, where t 5 number of 2-month time periods Cancer Growth for b. Different Doubling Times 250 Projected size in cu. cm
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Heavy smoker Asbestos worker
200 150 100 50
0
5 10 Number of 2-month time periods
15
The faster-growing cancer of the heavy smoker if untreated gets dangerously large very quickly after about 6 time periods (1 year). The slower-growing cancer of the asbestos worker after 12 time periods (2 years) is still relatively small compared to that of the smoker (the graph does not yet show an exponential curve). c. 1 year 5 6 time periods The tumor size for the smoker after 1 year (or 6 time periods) is S(6) 5 0.5 ? 26 5 32 cubic centimeters 1 Volumesmoker 5 32 5 A43B pr3 1 324p? 3 5 r3 7.64 < r3 Taking the cube root of both sides, r < 1.97 cm, or a tumor diameter of about 3.94 cm. The tumor size for the asbestos worker after 1 year (or 6 time periods) is A(6) 5 0.5 ? 26/4 < 1.41 cubic centimeters 1 ? 3 Volumeasbestos 5 1.41 5 A43B pr3 1 1.41 5 r3 4p 0.34 < r3 Taking the cube root of both sides, r < 0.70 cm, or a tumor diameter of about 1.4 cm. 27. a. Let M(t) 5 amount of methane gas in the atmosphere (in parts per billion), where t 5 number of years since 1850. The initial value in 1850 is 750 parts/billion. In 2005 the methane levels were at 1750 parts/billion, so (1/155)t M(t) 5 750 ? A1750 < 750(1.0055)t 750 B b. Estimates from graph: Around 1980 methane gas reaches 1500 parts per billion, which is double the 1850 level of
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CH. 6 Algebra Aerobics Solutions for Section 6.1
about 750 parts per billion. So the doubling time using this method is about 130 years. Using the “rule of 70” and a growth rate of 0.55% from part (a), we have 70/0.55 ≈ 127 years. So the estimates are pretty close. c. Since 1850 the amount of methane gas in the atmosphere has been growing exponentially at a rate of about 0.55% per year. Between 1850 and 1980, the amount of methane gas in the atmosphere doubled.
CHAPTER 6
CH. 6 Algebra Aerobics Solutions for Section 6.1
3. a. False b. False c. True; Rule 3
d. True; Rule 3 and that log 10 5 1 e. False f. False
2x 2 1 2x 2 1 1/2 2x 2 1 1 4. a. log 3 x 1 1 5 log A x 1 1 B 5 2 log A x 1 1 B 5 12 [logs2x 2 1d 2 logsx 1 1d] xy
b. log z 5 logsxyd 2 log z 5 log x 1 log y 2 log z x(x 1 1) 1/2
c. log x(x"x2 11) 21 5 log (x 2 1) 2
5 log x(x 1 1) 1/2 2 log(x 2 1) 2
Section 6.1
5 log x 1 log(x 1 1) 1/2 2 2 log(x 2 1) 5 log x 1 12 log(x 1 1) 2 2 log (x 2 1)
Algebra Aerobics 6.1a When t 5 0, M 5 250(3)0 5 250(1) 5 250. When t 5 1, M 5 250(3)1 5 250(3) 5 750. When t 5 2, M 5 250(3)2 5 250(9) 5 2250. When t 5 3, M 5 250(3)3 5 250(27) 5 6750.
2. a. 4 , t , 5
b. 8 , t , 9
d. log
x 2 sy 2 1d y 3z
5 log x 2 1 logsy 2 1d 2 [log y 3 1 log z] 5 2 log x 1 logsy 2 1d 2 3 log y 2 log z 5. 13 [log x 2 log(x 1 1)] 5 13 log x 1x 1 1/3 5 log Ax 1x 1B
3. To read each value of x from this graph, draw a horizontal line from the y-axis at a given value until it hits the curve. Then draw a vertical line to the x-axis to identify the appropriate value of x.
y=7
–2
3 x 5 log 3 x 1 1
6. a. log x 5 3 1 103 5 x, x 5 1000 b. log x 1 log 5 5 2 1 log 5x 5 2 1 102 5 5x 1 x 5 20 c. log x 1 log 5 5 log 2 1 log 5x 5 log 2 1 5x 5 2 1 x 5 2/5
y 10
–2
5 log x 2 sy 2 1d 2 log y 3z
2
x
a. horizontal line y 5 7, from y-axis to curve, then down to x-axis 1 x < 1.8 b. horizontal line y 5 0.5, from y-axis to curve, then down to x-axis 1 x < 20.5. 4. a. 2 , x , 3
b. 21 , x , 0
5. a. 23 , 13 , 24 b. 34 , 99 , 35
c. 521 , 0.24 , 50 d. 103 , 1500 , 104
Algebra Aerobics 6.1b 1. a. log(105/107) 5 log 1022 5 22 log 10 5 22(1) 5 22 and log 105 2 log 107 5 5 log 10 2 7 log 10 5 5 2 7 5 22. So log(105/107) 5 log 105 2 log 107. b. log(105 ? (107)3) 5 log(105 ? 1021) 5 log 1026 5 26 log 10 5 26 and log 105 1 3 log 107 5 5 log 10 1 3(7) log 10 5 5 1 (3 ? 7) 5 26. So log[105 ? (107)3] 5 log(105) 1 3 log(107). 2. a. Rule 2: log 3 5 log 15 5 5 log 15 2 log 5 b. Rule 3: log 1024 5 log(210) 5 10 log 2 c. Rule 3: log "31 5 log A311/2B 5 12 log 31 d. Rule 1: log 30 5 log(2 ? 3 ? 5) 5 log 2 1 log 3 1 log 5 e. Rule 2: log 81 2 log 27 5 log A81 27 B 5 log 3 or Rule 3: 4 log 3 2 3 log 3 5 log 3
d. log x 2 log 2 5 1 1 log 2x 5 1 1 101 5 2x 1 x 5 20 e. log x 2 logsx 2 1d 5 log 2 1 log x 2x 1 5 log 2 1 x x 2 1
5 2 1 x 5 2(x 2 1) 1 x 5 2
f. logs2x 1 1d 2 logsx 1 5d 5 0 1 1 1 log 2x x 1 5 5 log 1 1 2x 1 1 x 1 5
5 1 1 2x 1 1 5 x 1 5 1 x 5 4
7. log 103 2 log 102 5 3 log 10 2 2 log 10 5 3 2 2 5 1 log 103 log 102
3 log 10 5 2 log 10 5 32 5 1.5
Since 1 2 1.5 1 log 103 2 log 102 2
log 103 2
log 10
Algebra Aerobics 6.1c 1. a. 60 5 10 ? 2t 1 6 5 2t (dividing both sides by 10) log 6 5 log 2t (taking the log of both sides) log 6 5 t log 2 (using Rule 3 of logs) log 6/log 2 5 t 0.7782/0.3010 < t or t < 2.59 b. 500(1.06)t 5 2000 1 (1.06)t 5 2000 500 1 (1.06)t 5 4 1 log(1.06)t 5 log 4 1 log 4
t(log 1.06) 5 log 4 1 t 5 log 1.06 5 0.6021 0.0253 < 23.8 t t c. 80(0.95) 5 10 1 (0.95) 5 1/8 5 0.125 1 log (0.95)t 5 log 0.125 1 t log 0.95 5 log 0.125 1 log 0.125 t 5 log 0.95 < 20.9031 20.0223 < 40.5
Ch. 6
1. a. b. c. d.
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CH. 6 Exercises Solutions for Section 6.1
2. a. 7000 5 100 ? 2t 1 70 5 2t 1 log 70 5 log 2t 1 1.845 0.301
log 70 5 t log 2 1 t 5 log 70/log 2 < < 6.13 It will take 6.13 time periods or approximately (6.13)(20) min 5 122.6 min (or a little more than 2 hours) for the bacteria count to reach 7000. b. 12,000 5 100 ? 2t 1 120 5 2t 1 log 120 5 log 2t 1 log 120 5 t log 2 1 t 5 log 120/log 2 < 6.907 time periods or approximately (6.907)(20) min 5 138.14 min (or a little more than 214 hrs.) for the bacteria count to reach 12,000. 3. Using the rule of 70, since R 5 6% per yr, then 70/R 5 70/6 < 11.7 yr. More precisely, we have: 2000 5 1000(1.06)t 1 2 5 1.06t 1 log 2 5 log 1.06t 1 log 2 5 t log 1.06 1
Ch. 6
log 2
t 5 log1.06 < 0.3010 0.0253 < 11.9 yr 4. 1 5 100(0.976)t 1 0.01 5 (0.976)t 1 log 0.01 5 log (0.976)t 1 log 0.01 5 t log 0.976 1 22 t 5 log 0.976 5 20.0106 < 189 years, or almost 2 centuries! log 0.01
5. a. 30 5 60(0.95)t 1 0.5 5 (0.95)t 1 log(0.5) 5 log(0.95t) 1 log 0.5 5 t log 0.95 1 t 5 log 0.95 < 20.3010 20.0223 < 13.5 years for the initial amount to drop in half. b. 16 5 8(1.85)t 1 2 5 (1.85)t 1 log 2 log 2 5 log(1.85)t 1 log 2 5 t log 1.85 1 t 5 log 1.85 1 log 0.5
CH. 6 Exercises Solutions for Section 6.1
9. a. 12
b. 12
c. 12
11. If w 5 log(A) and z 5 log(B), then 10w 5 A and 10z 5 B. Thus A/B 5 10w/10z 5 10w2z and therefore log(A/B) 5 log(10w2z) 5 w 2 z 5 log(A) 2 log(B), as desired. 13. a. log Q
K3 R sK 1 3d 2
c. log AT 4 ? "T B 5 log A "T 9 B
b. log Q
s3 1 nd 5 R m
d. log a (xy2 ) 3 b 5 log A[xy2]28/3B
3
"xy2
15. Let w 5 log(A). Then 10w 5 A and thus Ap 5 (10w)p 5 10wp, and then log(Ap) 5 w ? p 5 p ? w 5 p ? log(A), as desired. 17. a. b. c. d. e.
1000 999 105/3 < 46.42 1/9 No solution, since x can not be negative.
19. a. Solve 300 5 100 ? 1.03t to get t 5 log(3)/log(1.03) < 37.17 years. b. Similarly, log(3)/log(1.07) < 16.24 years. 21. a. The graphs of ƒ(x) 5 500 ? (1.03)x and g(x) 5 4500 are in the accompanying diagram. b. x 5 75 is a good eyeball estimate. y 7000
t < 0.3010 0.2672 5 1.13 years for the initial amount to double.
g (x) = 4500
c. 500 5 200(1.045)t 1 2.5 5 (1.045)t 1 log 2.5 5 log(1.045)t 1 log 2.5 5 t log 1.045 1
t 5 log 0.983 1 t < 20.3010 20.0074 1 t < 40.6 days b. 0.25 5 0.5(0.92)t 1 0.5 5 (0.92)t 1 log 0.5 5 t log 0.92 1 log 0.5 t 5 log 0.92 1 t < 20.3010 20.0362 1 t < 8.3 hours c. 0.5A0 5 A0(0.89)t 1 0.5 5 (0.89)t 1 log 0.5
log 0.5
Exercises for Section 6.1 1. Student estimates will vary for each interest rate. a. About 38 years. b. About 16 years. 3. Eyeball estimates will vary. One set of guesses is: a. 1.3 hrs. b. 2.4 hrs. c. 5 hrs. 5. a. 0.001 7. a. 2 ? log 3
b. 106
c. 1
d. 10
b. 2 ? log 3 1 log 2
0 0
e. 0.1 c. 3 ? log 3 1 log 2
50
x 100
c. log(4500) 5 log(500) 1 x ? log(1.03) or x 5 [log(4500) 2 log(500)]/log (1.03) 5 log(9)/log(1.03) <74.33 d. The eyeball estimate and the logarithm-computed answer are very close. 23. a. b. c. d.
log 0.5 5 t log 0.89 1 t 5 log 0.89 1 t < 20.3010 20.0506 1 t < 5.9 years
(74.33, 4500)
f(x) = 500 • 1.03x
log 2.5
t 5 log 1.045 1 t < 0.3979 0.0191 1 t < 20.8 years for the initial amount to increase from 200 to 500. 6. a. 60 5 120(0.983)t 1 0.5 5 (0.983)t 1 log 0.5 5 t log 0.983 1
d. 2
Doubling time: x 5 log(2)/log(4) 5 0.5 years Half-life 5 log(0.5)/log(0.25) 5 0.5 years Doubling time 5 log(2)/log(4) 5 0.5 years Half life 5 log(0.5)/log(0.25) 5 0.5 years
25. a. N(t) 5 90(1.002)t, where N(t) is the number of articles (in thousands) for t days after October 23. b. The number of articles has doubled when N(t) 5 180 (thousand articles). So setting 180 5 90(1.002)t and solving for t, we get 2 5 (1.002)t 1 log 2 5 t log(1.002) 1 t 5 (log 2)/log(1.002) < 347 days, or 347/30 < 11.6 months or about one year. So if the rate of growth continues, every 12 months the number of articles will more than double. 27. a. t 5 log(2)/log(1.5) < 1.71 20-minute time periods or about 34 min. b. t 5 log(10)/log(1.5) < 5.68 20-minute time periods or about 114 min.
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CH. 6 Algebra Aerobics Solutions for Section 6.2
29. a. B(t) 5 B0(0.5)0.05t, where t is measured in minutes and B(t) and B0 are measured in some weight unit. None is specified in the problem. b. In one hour, t 5 60 and thus B(60) 5 B0 (0.5) 0.05?60 5 0.125 ? B0, or 12.5% is left. c. If its half-life is 20 minutes, then its quarter-life is 40 minutes. d. Solving 0.10 5 0.50.05t for t, we get 0.05t ? log(0.5) 5 log(0.10) or t 5 log(0.1)/[log(0.5) ? 0.05] < 66 minutes. 31. a. S 5 300 ? 0.9W if 0 # W # 10, where W is measured in weeks and S is measured in dollars. b. Solving S 5 150 for W means solving 0.5 5 0.9W for W 1 W 5 log(0.5)/log(0.9) < 6.6 weeks. The selling price when it would be given to charity is at W 5 10, when S 5 300 ? 0.910 < $104.60
log(2.952)
t < log(1.036) < 30.6 years after 1970, or sometime in late 2000. During that year U.S. fish production actually dropped to a little over 3000 million pounds, almost a million pounds less than the model predicted. b. Americans are actually increasing their fish and shellfish consumption, probably spurred on by doctors’ and nutritionists’ advice that “eating fish is good for you.” However, the amount of imported fish has grown substantially, probably cutting into U.S. fish production.
4. a. b. c. d. e. 5.
645
5e0.03t 5 5(e0.03)t < 5(1.030)t 3500e0.25t 5 3500(e0.25)t < 3500(1.284)t 660e1.75t 5 660(e1.75)t < 660(5.755)t 55,000e20.07t 5 55,000(e20.07)t < 55,000(0.932)t 125,000e20.28t 5 125,000(e20.28)t < 125,000(0.756)t n
1/n
1 1 1/n
(1 1 1/n)n
1 100 1000 1,000,000 1,000,000,000
1 0.01 0.001 0.000 001 0.000 000 001
2 1.01 1.001 1.000 001 1.000 000 001
2 2.704 813 829 2.716 923 932 2.718 280 469 2.718 281 827
The values for (1 1 1/n)n come closer and closer to the irrational number we define as e and are consistent with the value for e in the text of 2.71828. 6. a. b. c. d. e.
about $12,000 about $18,000 about $33,000 about 6 years 1 year: A 5 $11,255; 5 years: $18,061; 10 years: $32,620. doubling time: 1.034t 5 2 1 4t log 1.03 5 log 2 1 log 2
0.3010 t 5 4 log 1.03 1 t 5 4s0.0128d 1 t < 5.9 years 4?(18) 7. a. 8000 ? A1 1 0.08 5 $33,289 4 B
b. 8000 ? e0.08?(18) 5 $33,766 c. 8000 ? (1.084)18 5 $34,168 4?(30) 8. a. 8000 ? A1 1 0.08 5 $86,121 4 B
b. 8000 ? e0.08?(30) 5 $88,185 c. 8000 ? (1.084)30 5 $89,943
Section 6.2 Algebra Aerobics 6.2 1. a. $1000(1.085) 5 $1085 b. $1000 A1 1 0.085 4 B 5 $1087.75 0.085 5 $1088.72 c. $1000e 4
2. a. e0.04 5 1.0408 1 4.08% is effective rate b. e0.125 5 1.133 1 13.3% is effective rate c. e0.18 5 1.197 1 19.7% is effective rate 3. a. Principal 5 6000; nominal rate 5 5%; effective rate 5 5% since 1.051 5 1.05; number of interest periods 5 1 b. Principal 5 10,000; nominal rate 5 8%; effective rate < 8.24% since 1.024(1) < 1.0824; number of interest periods 5 4 c. Principal 5 500; nominal rate 5 12%; effective rate < 12.68% since 1.0112(1) < 1.1268; number of interest periods 5 12 d. Principal 5 50,000; nominal rate 5 5%; effective rate < 5.06% since 1.0252(1) < 1.0506; number of interest periods 5 2 e. Principal 5 125; nominal rate 5 7.6%; effective rate < 7.90% since e0.076 (1) < 1.0790; interest is continuously compounded
9. a. continuous growth rate: 0.6 or 60%; annual effective growth rate < 82.2% since e0.6(1) < 1.822, which is the growth factor. So 1.822 2 1 5 0.822 is the growth rate. b. continuous growth rate: 2.3 or 230%; annual effective growth rate < 897% since e 2.3(1) < 9.97, which is the growth factor. So 9.97 2 1 5 8.97 is the growth rate. 10. a. continuous decay rate: 0.055 or 5.5%; annual effective decay rate < 5.35% since e20.055 < 0.946, which is the decay factor. So 1 2 0.946 5 0.054 is the decay rate. b. continuous decay rate: 0.15 or 15%; annual effective decay rate < 13.9% since e20.15 < 0.861, which is the decay factor. So 1 2 0.861 5 0.139 is the decay rate.
Exercises for Section 6.2 1. a. 1.0353 b. 1.0355 c. 1.0356 d. 1.0356 e. These values represent the growth factors for compounding semi-annually, quarterly, monthly, and continuously. As the number of compoundings increase, the results come closer and closer to continuous compounding. 3. In general, if the nominal rate is 8.5%, then Ak(n) 5 10,000 ? (1 1 0.085/k)kn gives the value of $10,000 after n years if the interest is compounded k times per year and
Ch. 6
33. a. If we set 4000 1355(1.036)t then 4000/1355 5 (1.036) t 1 2.952 < 1.036t 1 log(2.952) < t log (1.036) 1
CH. 6 Exercises Solutions for Section 6.2
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CH. 6 Exercises Solutions for Section 6.2
Ac(n) 5 10,000 ? e0.085n gives that value if the interest is compounded continuously. a. annually: (1 1 0.085/1)1 5 1.0850, and thus the effective rate is 8.50%. b. semi-annually: (1 1 0.085/2)2 < 1.0868, and thus the effective rate is 8.68%. c. quarterly: (1 1 0.085/4)4 < 1.0877, and thus the effective rate is 8.77%. d. continuously: e0.085n < 1.0887, and thus the effective rate is 8.87%.
Ch. 6
5. a. A(t) 5 25,000 ? (1 1 0.0575/4)4t, where t 5 number of years b. B(t) 5 25,000 ? e0.0575t, where t 5 number of years c. A(5) 5 $33,259.12 and B(5) 5 $33,327.26 d. The effective rate for compounding quarterly is 4 Q1 1 0.0575 4 R 2 1 < 0.05875, and the effective rate for compounding continuously is e0.0575 2 1 < 0.05919. Thus continuous compounding has a slightly greater effective rate. 7. a. b. c. d. e.
e0.045 < 1.046 1.0680 , 1.0704 < e0.068 1.2690 . e0.238 < 1.2687 e20.10 < 0.9048 . 0.9000 0.8607 < e20.15
9. a. U(x) 5 10 ? (1/2)x/5, where x is measured in billions of years. b. 5 5 10 ? (1/2)x/5, and thus 1/2 5 (1/2)x/5 1 1 5 x/5 1 x 5 5 billion years. 11. a. t 5 log(2)/log(1.12) < 6.12 years b. t 5 log(2)/log[(1 1 0.12/4)4] < 5.86 years. 13. a. The 10-year decay factor is 0.85. The yearly decay factor is 0.851/10 < 0.9839. The yearly decay rate is 1 2 0.9839 < 0.0161 or 1.61%. b. g(t) 5 1.5 ? 0.9839t, where g(t) is measured in millions and t in years. c. h(t) 5 1.5 ? e–0.01625t, where h(t) is measured in millions and t in years. d. g(20) < 1.0842 million and h(20) < 1.0838 million. The h function decays a bit faster than the g function, but they are quite close. 15. a. P(t) 5 500 ? (1.0202)t; the continuous growth rate is 2% and the effective annual growth rate is 5 2.02%. b. N(t) 5 3000 ? (4.4817)t; the continuous growth rate is 150% and the effective growth rate is 348.17%. c. Q(t) 5 45 ? (1.0618)t; the continuous growth rate is 6% and the effective growth rate is 6.18%. d. G(t) 5 750 ? 1.0356t; the continuous growth rate is 3.5% and the effective growth rate is 3.56%. 17. a. 0.0343 5 er 2 1, and thus r 5 log(1.0343)/log(e) < 0.0337, so the continuous nominal interest rate is 3.37%. b. 0.046 5 er 2 1, and thus r 5 log(1.046)/log(e) < 0.04497, so the continuous nominal interest rate is 4.497% or 4.5% rounded off.
CH. 6 Algebra Aerobics Solutions for Section 6.3
Section 6.3 Algebra Aerobics 6.3 1. a. ln e2 5 2 ln e 5 2(1) 5 2 b. ln 1 5 0 c. ln 1e 5 ln 1 2 ln e 5 0 2 1 5 21 d. ln e12 5 ln 1 2 ln e2 5 ln 1 2 2 ln e 5 0 2 2(1) 5 0 2 2 5 22 e. ln "e 5 ln e1/2 5 21 ln e 5 12 (1) 5 1/2 2. a. ln "xy 5 ln(xy) 1/2 5 12 ln(xy) 5 12 [ln x 1 ln y]
B 5 lns3x 2 d 2 lnsy 3 d b. ln A3x y3 5 ln 3 1 2 ln x 2 3 ln y 2
c. ln A sx 1 yd 2 sx 2 yd B 5 lnsx 1 yd 2 1 lnsx 2 yd 5 2 lnsx 1 yd 1 lnsx 2 yd sx 1 2d 1/2
"x 1 2 1/2 d. ln xsx 2 1d 5 ln xsx 2 1d 5 ln(x 1 2) 2 ln x(x 2 1)
5 12 lnsx 1 2d 2 ln xsx 2 1d 5 12 lnsx 1 2d 2 [ln x 1 lnsx 2 1d] 5 12 lnsx 1 2d 2 ln x 2 lnsx 2 1d 3. a. ln x(x 2 1) sx 1 1d
b. ln x 2 c. ln x 2 2 ln y 3 5 ln xy 3 d. lnsx 1 yd 1/2 5 ln "x 1 y e. ln x 2 2 ln(2x 2 1) 5 ln x 2 ln(2x 2 1)2 5 ln
x 2 s2x 2 1d
4. growth factor 5 er 5 1 1 0.064 5 1.064 1 ln er 5 ln 1.064 1 r 5 ln 1.064 5 0.062 or 6.2% 5. 50,000 5 10,000 e0.078t 1 5 5 e0.078t 1 ln 5 5 0.078t 1 t 5 ln 5/0.078 < 20.6 yr 6. a. ln e x11 5 ln 10 1 (x 1 1)ln e 5 ln 10 1 (x 1 1)(1) < 2.30 1 x < 21 1 2.30 1 x < 1.30 b. ln e x22 5 ln 0.5 1 (x 2 2) ln e 5 ln 0.5 1 x 2 2 < 20.69 1 x < 2 2 0.69 1 x < 1.31 7. a. b. c. d. e. f.
true, since ln 81 5 ln 34 5 4 ln 3 false; ln 7 5 ln 14 2 5 ln 14 2 ln 2 true, since ln 35 5 ln(5 ? 7) 5 ln 5 1 ln 7 false; 2 ln 10 5 ln 102 5 ln 100 true, since ln e1/2 5 21 ln e 5 12 false: 5 ln 2 5 ln 25 5 ln 32
8. a. ln 2 1 ln 6 5 x 1 ln(2 ? 6) 5 x 1 ln 12 5 x 1 x < 2.48 b. ln 2 1 ln x 5 2.48 1 ln 2x 5 2.48 1 2.48
e2.48 5 2x 1 x 5 e 2 < 5.97 c. ln(x 1 1) 5 0.9 1 e0.9 5 x 1 1 1 x 5 e0.9 2 1 < 1.46 d. ln 5 2 ln x 5 20.06 1 ln 5x 5 20.06 1 5 < 5.3 e20.06 5 5x 1 x 5 e20.06
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CH. 6 Exercises Solutions for Section 6.3
CH. 6 Exercises Solutions for Section 6.4
Exercises for Section 6.3 1. a. b. c. d. e. f.
ln(A ? B) 5 ln(A) 1 ln(B) or Rule 1 ln(A/B) 5 ln(A) 2 ln(B) or Rule 2 ln(Ap) 5 p ? ln(A) or Rule 3 Rules 1 and 3 Rules 1 and 3 Rules 2 and 1
3. a. 10 n 5 35
b. ex 5 75
5. a. x 5 5 ? 2 5 10 b. x 5 24/2 5 12 c. x 5 11
d. N 5 N0 ? e2kt
c. e3/4 5 x
d. x 5 8 ? 36 5 288 e. x 5 64/9 < 7.1 f. x 5 16/8 5 2
647
b. The graph of log x reaches a y value of 1 when x 5 10 while the graph of ln x reaches a value of 1 when x 5 e (or approximately 2.7). The graph of log x reaches 2 at x 5 100 while the graph of ln x reaches 2 at x 5 e2 < 7.4. c. For 0 , x , 1, both graphs lie below the x-axis and approach the y-axis asymptotically as x gets closer to 0. d. To the right of x 5 1, both graphs lie above the x-axis, with the ln x graph rising slightly faster than the log x graph, and thus staying above the log x graph. 4. f and its inverse f 21 along with the dotted line for y x: y 3
f
–1
(x) f(x)
–3
3
x
–3 3 4 9. a. ln Q "sx 1 1dsx 2 3dR b. ln Q R R
"P
c. ln QNN2R 0
11. a. r 5 ln(1.0253) < 0.025 c. x 5 ln(0.5)/3 < 20.231 b. t 5 ln(3)/0.5 < 2.197 13. a. x 5 ln(10) < 2.303 b. x 5 log(3) < 0.477 c. x 5 log(5)/log(4) < 1.161 d. x 5 e5 < 148.413 e. x 5 e3 2 1 < 19.086 f. no solution [ln(24/3) not defined]
5. a. b. c. d. e. f. g. h.
log 103 3 log 1025 5 25 3 log 100.09 5 3(0.09) 5 0.27 10log 3.4 5 3.4 ln e5 5 ln e0.075 0.07 ln e3.02 1 ln e20.27 5 3.02 2 0.27 5 2.75 eln 0.9 5 0.9
15. Let w 5 ln(A) and z 5 ln(B). Thus ew 5 A and ez 5 B. Therefore A ? B 5 ew1z and therefore ln(A ? B) 5 w 1 z 5 ln(A) 1 ln(B), as desired.
6. Acidic. If 4 5 2log[H1], then 24 5 log[H1]. So [H1] 5 1024. Since the pH is 3 less than pure water’s, it will have a hydrogen ion concentration 103 or 1000 times higher than pure water’s.
17. t 5 [ln(100,000) 2 ln(15,000)]/0.085 < 22.3 years
10212 R 5 10 log(1.5 ? 104) 7. N 5 10 log Q1.5 ? 216
19. 1.0338 5 er, and thus the nominal continuous interest rate r 5 ln(1.0338) < 0.0332 5 3.32%. 21. a. r 5 ln(1.025) < 0.0247 b. r 5 ln(0.5) < 20.6931 23. a. growth; 37% b. growth; 115%
c. r 5 ln(1.08) < 0.0770
c. decay; 19% d. decay; 120%
e. growth; 56% f. decay; 29%
10
5 10(log 1.5 1 log 104) < 10(0.176 1 4) 10(4.176) < 42 dB 8. Multiplying the intensity by 100 5 102 corresponds to adding 20 to the decibel level. Multiplying the intensity by 10,000,000 5 107 corresponds to adding 70 to the decibel level.
Exercises for Section 6.4
Section 6.4
1. Since log(5x) 5 log(5) 1 log(x) and log(5) . 0, the graph of log(5x) (graph A) is above the graph of log(x) (Graph B).
Algebra Aerobics 6.4 1. a. Since log x2 5 2 log x (by Rule 3 of logarithms), the graphs of y 5 log x2 and y 5 2 log x will be identical (assuming x . 0). 2. The graph of y 5 2ln x will be the mirror image of y 5 ln x across the x-axis. y 2
y = ln x
3. a. A 5 log(x), B 5 log(x 2 1), and C 5 log(x 2 2); so ƒ(x) matches A, g(x) matches B, and h(x) matches C. b. log(1) 5 0; thus f (1) 5 g(2) 5 h(3) 5 0. c. f has 1 as its x-intercept; g has 2 and h has 3. d. The graph moves from crossing the x-axis at x 5 1 to crossing it at x 5 k. Assuming that k . 0, then ƒ(x 2 k) is the graph of f moved k units to the right.
x –2 –2
y = –ln x
10
3. a. The graphs are very similar (see Figure 6.5). They intersect at (1, 0), which is the x-intercept of each graph.
5. a. The graphs of f and h are mirror images of each other across the y-axis, as are the graphs of g and k. b. The graphs of f and g are mirror images of each other across the x-axis, as are the graphs of h and k.
Ch. 6
7. a. 12 sln 4 1 ln x 1 ln yd c. ln s3d 1 34 ln sxd b. 13 sln 2 1 ln xd 2 ln s4d
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CH. 6 Algebra Aerobics Solutions for Section 6.5
7. The table for log3(x) is: x 1/9 1/3 y 22 21
1 0
CH. 6 Exercises Solutions for Section 6.5
Exercises for Section 6.5
The table for log4(x) is: x 1/16 1/4 y 22 21
3 9 1 2
1 0
4 1
16 2
1. a. N 5 10 e(ln 1.045)t 5 10 ? e0.0440t b. Q 5 5 ? 1027 ? e(ln 0.072)A 5 5 ? 1027 ? e22.631A c. P 5 500 ? e(ln 2.10)x 5 500 ? e0.742x 3. a. decay
Graph of y = log3(x) y 3
Graph of y = log4(x) y 3 y = log4(x)
y = log3(x)
(16, 2)
(9, 2) (3, 1)
(4, 1) x 10
0 (1, 0) 1 ( , –1) 3
0 (1, 0) 1 ( , –1) 4
1 ( , –2) 9
(
Ch. 6
–3
x 20
1 , –2) 16
–3
9. dB 5 10 ? log(I/I0) 5 28 implies that I/I0 5 102.8 and thus I 5 10213.2 watts/cm2, and dB 5 92 has I/I0 5 109.2 and thus I 5 1026.8 watts/cm2. (Note that these answers assume, of course, that I0 5 10216 watts/cm2.) 11. If I is the intensity of one crying baby, then 5I is the intensity of five crying babies. Thus the perceived noise in decibels is 10 ? log[5(I/I 0)] 5 10 ? log(5) 1 10 log(I/I 0) < 6.99 1 10 ? log(I/I0) < 7 1 noise of one baby crying. Thus quintuplets crying are about 7 decibels louder than one baby crying. 13. Total volume 5 214 5 94 cups or equivalently 9 quarter cups. So lemon juice is 19th of the mixture volume and water is 8 ths 9 of the volume. Since the lemon juice has a pH of 2.1, its hydrogen ion concentration is 10 22.1 (moles per liter). Similarly, since the tap water has a pH of 5.8, its ion concentration is 1025.8. So the hydrogen ion concentration of the mixture 5 [H1] 5 A19B ? 1022.1 1 A89B ? 1025.8 1 pH 5 2log[ A19B ? 1022.1 1 A89B ? 1025.8] < 3.05. Thus the mixture is slightly less acidic than orange juice (with a pH of 3).
b. decay
c. growth
5. a. t 5 5 years; P 5 P0e0.1386t b. t 5 25 years; P 5 P0e0.0277t c. t 5 1/2 year; P 5 P0e1.3863t 7. If 200 5 760 ? e20.128h, then h 5 ln(200/760)/(20.128) < 10.43 km. 9. The half-life is ln(1/2)/(2r) 5 [ln(1) 2 ln(2)]/(2r) 5 [0 2 ln(2)]/(2r) 5 ln(2)/r 5 100 ? ln(2)/R < 69.3137/R, which is approximately 70/R. 11. a. f goes with C
b. g goes with B
c. h goes with A
13. a. It is a 2% nominal continuous decay rate. b. 2500 ? e20.02t 5 2500 ? 0.5t/n 1 20.02t 5 (t/n) ? ln(0.5) 1 n 5 ln(0.5)/(20.02) < 34.66. c. It represents the half-life in whatever time units t is measured in. 15. a. Since Q(8000) 5 0.5Q0, the half-life is 8000 years. b. The annual decay rate is 1 2 0.51/8000 < 0.000 086 639 6. c. Solving er 5 0.51/8000 for r gives: r < 20.000 086 643 4 as the nominal continuous decay rate. 17. Answers from students will vary. Look for an exponential curve that goes roughly through the middle of each cluster. One such curve contains the points (500, 1000) and (2000, 300). Then the best-fit exponential through these two points is y 5 1494(0.9992)d, where d measures depth in meters and y measures species density in an unknown unit. (Note that in base e, one gets y 5 1494e–0.0008d.) 19. Given:
.
.
n 5 n0 ? eln(2) t/T and 11 5 n0 ? eln(2) 2/T and 30 5 n0 ? eln(2) ? 22/T.
Section 6.5
Therefore:
ln(11) 5 ln(n0) 1 2 ? ln(2)/T and ln(30) 5 ln(n0) 1 22 ? ln(2)/T
Algebra Aerobics 6.5
Thus
ln(30) 2 ln(11) 5 ln(2)[22 2 2]/T
1
T 5 20 ? ln(2)/[ln(30) 2 ln(11)] < 13.863/1.003 < 13.8 seconds, the reactor period
1. a. decay b. decay
c. growth d. growth
e. decay f. decay
2. a. ek 5 1.062 1 k 5 ln 1.062 < 0.060 1 y 5 1000e0.06t b. ek 5 0.985 1 k 5 ln 0.985 < 20.015 1 y 5 50e20.015t 3. a. continuous nominal rate < 5.45% since ln 1.056 < 0.545; effective rate 5 5.6% since eln 1.056 5 1.056 b. continuous nominal rate < 3.34% since ln 1.034 < 0.0334; effective rate 5 3.4% since eln1.034 5 1.034 c. continuous nominal rate < 7.97% since ln 1.083 < 0.0797; effective rate 5 8.3% since eln1.083 5 1.083 d. continuous nominal rate < 25.85% since ln 1.295 < 0.2585; effective rate 5 29.5% since eln1.295 5 1.295 4. a. growth factor 5 e 0.08 < 1.083 b. decay factor 5 e20.125 < 0.883
If we substitute 13.8 for T in the equation 11 5 n0 ? e(2 ?ln 2)/T, we get 11 5 n0 ? e(2? ln 2)/13.8 < n0 ? e0.100456 < n0 ? 1.1057. So n0 < 9.95 or 10 neutrons. 21. a. Newton’s Law here is of the form T 5 75 1 Ce2kt where A 5 758, the ambient temperature, and T is the temperature of the object at time t. When t 5 0, T 5 1608, so 160 5 75 1 Ce0 1 C 5 85. So the equation becomes T 5 75 1 85e2kt. When t 5 10, T 5 1008 so 100 5 75 1 85e210k 1 e210k 5 25 85 < 0.2941 1 ln(e210k ) 5 ln(0.2941) 1 210k < 21.224 1 k < 0.1224. So the Law of Cooling in this situation is: T 5 75 1 85e20.1224t
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CH. 6 Algebra Aerobics Solutions for Section 6.6
b.
CH. 6 Exercises Solutions for Section 6.6
5. a. b. c. d.
Temperature of Hot Tea Temperature, T (in degrees F)
180 160 140
T = 75 + 85e
–0.1224t
120 (10, 100)
100
(20, 82)
a < 2.00 since 100.301 < 2 C < 524.81 since 102.72 < 524.81 a < 0.75 since 1020.125 < 0.75 C < 100,000 since 105 5 100,000
6. a. y 5 100.301 ? 100.477x 1 y 5 2 ? 3x b. y 5 103 ? 100.602x 1 y 5 1000(4) x c. y 5 101.398 ? 1020.046x 1 y 5 25 ? s0.90x d
80 60 T = 75
40
649
20 0 0
5
10
15
20
7. Exponential functions for both since the graphs on semi-log plots are approximately linear.
25
c. When t 5 20 minutes, then the temperature of the tea T 5 75 1 85e20.1224 ? 20 < 828.
8. Using the point (0, 4), the vertical intercept, and (2, 5), the 4 5 0.5. So log y 5 4 1 0.5x 1 y 5 10410.5x 1 slope is 5 2 2 y 5 104 ? 100.5x < 104 s3.16d x.
Section 6.6 1. a. The graph of y 5 3x 1 4, a linear function, is a straight line on a standard linear plot. The graph of y 5 4 ? 3x, an exponential function, is a straight line graph on a semi-log plot. The equation log y 5 (log 3) ? x 1 log 4 is equivalent to y 5 4 ? 3x, whose graph is a straight line on a semi-log plot. b. The graph of y 5 3x 1 4 has slope 5 3 and vertical intercept 5 4. The equations y 5 4 ? 3x and log y 5 (log 3) ? x 1 log 4 are equivalent; their graphs have slope 5 log 3 and vertical intercept 5 log 4 on a semi-log plot. 2. a. log y 5 log C5(3) xD 1 log y 5 log 5 1 x(log 3) b. log y 5 log C 1000(5) x D 1 log y 5 log 1000 1 x log 5 5 3 1 x log 5 c. log y 5 log C10,000(0.9) D 1 log y 5 log 10,000 1 x log 0.9 5 4 1 x log 0.9 x
d. log y 5 log C5 ? 106 (1.06) xD 1 log y 5 log 5 1 log 106 1 x log 1.06 5 log 5 1 6 1 x log 1.06 3. a. log y 5 log 7 1 slog 2dx 1 log y 5 logs7 ? 2x d 1 x 10log y 5 10log 7?2 1 y 5 7 ? 2x b. log y 5 log 20 1 slog 0.25dx 1 log y 5 log (20 ? 0.25x ) 1 x 10log y 5 10log s20?0.25 d 1 y 5 20 ? 0.25x x
c. log y 5 6 1 Alog 3B ? x 1 10log y 5 1061log s3 d 1 x
y 5 106 ? 10log s3 d 1 y 5 106 ? 3x d. log y 5 6 1 log 5 1 Alog 3B ? x 1 log y 5 6 1 log 5 1 log (3x ) 1 x x 10log y 5 1061log 51log 3 1 y 5 106 ? 10log 5 ? 10log 3 1 y 5 106 s5d3x 5 5 ? 106 s3d x 4. a.
i. ii. iii. iv. b. i. ii. iii. iv.
slope 5 log 5; vertical intercept 5 log 2 slope 5 log 0.75; vertical intercept 5 log 6 slope 5 log 4; vertical intercept 5 0.4 slope 5 log 1.05; vertical intercept 5 3 1 log 2 y 5 2 ? 5x y 5 6 ? s0.75d x y 5 100.4(4)x y 5 103 ? 2 ? 1.05x 5 2000s1.05d x
1. a. goes with f. b. goes with h. 3. a. b. c. d.
Ch. 6
Exercises for Section 6.6
Algebra Aerobics 6.6
c. goes with e. d. goes with g.
log(y) 5 4.477 1 0.301x log(y) 5 3.653 1 0.146x log(y) 5 6.653 2 0.155x log(y) 5 3.778 2 0.244x
5. a. goes with Graph B. b. goes with Graph C.
c. goes with Graph A. d. goes with Graph D.
7. Here are the average rates of change: a. x
Y 5 log(x)
Avg. Rt. of Ch.
b. x
0 10 20 30 40
2.30103 4.30103 4.90309 5.25527 5.50515
n.a. 0.20000 0.06021 0.03522 0.02499
0 4.77815 10 3.52876 20 2.27938 30 1.02999 40 20.21945
Y 5 log(x)
Avg. Rt. of Ch. n.a. 20.12494 20.12494 20.12494 20.12494
a. This is not exponential; the average rate of change between consecutive points keeps decreasing. b. The average rate of change over each decade is 20.12494. Thus the plot of Y vs. x is linear and therefore y is exponential. The exponential function is y 5 60,000(0.75)x. 9. a. The scatter plot of points (x, log(y)), where Y 5 log(y), is nearly linear; thus the growth is very close to being exponential. b. The equation of the best-fit line is given in the graph and the corresponding exponential equation is approximately y 5 38.788 ? 1.3137x. The daily growth rate is 31.37%. 11. a. Approximately 12.2 micrograms/liter. b. Yes, because its log graph is a straight line with negative slope. The decay factor is 1020.0181 < 0.959, and thus the decay rate is 1 2 0.959 5 0.041 or 4.1%. c. The exponential model is approximately y 5 12.2 ? 0.959 x. 13. a. To find out the average health care costs per person, one would divide the total costs by the population estimate for each year. (See Table 2.1 for U.S. population counts.)
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CH. 6 Check Your Understanding
CH. 6 Review: Putting It All Together
b. On a semi-log plot (where years are reinitialized at 1960) the data look rather linear, so it is reasonable to model the growth with an exponential function.
Expenditures (in billions of $)
U.S. Health Care Expenditures from 1960 to 2005 10000 1000 100 10 1 0
10
20
30
40
50
Ch. 6
Years since 1960
c. The graph shows the best-fit exponential, whose equation is H(t) 5 31(1.102)t, where H(t) is the health care costs (in billions) t years after 1960. In 2005, when t 5 45, H(45) 5 31(1.102)45 < 2450 billion or 2.45 trillion dollars. The estimate exceeds the actual cost in 2005 of 2016 billion by 434 billion dollars.
Ch. 6: Check Your Understanding 1. True
8. False
15. True
22. False
2. True
9. False
16. False
23. True
3. False
10. False
17. False
24. False
4. False
11. True
18. True
25. True
5. True
12. False
19. False
26. False
6. False
13. True
20. False
27. True
7. True
14. False
21. False
28. False
b. 100 5 200 ? 0.92t 1 0.5 5 0.92t 1 log 0.5 5 t log 0.92 1 (log 0.5)/(log 0.92) 5 t 1 t < (20.301)/(20.036) < 8.4 months (or about 250 days) for the quantity to halve. 3. Expressions (a) and (e) are equivalent and (b), (c), and (f) are equivalent. Expression (d) does not match any other. 5. a. After one time period (20 minutes), A(1) 5 325 ? (0.5)1 5 162.5 mg (or half the original amount). After two time periods (40 minutes), A(2) 5 325 ? (0.5)2 5 81.25 mg (or one-quarter of the original amount). b. Using the graph, it appears that after about 1.7 time periods (34 minutes) A(t) 5 100. Using the equation, we have 100 5 325 ? (0.5)t 1 100/325 5 (0.5)t 1 0.3077 < (0.5) t 1 log (0.3077) < t log(0.5) 1 t < log (0.3077)/log(0.5) 1 t < 1.70 time periods, or 1.7 ? 20 5 34 minutes. So our estimate was accurate. c. B(t) 5 81 ? (0.5)t. So A(t) and B(t) have different initial amounts, but the decay rate is the same. 7. a. Density for India 5 (1.08 ? 109)/(1.2 ? 106) 5 0.9 ? 103 5 900 people per square mile. Density for China 5 (1.30 ? 109)/(3.7 ? 106) < 0.35 ? 103 5 350 people per square mile. So India’s population density is about 900/350 < 2.6 times larger than China’s. b. C(x) 5 1.30(1.006)x and I(x) 5 1.08(1.016)x, where C(x) and I(x) are in billions and x 5 years since 2005. c.
Projected Populations of India and China Population (in billions)
650
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1.8000 1.7000
India China
1.6000 1.5000 1.4000 1.3000 1.2000 1.1000 1.0000 0
10
20
30
Years since 2005
29. True 30. True 31. Possible answer: y 5 12 slog x 2 log 3d 32. Possible answer: y 5 50.3(1.062)t 33. Possible answer: y 5 100e20.026t 34. Possible answer: F(t) 5 3.3(0.989)t, t 5 number of years since 1966, F(t) in millions 35. Possible answer: y 5 100(1.20)x 36. False
37. False
38. True
39. True
40. True
Ch. 6 Review: Putting It All Together 1. a. 100 5 50 ? 1.16t 1 2 5 1.16t 1 log 2 5 t log 1.16 1 (log 2)/(log 1.16) 5 t 1 t < 0.301/0.064 < 4.7 months (or about 141 days) for the quantity to double.
Note that India’s exponential growth rate is so much higher than that for China, the graph for the Chinese population appears almost linear in comparison. d. The projected populations are the same roughly 18 years after 2005, or in 2023. Using the models, we need the point at which C(x) 5 I(x) 1 1.30(1.006)x 5 1.08(1.016)x 1 1.204(1.006)x < (1.016)x 1 log(1.204) 1 x log(1.006) < x log(1.016) 1 log(1.204) < x [log(1.016) 2 log(1.006)] 1 0.081 < xlog[(1.016)/(1.006)] 1 0.081 < 0.0043x 1 x < 18.8 years after 2005, that is, in late 2023 our model predicts the two populations will be the same. Evaluating C(18.8), we get 1.30(1.006)18.8 < 1.455 billion people. (To double-check you could calculate I(18.8) < 1.456 billion, with the difference due to rounding.) 9. a. A(x) 5 10,000(1.07)x, where A(x) is the dollar amount after x years. b. A(18) 5 10,000(1.07)18 < $33,799 c. 10,000(1.03)18 < $17,024, the difference 5 $33,799 2 $17,024 5 $16,775 d. Investing $10,000 at 4% for 18 years gives $10,000(1.04)18 < $20,258, which is not equivalent to the
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CH. 6 Review: Putting It All Together
result in part (c). However it does represent the amount in the account after 18 years in current dollars (dollars that have been adjusted for inflation). 11. a. t 5 log(2.3) < 0.362 b. t2 ? 4 5 102 1 t2 5 25 1 t 5 5 (Note: t 5 25 is not a solution here, since log (25) is not defined.) c. 2 5 e0.03t 1 ln 2 5 0.03t 1 t 5 (ln 2)/0.03 < 0.693/0.03 < 23.1 d. ln [(2t 2 5)/(t 2 1)] 5 0 1 (2t 2 5)/(t 2 1) 5 1 1 2t 2 5 5 t 2 1 1 t 5 4 13. a. 1. 5 5 eln 1.5 < e0.405 b. 0.7 5 eln 0.7 < e –0.357 c. 1 5 e0
651
CHAPTER 7 Section 7.1 Algebra Aerobics 7.1 1. a. S(r) 5 4pr 2; S(2r) 5 4p(2r)2 5 4 ? (4pr 2) 5 4S(r); S(3r) 5 4p(3r)2 5 9 ? (4pr 2) 5 9S(r) b. When the radius is doubled, the surface area is multiplied by 4. When the radius is tripled, the surface area is multiplied by 9. 2. a. V(r) 5 (4/3)pr 3; V(2r) 5 (4/3)p(2r)3 5 8 ? ((4/3)pr 3) 5 8V(r); V(3r) 5 (4/3)p(3r)3 5 27 ? ((4/3)pr 3) 5 27V(r) b. When the radius is doubled, the volume is multiplied by 8. When the radius is tripled, the volume is multiplied by 27. 3. The volume grows faster than the surface area, so as the radius of a sphere increases, the ratio of (surface area)/volume decreases.
17. a. Matches Graph B since when x 5 1, 2 ln x 5 2 ln 1 5 2 ? 0 5 0. So the horizontal intercept is (1, 0). b. Matches Graph A since when x 5 1, 2 1 ln x 5 2 1 ln 1 5 2 1 0 5 2. So the graph passes through (1, 2). c. Matches Graph D since when x 5 0, ln (x 1 2) 5 ln 2 < 0.693 . 0. So the vertical intercept is above the origin at approximately (0, 0.693). d. Matches Graph B since the functions y1 5 2 ln x and y4 5 ln(x2) are the same. (Note: There is no match to Graph C.)
4. a. C(r) 5 2pr; C(2r) 5 2p(2r) 5 2(2pr) 5 2C(r); C(4r) 5 2p(4r) 5 4(2pr) 5 4C(r). So when the radius is doubled, the circumference is multiplied by 2. When the radius quadruples, the circumference is multiplied by 4. b. A(r) 5 pr 2; A(2r) 5 p(2r)2 5 4(pr 2) 5 4A(r); A(4r) 5 p(4r)2 5 16(pr 2) 5 16A(r). So when the radius is doubled, the area is multiplied by 4. When the radius is quadrupled, the area is multiplied by 16. c. The area grows faster than the circumference. So as the radius of the circle increases, the ratio of circumference/ area decreases.
19. a. Using the table in the text, five orders of magnitude. b. Using the function definition, if 30 5 10 log (I30/I0), where I30 5 intensity level corresponding to 30 decibels, and 80 5 10 log (I80/I0), where I80 5 intensity level corresponding to 80 decibels. Subtracting the two equations gives:
5. a. The volume of the sphere is equal to the volume of the cube when 43pr 3 5 r 3, which is true only if r 5 0. b. The volume of the cube is greater than the volume of the sphere if r 3 . 43pr 3 1 1 . 43p, which is never true. The volume of the cube is less than the volume of the sphere for all r . 0 since r 3 , 43pr 3 1 1 , 43p.
80 2 30 5 10 log(I80/I0) 2 10 log(I30/I0) 1 50 5 10 log[(I80/I0 )/(I30/I0)] 1 5 5 log [(I80/I0) ? (I0/I30)] 1 5 5 log[(I80/I30)] 1 105 5 I80/I30,
6. No, since if the radius is doubled from 5 to 10, the volume is increased by a factor of 4. That is, given V 5 pr 2h, if r 5 5 then V 5 p52 ? 25 5 625p cubic feet and if r 5 10 then V 5 p102 ? 25 5 2500 p cubic feet, which is a factor of 4 larger.
so I80 is five orders of magnitude larger than I30.
7. Yes; since V 5 pr 2h, if the height is doubled, the volume is doubled. That is, for this example, if r ⫽ 5, then V 5pr 2h 1 V 5 p122 ? 5 5 720p cubic feet and if r ⫽ 10, then V 5 p122 ? 10 5 1440 p cubic feet, which is twice the volume.
21. If R 5 6.9, we have 6.9 5 log(A/A0) 1 106.9 5 A/A0, so the earthquake’s amplitude was 6.9 orders of magnitude (or almost 10 million times) larger than that of the base amplitude, A0. 23. a. The data appear to be exactly linear, and two estimated points on the best-fit line are (0, 0.3) and (10, 5). The slope 5 (5 2 0.3)/(10 2 0) 5 0.47. The equation is then Y 5 0.3 1 0.47x. b. Substituting log y for Y gives log y 5 0.3 1 0.47x 1 y 5 100.310.47x 1 y 5 100.3100.47x < 2 ? 3x. c. The function is exponential, suggesting that the graph of exponential functions is a straight line on a semi-log plot (where the log scale is on the vertical axis).
8. a. i. The area B of the cylinder base 5 pr 2 1 V 5 spr 2 dh 5 pr 2h ii. The area B of the triangular prism base 5 12ab 1 V 5 A12 abB h 5 12 abh b. If the height is doubled, the volume is doubled since (pr 2 )2h 5 2(pr 2 )h 5 2V; A12 abB 2h 5 2 A12 abB h 5 2V c. If all dimensions are doubled, the volume is increased by a factor of 8 since sps2rd 2 d ? 2h 5 (p4r 2)2h 5 8(pr 2)h 5 8V; A12 s2ads2bd B (2h) 5 23 ? A12 abhB 5 8V
Ch. 7
15. a. Matches Graph C b. Matches Graph D c, d. Both match Graphs A and B since y 5 100(10)x 1 log y 5 log 100 1 x log 10 1 log y 5 2 1 x.
CH. 7 Algebra Aerobics Solutions for Section 7.1
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CH. 7 Exercises Solutions for Section 7.1
CH. 7 Exercises Solutions for Section 7.2
b. gs5d 5 5s5d 3 5 5s125d
Exercises for Section 7.1 1. a. l 5
V wh
c. w 5
P 2 2l 2
5 625
2 d. h 5 S 24x2x
b. b 5 2A h
3. a. r/2; increasing; as the radius increases, the area increases faster than the circumference. 4 b. 3r ; increasing; as the radius increases, the volume of the sphere increases faster than the area of its radial cross-section. c. 23r 2; increasing; as the radius increases, the volume of the sphere increases faster than the length of any of its great circles.
Ch. 7
5. a. S < 1.26 ? 10219 m2 b. V < 4.19 ? 10230 m3 7. a. b. c. d.
c. S/V < 3.0 ? 1010 m21 d. ratio 5 3/r; decreases
S is multiplied by 16; V is multiplied by 64. S is multiplied by n2; V is multiplied by n3. S is divided by 9; V is divided by 27. S is divided by n2; V is divided by n3.
9. a. V 5 2632.5 cm3 b. quadrupled c. Quadruple either the length or the width or double both. 11. a. The volume doubles if the height is doubled. b. The volume is multiplied by 4 if the radius is doubled. 13. a. V 5 3pr 3; S 5 8pr 2 b. Volume eventually grows faster since V/S 5 3r/8; as r S 1 ` , V/S increases without bound. 15. The ratio of area to circumference is r/2 and its graph is labeled h(x); the inverse ratio is 2/r and its graph is labeled j(x).
Section 7.2 Algebra Aerobics 7.2 1.
Power Independent Dependent Constant of Function Variable Variable Proportionality Power a. b. c. d. e.
yes yes no no yes
r z — — x
p 1 — — 3
A y — — y
2 5 — — 5
2. a. y is directly proportional to x2. b. y is not directly proportional to x2. c. y is not directly proportional to x. 3. gsxd 5 5x 3 a. gs2d 5 5s2d 3 5 5s8d 5 40
gs10d 5 5s10d 3 5 5s1000d
c. gs2xd 5 5 5 5
5 5000
So gs10d is eight times larger than gs5d
5s2xd 3 5s8x 3 d 8s5x 3 d 8gsxd So gs2xd is eight times larger than gsxd
d. g A12xB 5 5 A12xB 3 5 5 A18x3B 5 18 s5x3 d 5 18gsxd
So g A12xB is one-eighth the size of g(x)
4. a. hs2d 5 0.5s2d 2 5 2 and hs6d 5 0.5s6d 2 5 18, an increase by a factor of 9. b. hs5d 5 0.5s5d 2 5 12.5 and hs15d 5 0.5s15d 2 5 112.5, an increase by a factor of 9. c. hsxd will increase by a factor of 9 since h(3x) 5 0.5(3x)2 5 9 ? [0.5x 2] 5 9 ? hsxd . d. hsxd will decrease by a factor of 9 since h A 13 xB 5 0.5 A13 xB 2 5 19 ? [0.5x 2] 5 19 ? hsxd . 5. a. V 5 k ? r 3 b. V 5 k ? l ? w ? h c. f 5 k ? c 6. a. y is equal to 3 times the fifth power of x. b. y is equal to 2.5 times the cube of x. c. y is equal to one-fourth of the fifth power of x. 7. a. y is directly proportional to x 5 with a proportionality constant of 3. b. y is directly proportional to x 3 with a proportionality constant of 2.5. c. y is directly proportional to x 5 with a proportionality constant of 1/4. 2 8. a. P 5 aR2 1 Pa 5 aRa 1 Pa 5 R2 1 "Pa 5 "R2 1 R 5 "Pa
b. V 5 A13B pr 2h 1 3sVd 5 3 A13B pr 2h 1 2 3V 5 pr 2h 1 3V2 5 pr 2h 1 h 5 3V2 pr
pr
pr
9. a. f s2xd 5 0.1s2xd 3 5 0.8x 3;
2f sxd 5 2s0.1x3 d 5 0.2x3
b. f s3xd 5 0.1s3xd 3 5 2.7x3;
3f sxd 5 3s0.1x3 d 5 0.3x3
c. f s2 ? 5d 5 0.1s10d 3 5 100; 2f s5d 5 2s0.1 ? 53 d 5 25
Exercises for Section 7.2 gs4d 5 5s4d 5 5s64d 5 320 So gs4d is eight times larger than gs2d 3
1. a. f(2) 5 20; f(22) 5 20 c. h(2) 5 220; h(22) 5 220 b. g(2) 5 40; g(22) 5 240 d. k(2) 5 240; k(22) 5 40 3. a. k 5 1/2; P 5 288 b. k 5 14; M 5 84
c. k 5 4p; T 5 1728p 3 d. k 5 "18; N 5 6
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CH. 7 Exercises Solutions for Section 7.2
CH. 7 Algebra Aerobics Solutions for Section 7.3
5. a. y 5 15 x;4 b. p 5 6 "s;24
c. h’s value is multiplied by 9. d. h’s value is divided by 9.
c. A 5 43 πr 3; 64p d. P 5 12 m 2; quartered, i.e., P is divided by 4. 7. a. b. c. d.
Y 5 kX3 k 5 1.25 Increased by a factor of 125 Divided by 8
e. X 5
c. f ssd 5 4s 3
3 proportional to "Y .
d. f s3sd 5 4s3sd 3 5 4s27ds 3 5 108s 3 2. a. gs2d 5 24s2d 3 5 24s8d 5 232 b. gs22d 5 24s22d 3 5 24s28d 5 32
b. M is multiplied by 2p.
9. a. L is multiplied by 32.
11. In all the formulas below, k is the constant of proportionality. c. w 5 k ? x2 ? y1/3 a. x 5 k ? y ? z 2 b. V 5 k ? l ? w ? h d. V 5 k ? h ? r 2
c. g A12 tB 5 24 A12 tB 5 24 A18B t3 5 212 t3 d. gs5td 5 24s5td 3 5 24s125dt 3 5 2500t 3 3
3. a. f s4d 5 3s4d 2 5 48
d. 2f ssd 5 2s3s 2 d 5 6s 2
3 2 s s h. f A4B 5 3 A4B 5 16s
4. a.
D
Deflection (in)
b. f s24d 5 3s24d 2 5 48 c. f ssd 5 3s 2
2 2 e. f s2sd 5 3s2sd 5 12s 2 2 f. f s3sd 5 3s3sd 5 27s 3 2 s s 2 g. f A2B 5 3 A2B 5 4 s
x
f(x) 5 4x2
g(x) 5 4x3
24 22 0 2 4
64 16 0 16 64
2256 232 0 32 256 y
Dsafe
10 f (x) = 4x 2
Ddeflection L 25
0
g (x) = 4x 3
The graphs of Ddeflection and Dsafe (with the deflections measured in inches and the plank length L measured in feet) are given in the accompanying diagram. c. The safety deflections are well above the actual deflections for all values of L between 0 and 20. It would cease to be safe if the plank were longer than about 22 ft. (This is the L value where Dsafe and Ddeflection meet.) 17. In all the formulas below, k is the constant of proportionality. d. R 5 k ? [O2] ? [NO]2 a. d 5 k ? t 2 b. E 5 k ? m ? c 2 e. v 5 k ? r 2 c. A 5 k ? b ? h V is quadrupled; is multiplied by 9 V is doubled; is tripled V is multiplied by n2 V is multiplied by n
21. a. h(2) 5 2; h(6) 5 18; the latter is nine times the former. b. h(5) 5 12.5; h(15) 5 112.5; the latter is nine times the former.
x 4
–4
Length (ft)
19. a. b. c. d.
2
–10
b. As x S 1` , both f sxd and gsxd S 1` . c. As x S 2` , f sxd S 1` but gsxd S 2` . d. The domain of both functions is all the real numbers, and the range of g is also all the real numbers; however, the range of f is only all the nonnegative real numbers. e. They intersect at the origin and at the point (1, 4). f. All values greater than x 5 1. 5.
y 16 12 8 4
g f x
–2
–1 –4 –8 –12 –16
1
2
a. f sxd 5 gsxd for x 5 0 and x 5 1. b. f sxd . gsxd for 0 , x , 1 and for x , 0 c. f sxd , gsxd for x . 1
Ch. 7
13. a. C 5 kAt b. k 5 0.06 dollars per sq. ft, per in. c. The area of the four walls (without the door) 5 2(15 ? 8) 1 2(20 ? 8) 5 240 1 320 5 560 sq. ft. The area of the door is 3 ? 7 5 21 sq. ft. Thus the total wall area 5 560 – 21 5 539 sq. ft. d. $129.36; $194.04 1.6
Algebra Aerobics 7.3 b. f s22d 5 4s22d 3 5 4s28d 5 232
Y k . So X is not directly proportional to Y. It is directly
15. a. and b.
Section 7.3 1. a. f s2d 5 4s2d 3 5 4s8d 5 32
3 3
653
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6.
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CH. 7 Exercises Solutions for Section 7.3
a.
y2
CH. 7 Exercises Solutions for Section 7.3
b.
y 3 2
down. The larger the absolute value of a, the narrower the opening of the graph.
y 3
y1
2 1
1
y
–3 –2 –1 –1 –2
1
2
5
x
x 3
–3 –2 –1 –1 –2
y1 –3
1
2
3
y = x2
y = 0.5x2
y = 2x2
–3
y2
x –5
Ch. 7
7. a. The graphs are similar; both have similar end behavior; they intersect at the origin; for positive values of x, y1 , y2; for negative values of x, y1 . y2. b. The graphs are similar; both have similar end behavior; they intersect at the origin; for all nonzero values of x, y1 . y2. 8. a. p is even and a is positive; a 5 3 since (1, 3) is on the graph of the function. b. p is odd and a is negative; a 5 22 since (1, 22) is on the graph of the function. 9.
a.
b.
y 500
y 1250
9
y=x
10
–3 –2
1
2
3
x –2
2
–500
–500
10. a. y 5 x and y 5 4x and y 5 24x:
y = –x2
y = –0.5x2
y = –2x2 –5
3. f goes with Graph C; g goes with Graph A; h goes with Graph D; j goes with Graph B. The graphs in B and D are mirror images of each other across the x-axis; the graph in D is steeper than that in C, the graph in A is flatter than those in C and D for 21 , x , 11, but the Graph in A is steeper as x S 6 `. 5. a. g(x) 5 6 ? x4 b. h(x) 5 0.5 ? x4 c. j(x) 5 22x4
y=x
x
5
7. a. b. c. d. e.
ƒ(0) 5 g(0) 5 0 If 0 , x , 1, then ƒ(x) . g(x) ƒ(1) 5 g(1) 5 1 If x . 1, then ƒ(x) , g(x) If x , 0, then ƒ(x) . g(x) y 5
y 20
f(x) = x2
y = 4x y=x
(1, 1)
x –5
5
x –5
–20
y = –4x
g(x) = x5 –5
b. y 5 x 4 and y 5 0.5x 4 and y 5 20.5x4 y 700
y = x4
y = 0.5x4 x –5
5 y=
d. Yes, f(2x) 5 f(x) e. 2` f. 2`
9. a. n is even b. k , 0 c. Yes, f (22) 5 f (2) 11.
j(x) = –2x3
y 150 j(x) = x3
–0.5x4
–700
5
(0, 0)
g (x) = –x3
h(x) = 0.5x3 x
–5
Exercises for Section 7.3 1. The graphs of the six functions are labeled in the accompanying diagram. They are all parabolas since they are graphs of functions of the form y 5 ax2, for various values of a. The differences are due to the value of a. If a . 0, then the graph is concave up. If a , 0, then the graph is concave
h(x) = 0.5x3
5 j(x) = x3
–150
g (x) = –x3 j(x) = –2x3
a. The constant of proportionality for f is 1; for g it is –1; for h it is 1/2; for j it is –2. b. g’s graph is the reflection of ƒ’s graph across the x-axis.
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CH. 7 Algebra Aerobics Solutions for Section 7.4
CH. 7 Exercises Solutions for Section 7.4
655
c. j’s graph is both a stretch and a reflection of ƒ’s graph across the x-axis. d. h’s graph is a compression of ƒ’s graph.
b. y 5 4x dominates y 5 x 3. In this case, 4x . x 3 for all values of x, but as x S 1` , the values grow farther and farther apart.
13. g(x) 5 24x2. Its graph is a reflection of the graph of ƒ(x) across the x-axis.
2. Graphs A and D are likely to be power functions; Graphs B and C are likely to be exponential functions. 3. a. y 5 2x eventually dominates. b. y 5 s1.000 005d x eventually dominates.
y 5
f(x) = 4x2
Exercises for Section 7.4 1. a. The two graphs intersect at (2, 16) and (4, 256). Thus, the functions are equal for x 5 2 and x 5 4. (See the accompanying figure.)
x –5
5 g(x) = –4x2
500
–5
y y = 4x
15. a. The graphs for the four functions are given in the diagram. All go through (0, 0) and (1, 1). The smaller the fractional power, the bigger the y value when 0 , x , 1.
(4, 256) y=x
1
y (2, 16) 0
x 0
b. As x increases, both functions grow. For 0 # x , 2 we have that x4 , 4x; at x 5 2 both have a y value of 16. From 2 , x , 4, we have 4x , x4. For x 5 4 they are again equal. For x . 4 we have that x4 , 4x. As x keeps on increasing 4x will continue to grow faster than x4. c. The graph of y 5 4x dominates.
y4 = x1/4 y3 = x1/3 y2 = x1/2
y1 = x 0
x 0
0.5
1
b. The graphs for the four functions are given in the diagram. All graphs go through (0, 0) and (1, 1). The higher the power of x, the smaller the y value when 0 , x , 1. Notice that over the interval [0, 1] the powers #1 all have y values above or on the graph of y 5 x, but the powers $1 all would have y values below or on the graph of y 5 x. 1
y1 = x
y
5
3. a. b. c. d. e. f.
f (x) . g(x) as x S 2` f (x) , g(x) as x S 1` f (x) . g(x) for x in (2`, 21) f (x) , g(x) for x in (20.5, 1) f (x) . g(x) for x in (1, 6) g(x) . f (x) for x in (7, 1`)
5. Graphs of y 5 2x and y 5 x2 y2 = x2
35
y3 = x3
y y = 2x
y4 = x4
0
y = x2
x 0
0.5
(4, 16)
1 (2, 4) 0
Section 7.4
Graphs of y 5 3x and y 5 x3
Algebra Aerobics 7.4 1. a.
x 0
x
y 5 4x
y 5 x3
0 1 2 3 4 5
1 4 16 64 256 1024
0 1 8 27 64 125
y 1,000
y = 4x
500 y = x3 0 0
x 10
5
Ch. 7
4
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CH. 7 Algebra Aerobics Solutions for Section 7.5
In the top figure, the graphs of y 5 2x and y 5 x2 intersect at x 5 2 and x 5 4. The graph of y 5 2x lies above the graph of y 5 x2 for 0 # x , 2 and for x . 4. In the bottom figure, the graphs of y 5 3x and y 5 x3 intersect at x 5 3 and at x < 2.478. The latter intersection point can only be seen if we use a graphing device and zoom in. The graph of y 5 3x is above the graph of y 5 x3 over the intervals (–`, 2.478) and (3, 1`). It is below that graph over the interval (2.478, 3). 7. 3 ? 2x . 3 ? x2 if 0 , x , 2 and if x . 4. Also, 3 ? 2x , 3 ? x2 if 2 , x , 4.
Ch. 7
9. a. h(x) goes with C b. i(x) goes with A c. j(x) goes with B
CH. 7 Exercises Solutions for Section 7.5
3 16 x 4 x4 b. g A2x B 5 sx/2d 4 5 3 4 A2 B 5 3 4 16 5 3 ? 4 x
5 16 A x34B , so g A2x B 5 16sgsxdd. 8. h(x) 5 2x23 2 2 1 1 5 27 ? A2x23B 5 27 ? h(x), a. h(3x) 5 2 (3x) 3 5 2 27x3 or one-twenty-seventh of the original amount. 2 2 27 2 5 b. h( 3x ) 5 2 (x/3) 3 5 2 3 x /27 5 22 ? x3 5 27 ? A2x3 B 27h(x), or 27 times the original amount.
9. a. x 5 kyz 2
b. a 5 kbc d
2 c. a 5 kb de"c
3
10. a. x 5 k ; If x 5 3 and y 5 4, then 3 5 k 1 3 5 2k 1 "y "4 k 5 6, so x 5 6 . "y
11. The functions ƒ(ax) and g(ax) are the functions ƒ(x) and g(x), respectively, where the input x has been multiplied by a. Since ƒ(ax) 5 (ax)n 5 an ? ƒ(x), then ƒ(ax) is ƒ(x) multiplied by an. Since g(ax) 5 bax 5 (bx )a 5 [g(x)]a, then g(ax) is g(x) raised to the ath power. 0.5 13. Linear function: m 5 324 2 2 1 5 10.5 1 y 5 10.5x 210.
So if y 5 400, then x 5
6 5 20 5 0.3.
b. x 5 k ? y ? z; If x 5 3, y 5 4 and z 5 0.5, then 3 5 k ? 4 ? (0.5) 1 k 5 1.5, so x 5 1.5 ? y ? z. So if x 5 6 and z 5 2, then 6 5 1.5ys2d or y 5 2. c. x 5
ky2 3
; If x 5 6, y 5 6 and z 5 8, then
"z k(6) 2
2 1 1 y x 5 1 6 5 36k 1 k 5 so 2 3 3 3 . "8 "z So if y 5 9 and z 5 0.027, then
65
Exponential function: y 5 18 ? 4x Power function: y 5 0.5x3 15. a. True b. True
6 "400
x 5 13
c. False d. False
3
92 3 "0.027
81 5 13 ? 0.3 5 90.
Exercises for Section 7.5
Section 7.5 Algebra Aerobics 7.5 1. a. x is directly proportional to y and z. b. y is inversely proportional to the square of x c. D is directly proportional to the square root of y and inversely proportional to the cube of z. 2. (a) and (d) represent direct proportionality; (b) and (c) represent inverse proportionality; (e) does not represent direct or inverse proportionality. It is an exponential function where x, the input, becomes an exponent in the output. 3. The volume will increase by a factor of 4, from 1 ft3 to 4 ft3. (See Table 7.6. in Section 7.5.) 4. a. i.
15 x3
ii.
b. i. 1.5x
22
210 x4
ii. 26x
iii. 23
iii.
3.6 x 223
iv. x
22
2 x 2y 3
iv. 6x 23 y 24 z 21
5. t 5 kd 22 or t 5 k2 for some constant k d
1 6. Intensity at 4 ft is k2 5 16 k, but at 2 ft intensity is k 2 5 14 k. s2d 4 The light is four times as bright at a distance of 2 ft away than it is at 4 ft away.
7. gsxd 5 34 x
3 3 1 3 1 a. g(2x) 5 (2x) 4 5 16x4 5 16 Qx4 R , so gs2xd 5 16 gsxd.
1. a. b. c. d. e.
1 y changes from 143 5 4 to 443 5 16 decreases 4 y changes from 243 5 12 to (1/2) 3 5 32 increases y is inversely proportional to x3 and 4 is the constant of proportionality.
3. a. h(2x) 5 8xk 3 , 2k b. j(2x) 5 16x 4,
k 27k h(23x) 5 227x 3 and h(x/3) 5 x3 281k 2k j(23x) 5 81x 4 and j(x/3) 5 x4
5. a. P 5 0.016 f ; 0.0016 b. Q 5 54 ; 2/3 r2
16 c. S 5 wp ; 2
d. W 5 2 ; 1/6 3 "u
7. a. B’s value is divided by 16. b. Z’s value is divided by 2 p. 9. Given: I(x) 5 k/x2 and 4 5 I(6) 5 k/36; thus k 5 144 1 I(8) 5 144/64 5 2.25 watts per square meter and I(100) 5 144/10000 5 0.0144 watt per square meter. 11. I(d) 5 k/d2 and thus I(4)/I(7) 5 49/16 5 3.06. The light intensity will be more than 3 times as great. 13. a. b. c. d.
The volume becomes 1/3 of what it was. The volume becomes 1/n of what it was. The volume is doubled. The volume becomes n times what it was.
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CH. 7 Algebra Aerobics Solutions for Section 7.6
CH. 7 Exercises Solutions for Section 7.6
y
3. a. y 5 x 210
15. a. x 5 k ? z b. Solving 4 5 k ? 16 32 for k gives k 5 8.
b. y 5 x 211
y
y
10
c. x 5 8 ? 25/5 5 40 17. a. Let L 5 wavelength of a wave and t 5 time between waves. Using the fact that the speed of the waves is directly proportional to the square root of the wave’s 5 "L length, we then have t 5 L k and thus time is
10
–2 –2
657
2
–2
2
x
x –10
k ? "L
directly proportional to the square root of the wavelength of the wave.
4. a. y 5 x 22 and y 5 x 23 y
b. If the frequency of the waves on the second day is twice that of the first, then the waves will be four times as far apart as they were on the previous day. 19. a. v 5 kl b. 24 in. c. decrease
10
y = x –3
y = x –2
d. decrease e. It was doubled in length.
x 2
–2 y = x –3 –10
30
H
The graphs intersect at (1, 1). As x S 1` or S 2` , both graphs approach the x-axis, but y 5 x 23 is closer to the x-axis at each point after x 5 1 and before x 5 21 When x . 0, as x S 0, both graphs approach 1` , but y 5 x23 is steeper. When x , 0, as x S 0, the graph of y 5 x 22 approaches 1` , and the graph of y 5 x 23 approaches 2` . b. y 5 4x 22 and y 5 4x 23 y 10 y = 4x –2
x 2
–2
H = 87.19·P(–0.99) P 0
y = 4x –3
y = 4x –3
12
–10
Section 7.6 Algebra Aerobics 7.6 1 1 1. a. x12 ; (22) 2 5 4 or 0.25 1 1 b. x13 ; (22) 3 5 28 or 20.125 4 1 c. x43 ; (22) 3 5 22 or 20.5 1 d. 24 ; 24 5 24 28 5 2 or 0.5 x3 (22) 3 1 1 e. 2x1 4 ; 2(22) 4 5 32 < 0.031
f. 22 ; 22 5 41 5 0.25 x3 (22) 3
The graphs intersect at (1, 4). As x S 1` or x S 2` , both graphs approach the x-axis, but y 5 4x 23 is closer to the x-axis after x 5 1 and before x 5 21 than y 5 4x 22. When x . 0, as x S 0, both graphs approach 1` . When x , 0, as x S 0, the graph of y 5 4x 22 approaches 1` , and the graph of y 5 4x 23 approaches 2` . 5. Graph A matches f sxd ; matches hsxd .
Graph B matches gsxd ;
Exercises for Section 7.6 1. A table for r is given below.
g. 22 ; 22 5 281 5 20.125 x4 (22) 4
y 30
2 1 h. x24 ; (22) 4 5 8 5 0.125
2. a. Graph A: y 5 xa2 Graph B: y 5 ax3 b. Since the point (1, 4) lies on Graph A, 4 5 1a2 1 a 5 4 1 y 5 x42. Since the point (2, 4) lies on Graph B, 4 5 a(2) 3 1 a 5 0.5 1 y 5 0.5x 3
Graph C
x 6
–6
–30
x
R(x)
1 2 3 6 21 22 23 26
6 3 2 1 26 23 22 21
Ch. 7
21. a. H 5 k/P 5 k ? P –1, where k is a proportionality constant. b. The software gave H 5 87.19 ? P20.99, which is very close. Note that the fit is quite good, as can be seen from the graph of the data and the best-fit graph as given in the accompanying diagram.
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CH. 7 Exercises Solutions for Section 7.6
CH. 7 Exercises Solutions for Section 7.6
The domain of the abstract function is all real numbers x 2 0. For x . 0, as x S 0 we have R(x) S 1` and for x , 0, as x S 0 we have R(x) S 2`.
c. As x approaches 6`, the graph of y 5 x2 approaches 1`. As x approaches 6`, the graph of y 5 x22 approaches 0. y
3. a.
x
g(x) 5 5x
x
t(x) 5 1/x
22 21 0 1 2
210 25 0 5 10
22 21 0.5 1 2
21/2 21 2 1 1/2
16
y = x–2 y = x2
Ch. 7
(–1, 1)
x
h(x) 5 x/5
x
ƒ(x) 5 5/x
22 21 0 1 2
20.4 20.2 0.0 0.2 0.4
22 21 0.5 1 2
25/2 25 10 5 5/2
b. The graphs of g and h are both straight lines through the origin with positive slope. The y values of g’s graph are 25 times those of h. Graphs of g(x) 5 5x and h(x) 5 x/5
(1, 1)
–3
x
0
3
7. a. f goes with Graph A. Reason: f (x) values must be positive; rapid decrease in the values of f (x) when x . 0. b. g goes with Graph C. Reason: There must be positive and negative values of g(x). c. h goes with Graph D. Reason: h(x) values must be positive; when x . 0 and x S 1 ` a less rapid decrease in the values in graph D than in Graph A. d. j goes with Graph B. Reason: j(x) values must all be negative. b. h(x) 5 12 x23
9. a. g(x) 5 4x23
c. j(x) 5 23x23
11. a. If x . 1, then f (x) , g(x) b. If 0 , x , 1, then f (x) . g(x) c. If x , 0, then f (x) , g(x)
y 4 g(x) = 5x
h(x) = x/5 x 2
–2
2 g(x) = 1/x4
(1, 1) f(x) = 1/x
5
g(x) = 1/x4 x
–4 –2
Graphs f and t both have the x- and y-axes as asymptotes, i.e., they approach but never touch these axes, and both are confined to the first and third quadrants. The y values of the graph of f are five times the y values of the graph of t. Graphs of t(x) 5 1/x and f (x) 5 5/x y
2 f(x) = 1/x5 –2
13. a. n is even b. k , 0 c. f (21) , 0
d. yes e. 0 f. 0
15. The graph of f (x) 5 x23
5
3
f (x) = 5/x t(x) = 1/x
f (x) = x–3 x 5
–5
x –3
3
–5 –3
5. a. The graph of y 5 x2 decreases for x , 0 and increases for x . 0 (as can be seen in the accompanying diagram). The graph of y 5 x22 increases for x , 0 and decreases for x . 0 (as can also be seen from that diagram). b. The two graphs intersect at (21, 1) and (1, 1).
a. g has –1; h has 1/2 and k has –2. b. g’s graph is a reflection of f’s across the x-axis. c. j’s graph is a stretch and a reflection of f ’s across the x-axis. d. h’s graph is a compression of f’s.
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CH. 7 Algebra Aerobics Solutions for Section 7.7
CH. 7 Algebra Aerobics Solutions for Section 7.7
17. g(x) 5 24x22
659
a. y 5 10log 213 log x 1 y 5 10log 2 ? 103 log x 1 3
y 5 2 ? 10log x 1 y 5 2x3
y
b. y 5 1021x log 3 1 y 5 102 ? 10x log 3 1
5
x
y 5 100 ? 10log 3 1 y 5 100 ? 3x
f(x) = 4x–2
x –5
c. y 5 100.03111.25x 1 y 5 100.031 ? 101.25x 1 y 5 1.07 ? (101.25 ) x 1 y 5 1.07 ? (17.78) x
5 g(x) = –f (x)
d. y 5 102.45720.732 log x 1 y 5 102.457 ? 1020.732 log x 1 20.732
y 5 286.4 ? 10log x
–5
1 y 5 286.4x20.732
e. y 5 1020.28920.983x 1 y 5 1020.289 ? 1020.983x 1 1 1 1 19. a. f(2x) 5 (2x) 4 5 4 and –f (x) 5 2 4 x x 1 1 b. 2f (x) 5 x24 and f (2x) 5 (2x) 4 5 16x4 1 1 1 c. g(2x) 5 (2x) 5 5 2 5 and –g(x) 5 2 5 x x
f. y 5 1021.4710.654 log x 1 y 5 1021.47 ? 100.654 log x 1 0.654
y 5 0.034 ? 10log x 4. a.
1 y 5 0.034x0.654
i. y 5 4x3 is a power function; ii. y 5 3x 1 4 is a linear function; iii. y 5 4 ? 3x is an exponential function.
Ch. 7
1 1 d. 2g(x) 5 x25 and g(2x) 5 (2x) 5 5 32x5 e. h(x) 5 2f(x) 5 2x14
y 5 0.51 ? (1020.983 ) x 1 y 5 0.51 ? (0.104) x
f. k(x) 5 2g(x) 5 2x15 21. a. (0, 0) and (1/2, 1) b. (0, 0) and (1, 1)
d. (1, 1) e. (1, 4)
c. a 312, 2b and a2 312, 2b
Function
b. Type of Plot on Which Graph of Function Would Appear as a Straight Line
c. Slope of Straight Line
y 5 3x 1 4 y 5 4 ? (3x) y 5 4 ? x3
Standard linear plot Semi-log Log-log
m53 m 5 log 3 m53
Section 7.7 Algebra Aerobics 7.7 1. Let Y 5 log y and let X 5 log x. a. log y 5 log(3 ? 2x ) 1 log y 5 log 3 1 log 2x 1 log y 5 log 3 1 x log 2 1 log y 5 log 3 1 (log 2)x 1 Y 5 0.477 1 0.301x b. log y 5 log(4x3 ) 1 log y 5 log 4 1 log x3 1 log y 5 log 4 1 3 log x 1 Y 5 0.602 1 3X c. log y log y log y Y
5 5 5 5
log(12 ? 10 x ) 1 log y 5 log 12 1 log 10x 1 log 12 1 x log 10 1 log 12 1 (log 10)x 1 1.079 1 x (since log 10 5 1)
d. log y 5 log(0.15x22 ) 1 log y 5 log 0.15 1 log x22 1 log y 5 log 0.15 2 2 log x 1 Y 5 20.824 2 2X 2. a. y 5 100.06711.63 log x 1 y 5 100.067101.63 log x 1 1.63 y 5 100.06710log x 1 y < 1.167x1.63 1 y 5 10 10 1 b. y 5 10 y 5 102.135 (101.954 ) x 1 y < 136.458 ? 89.95x 2.13511.954x
2.135
1.954x
c. y 5 1021.96310.865x 1 y 5 1021.963100.865x 1 y 5 1021.963 (100.865 ) x 1 y < 0.011 ? 7.328x d. y 5 100.24720.871 log x 1 y 5 100.2471020.871 log x 1 20.871 1 y < 1.766 ? x20.871 y 5 100.24710logx 3. (a), (d), and (f ) represent power functions. The graphs of their equations on a log-log plot are linear. (b), (c), and (e) represent exponential functions. The graphs of their equations on a semi-log plot are linear.
5. This is a log-log plot, so the slope of a straight line corresponds to the exponent of a power function. Younger ages: slope 5 1.2 1 original function is of the form y 5 a ? x 1.2, where x 5 body height in cm and y 5 arm length in cm. Older ages: slope 5 1.0 1 original function is of the form y 5 a ? x 1, where x 5 body height in cm and y 5 arm length in cm. 6. Graph A: Exponential function because the graph is approximately linear on a semi-log plot Graph B: Exponential function because the graph is approximately linear on a semi-log plot Graph C: Power function because the graph is approximately linear on a log-log plot 7. a. Estimated surface area is 104 5 10,000 square cm. b. Since S ~ M2/3, then S 5 kM2/3, where log k is the vertical intercept of the best-fit line on a log-log plot. In this case the vertical intercept 5 log 10, so k 5 10. So the equation is S 5 10M2/3. When M 5 70,000g, then S 5 10 ? s70,000d 2/3 5 10 ? 1700 5 17,000 cm2. c. Our estimated surface area was 7000 square cm lower than the calculated area. d. Since 1 cm 5 0.394 in, then 1 cm2 5 (0.394 in)2 5 0.155 in2. So 17,000 cm2 5 (17,000)(0.155) 5 2635 in2. Since 1 kg 5 2.2 lb, a weight of 70 kg translated to pounds is 70 kg 5 (70 kg) A2.2 1b kg B 5 154 lb.
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CH. 7 Exercises Solutions for Section 7.7
8. a. power b. y 5 ax 20.23 c. When body mass increases by a factor of 10, heat rate is multiplied by 1020.23 or about 0.59. 9. Estimated rate of heat production is somewhere between 103 and 104 kilocalories per day—roughly between 2000–3000 kcal/day.
Exercises for Section 7.7 1. When x = 4, then 100(5)4 = 62,500 and 100(4)5 = 102,400. So the exponential function y = 100(5)x in part (a) matches Graph A. Since an exponential function is a straight line on a semilog plot, then the function also matches Graph D. The power function y = 100x5 in part (b) matches Graphs B and C, since a power function (Graph C) is a straight line on a log-log plot (Graph B).
5. a. y 5 4x2 b. y 5 2x4
Adult Mass in Grams
Species
y 5 x5
1 2 3 4 5 6
1 32 243 1024 3125 7776 y
y
+
+ +
x 120,000
Bird mass
ii. log y vs. x log (y) 3.5
Log-Log Scale
10,000 1000
10
+
+ + + +
0
100
x 120,000
+
–1 Mass of bird
x
1 1
2
3 4 5 6 8 10
iii. log y vs. log x
x
log (x)
y 5 x5
log (y)
1 2 3 4 5 6
0 0.3010 0.4771 0.6021 0.6990 0.7782
1 32 243 1024 3125 7776
0 1.5051 2.3856 3.0103 3.4949 3.8908
log (y) 3.5
+ + + +
0 log (y) 4
0.015 0.036 0.026 0.033 0.049 0.167
2000
0 ++ 0
100,000
b.
1700.0 165.4 94.5 34.0 14.0 0.6
Answers will vary; notable is the fact that the egg/adult mass ratio for the hummingbird is very high and it is very low for the ostrich. The other ratios are not far apart from each other. b. The plots (where x 5 adult mass, y 5 egg mass (both in grams)) are, respectively, i. y vs. x
c. y 5 1.25x4 d. y 5 0.5x3
x
Egg Mass Egg/Adult in Grams Ratio
Ostrich 113,380.0 Goose 4,536.0 Duck 3,629.0 Pheasant 1,020.0 Pigeon 283.0 Hummingbird 3.6
Log (mass of egg)
7. a.
9. a.
log (mass of egg)
Ch. 7
3. An exponential function is indicated since the graph is a straight line on a semi-log plot.
CH. 7 Exercises Solutions for Section 7.7
Egg mass
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Log-Log Scale
log (x)
+
5
–1 log (mass of bird)
3 2 1 log (x)
0 0
0.5
1.0
c. The graphs are exactly the same. The points lie on straight line and are plotted on equivalent scales.
c. The log vs. log scatter diagram in part b (iii) is the most linear. Using technology the best-fit straight-line equation is: log(y) 5 log(0.1918) 1 0.7719 ? log(x) or, in linear equation form: Y 5 20.717 1 0.7719X where Y 5 log(y) and X 5 log(x). d. Regrouping, using the laws of logarithms, we get: log(y) 5 log(0.1918 ? x0.7719). Thus, y 5 0.1918 ? x0.7719, where
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CH. 7 Review: Putting It All Together
x 5 mass of the adult and y is the corresponding mass of the egg, both measured in grams. e. If x 5 12.7 kg (or 12,700 grams) for the weight of an adult turkey, then y < 282 grams is the predicted weight of its egg. f. If the egg weighs 2 grams, then the adult bird is predicted to have an adult weight < 20.8 grams. 11. a. Since l1 and l2 are parallel lines they have the same slope but different vertical intercepts. Thus their equations can be written as log(y) 5 mx 1 log(b1) and log(y) 5 mx 1 log(b2), with log(b1) 2 log(b2). Solving each for y in terms of x, we have y 5 b1 ? 10mx and y 5 b2 ? 10mx with b1 2 b2. Thus they have the same power of 10 but differ in their y-intercepts. b. The equation of l3 is log(y) 5 m ? log(x) 1 log(b3) and the equation of l4 is log(y) 5 m ? log(x) 1 log(b4), with log(b3) 2 log(b4). Solving for y in terms of x in each gives: y 5 b3 xm and y 5 b4x m with b3 2 b4. Thus they have the same power of x but differ in their coefficients of x. Graph of y = 3x
y
80
2
y = 3x2
60 40
b. Two estimated points on the best-fit line are (1, 40) and s70 2 40d (3, 70) 1 slope 5 s3 2 1d 5 15. So the equation is of the form y 5 b 1 15X. Substituting in (1, 40) we get 40 5 b 1 (15 ? 1) 1 40 5 b 1 15 1 b 5 25. So the equation is y 5 25 1 15X. Substituting log x for X, we finally get y 5 25 1 15 log x. So HALE is a logarithmic function of per-capita health expenditure. c. A logarithmic model makes sense, since there will always be a ceiling in life expectancy, so the results of adding funding will increase HALE but at a slower and slower rate. d. It suggests that the functions are log functions.
Ch. 7: Check Your Understanding 1. True
6. False
11. False
16. False
2. False
7. False
12. True
17. False
3. False
8. True
13. True
18. True
4. False
9. True
14. True
19. True
5. True
10. False
15. False
20. False
20 0
b. 140 120 100 80 60 40 20 0
1
2
3
4
5
6
22. Possible answer: ƒ(x) 5 4x 8 or ƒ(x) 5 10x 6
Graph of y = 4 • 2x
y
23. g(x) 5 3.2x4 24. Possible answer: ƒ(x) 5 2x4 25. Possible answer: h(m) 5 m12 or h(m) 5 em x
0
c.
21. True
x
0
1
2
3
4
5
26. Possible answer: y 5 3x or y 5 3 1 x
6
Graph of y = 3x + 6
y
27. Possible answer: y 5 3x5
25
28. Possible answer: ksxd 5 21 x3
20 15
29. Possible answer: T(m) 5 3m4
10 5 x
0 0
1
2
3
4
5
6
15. a. A power function seems appropriate. b. The log-log graph is linear and we get log(O) 5 0.75 ? log(m) 1 c, where O stands for oxygen consumption and m stands for body mass. Thus O 5 k ? m0.75, a power function, where k 5 10c. c. A slope of 3/4 means that if the body mass, m, is multiplied by 104 (510,000), then the oxygen consumption, O, is multiplied by 103 (51000). If m is multiplied by 10, then O is multiplied by 103/4 < 5.6. 17. a.
HALE versus Total Health Expenditure per Capita, 2002 90
HALE (years)
80
(3, 70)
70 60 50
(1, 40)
40 30 X = log x
20 1
2
3
Per-capita total health expenditure
4
30. False
33. False
36. True
39. True
31. True
34. False
37. False
40. False
32. False
35. True
38. True
Ch. 7 Review: Putting It All Together 1. a. V(x) 5 x(2x)(2x) 5 4x3. So V(X) 5 4X3 and V(2X) 5 4 (2X)3 5 8(4X3 ) 5 8V(X) 1 the volume is multiplied by 8. b. The surface area consists of the two top and bottom sides, which are both (2x)(2x) 5 4x2 in area, and the other four vertical sides, each (x)(2x) 5 2x2 in area. So the surface area S(x) 5 2(4x2) 1 4(2x2) 5 16x2. Hence S(X) 5 16X2 and S(2X) 5 16(2X)2 5 4(16X2) 5 4S(X) 1 when the value of x is doubled, the surface area is multiplied by 4. So the volume grows faster than the surface area. c. The ratio of (surface area)/volume 5 R(x) 5 S(x)/V(x) 5 (16x2)/(4x3) 5 4/x. If you double the value of x, from X to 2X, since R(X) 5 4/X, then R(2X) 5 4/(2X) 5 (1/2)(4/X) 5 (1/2)R(X). So if the value of x doubles, R(x), the ratio of
Ch. 7
13. a.
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CH. 7 Review: Putting It All Together
CH. 7 Review: Putting It All Together
b. VD(t) 5 15 – 0.5t; domain is 0 # t # 30; does not represent inverse proportionality. 15
VD (t)
0
t 30
7. a. The friend is wrong. b. If the height is increased by 50% (from 4 feet to 6 feet) or, equivalently, multiplied by 1.5 (since 1.5 ? 49 5 69), the width (29) and the length (29) must both also be multiplied by 1.5, giving 39 for the new width and length. Hence the new dimensions are 69 3 39 3 39, which give a volume of 54 cubic feet. The volume of the original block of ice 49 3 29 3 29 5 16 cubic feet. Since (new volume)/(old volume) 5 54/16 5 3.375, the new volume 5 3.375 ? (old volume) or, equivalently, a 237.5% increase in volume, almost five times what your friend predicted. c. 10% of 54 cubic feet 5 0.10 ? 54 5 5.4 cubic feet. So the volume of the melted sculpture would be 54 2 5.4 5 48.6 cubic feet. If h is the height of the melted sculpture, then the width and length are both 0.5h. So we have 48.6 5 (0.5h)2h 5 (0.5)2h3 1 h3 5 (48.6)/(0.5)2 5 194.4 1 h < 5.8 feet as the height of the melted sculpture. 9. a. t 5 kd/r (the k may be needed for unit conversions) b. D1 5 kD2 c. R 5 kV 2/W
15. a. From the previous problem, we have I 5 P/d2 where I is measured in foot-candles, P in candlepower, and d in feet. Setting I 5 4000 foot-candles and d 5 3 feet, we have 4000 5 P/(3)2 1 P 5 36,000 candlepower. b. From part (a) we know that the lamp has 36,000 candlepower. Setting P 5 36,000 and I 5 2000 footcandles, we have 2000 5 36,000/d 2 1 d 2 5 18 1 d < 4.24 feet or about 51 inches above the operating surface. 17. Graphs B and C are symmetric about the y-axis. Graph D is symmetric about the origin. Graphs A, B, and D all have the x-axis as a horizontal asymptote. Graphs B and D also have the y-axis as a vertical asymptote. 19. a. Y is a linear function of X. b. Using the estimated points (0, 0.3) and (1, 3.3), we have Y 5 0.3 1 3X. c. Substituting Y 5 log y and X 5 log x in Y 5 0.3 1 3X, we have log y 5 0.3 1 3 log x 1 log y 5 0.3 1 log(x3). Rewriting using powers of 10, we have 10log y 5 10(0.31log(x3) 1 y 5 100.3 ? 10logx 1 y 5 100.3 ? x3 1 y < 2x3. Hence y is a power function and y is directly proportional to x3. 3
21. To find the relationship, sketch a best-fit line to the data, where m 5 mass (in kg) and v 5 optimal flying speed (in meters per second). Since both axes use a logarithmic scale, we can replace the labels on the axes with log(m) 5 M and log(v) 5 V, respectively. The horizontal labels will now read 25, 24, 23, 22 21, 0 (5 log 1), which is now the beginning of the vertical axis, followed by 1, 2, 3, 5, 5, 6; and on the vertical axis 0, 1, 2, and 2.5 < log 300. V (= log v) 2.5 2
1
0 –5
–4
–3
–2
–1
Partridge Peregrine falcon Pheasant Blue heron Black vulture White stork Gannet Golden eagle Canada goose Wandering Mute swan albatross
5
–250
The abstract function is asymptotic to both the horizontal and vertical axes.
Sparrow hawk
t 0
n 30
–30
Ruby-throated Goldcrest hummingbird Magnolia warbler House wren Sand martin Barn swallow Skylark House sparrow Blue jay
VF (t)
P(n)
Bumblebee
15
250
Dung beetle
5. a. VF(t) 5 3t; domain is 0 # t # 5; represents direct proportionality.
13. a. P(n) 5 450/n for integer values of n between 0 and, say, 10. b. P(2) 5 450/2 5 $225/person; P(5) 5 450/5 5 $90/person. The more people, the lower the cost per person. c. If you take the function P(n) out of context and treat n as any real number (20), then the following graph shows the result.
Meat fly Honeybee
Ch. 7
3. Think of the cake as a cylinder, with a volume of pr 2h where r 5 the radius of the cake and h 5 the height. The radii of the 100 and 120 cakes are 50 and 60, respectively. If the corresponding heights are h1 and h2 , then the respective cake volumes are 25ph1 and 36ph2 cubic inches. The site claims that the volume of the 120 cake is twice that of the 100 cake (6 quarts vs. 3 quarts). That implies that 36ph2 5 2(25ph1) 1 h2/h1 5 50/36 < 1.4 1 h2 is about 40% larger than h1. So for the 120 cake to have twice the volume of the 100 cake, the height of the 120 cake would have to be 40% higher than the height of the 100 cake—which seems rather unlikely.
11. Graph C is symmetric across the y-axis. Graphs A, B, and D are rotationally symmetric about the origin. None of them has an asymptote.
Hornet
(surface area)/volume, is multiplied by 1/2—or equivalently, cut in two. This is again confirmation that the volume grows faster than the surface area. d. In general, as x increases, R(x)—the ratio of surface area to volume—will decrease. So as the rectangular solid gets larger, there will be relatively less surface area compared with the volume.
House fly
662
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Piper Warrior Beech Beech Bonanza Baron Beech King Air Learjet-31 F-16 Mig-23 Fokker F-28 F-14 Boeing 737 Airbus A310 Douglas Boeing 747 DC-10
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M (= log m) 0
1
2
3
4
5
6
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CH. 8 Algebra Aerobics Solutions for Section 8.1
Two points on the line with coordinates of the form (M, V) are (21, 1) and (5, 2). The slope is then (2 2 1)/(5 – (21)) 5 1/6. The vertical intercept is approximately at (0, 1.2). So the equation of the line would be V 5 1.2 1 (1/6)M. Substituting in, we get log (v) 5 1.2 1 (1/6)log(m) 5 1.2 1 log (m1/6). Rewriting both sides as powers of 10 and simplifying, we get: 1/6 1/6 v 5 10(1.2 1 log(m )) 5 101.2 ? 10log(m )) < 16m1/6. So v ~ m1/6 1 the cruising speed v (in meters per second) is directly proportional to the 1/6 power of body mass m (in kilograms). So Professor Bejan seems to be correct.
CHAPTER 8 Section 8.1 Algebra Aerobics 8.1 1. a. The vertex at (0, 0) is the minimum point. The focal point 1 1 1 1 . is 4a units above the vertex at A0, 12 5 4s3d 5 12 B b. The vertex at (0, 0) is the maximum point. The focal point 1 1 1 is 4a 5 4s26d 5 21 24 units below the vertex at A0, 224 B .
d. The vertex at (0, 0) is the maximum point. The focal point 1 1 is 4a 5 4s21/12d 5 23 units below the vertex at (0, 23). 2. a. A point on the rim is (15, 10), so if y 5 ax2, then 1 1 10 5 a(15)2, so a 5 2/45. The focal point is 4a 5 4s2/45d 5 45 8 5 5.625 ft from the vertex at (0, 0) on the back wall. 2 2 x b. y 5 45 c. You could not hear well if you were more than 15 ft on either side of the stage, because the sound would be traveling in straight lines from the parabolic wall. Also, the sound would not be good if the performer moved around and did not stay at the focal point.
3. a. g(2) 5 22 5 4 g(22) 5 (22)2 5 4 g(0) 5 (0)2 5 0 g(z) 5 (z)2 b. h(2) 5 2(2)2 5 24 h(22) 5 2(22)2 524 h(0) 5 02 5 0 h(z) 5 2z2 c. Q(2) 5 2(2)2 2 3(2) 1 1 5 29 Q(22) 5 2(22)2 2 3(22) 1 1 5 3 Q(0) 5 2(0)2 2 3(0) 1 1 5 1 Q(z) 5 2(z)2 2 3(z) 1 1 d. m(2) 5 5 1 2(2) 2 3(2)2 523 m(22) 5 5 1 2(22) 2 3(22)2 5211 m(0) 5 5 1 2(0) 2 3(0)2 5 5 m(z) 5 5 1 2z 2 3z2 e. D(2) 5 2(2 2 3)2 1 4 5 3 D(22) 5 2((22) 2 3)2 1 4 5 221 D(0) 5 2((0) 2 3)2 1 4 5 25 D(z) 5 2(z 2 3)2 1 4
663
f. k(2) 5 5 2 22 5 1 k(22) 5 5 2 (22)2 5 1 k(0) 5 5 2 02 5 5 k(z) 5 5 2 z2 4. a. f(x) i. concave down ii. maximum iii. axis of symmetry: x 5 0 iv. vertex: (0, 4) v. approx. horizontal intercepts: (22, 0), (2, 0) vertical intercept: (0, 4) b. g(x) i. concave up ii. minimum iii. axis of symmetry: x 5 1 iv. approx. vertex: (1, 22.25) v. horizontal intercepts: (22, 0), (4, 0) vertical intercept: (0, 22) c. h(x) i. concave up ii. minimum iii. axis of symmetry: x 5 2 iv. vertex: (2, 24) v. horizontal intercepts: (0, 0), (4, 0) vertical intercept: (0, 0) d. k(x) i. concave down ii. maximum iii. axis of symmetry: x 5 24 iv. vertex: (24, 22) v. no horizontal intercepts vertical intercept: (0, 210) 5. a. Equation is of the form y ax2 where a 0 1 1 focal length 4a . When x 12, focal length depth 1 1 1 1 1 2 2 4a a(12) 1 a 4 ? 122 1 a 5 2 ? 12 5 24 . 1 1 5 (1/6) 56 So focal length 5 4(1/24)
1 1 2 1 y 5 24 x. b. a 5 24
6. f(x) 5 x2 1 2x 2 15 h(x) 5 10x2 2 80x 1 150
g(x) 5 2x2 1 7x 1 5 j(x) 5 2x2 2 18
Exercises for Section 8.1 1. a. concave up; minimum; vertex at (21, 24); the line x 5 21 is axis of symmetry; (1, 0) and (23, 0) are horizontal intercepts; (0, 23) is vertical intercept. b. concave down; maximum; vertex at (2, 9); the line x 5 2 is axis of symmetry; (24, 0) and (8, 0) are horizontal intercepts; (0, 8) is vertical intercept. c. concave down; maximum; vertex at (25, 3); the line x 5 25 is axis of symmetry; (28, 0) and (22, 0) are horizontal intercepts; (0, 25) is vertical intercept. [Students may have a different y value for vertical intercept.]
Ch. 8
c. The vertex at (0, 0) is the minimum point. The focal point 1 1 is 4a 5 4s1/24d 5 6 units above the vertex at (0, 6).
CH. 8 Exercises Solutions for Section 8.1
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CH. 8 Algebra Aerobics Solutions for Section 8.2
3. The graphs of the three functions are in the accompanying diagram. 10
17. a. f (x) 5 x2 1 2x 2 3 b. P(t) 5 t2 2 3t 2 10
y y1 = x 2 y2 = (1/4)x 2
y3 = (1/12)x 2
–5
0
5
x
a. For y1: (0, 1/4); for y2: (0, 1); for y3: (0, 3) b. The graphs of all three functions are concave-up parabolas with vertex at (0, 0), but y1 rises faster than y2 and y3. c. Farther from the vertex and the graphs get wider. c. y 5 (3/4)x2
5. a. y 5 (1/24)x2 10
y y
15
(4, 12)
Ch. 8
the narrower the opening of the parabola, since focal 1 length 5 P 4a P when the parabola has the equation y 5 ax2. c. H(z) 5 2z2 2 z 1 2
19. a. P 5 2L 1 2W 5 200; Thus L 5 100 2 W and A 5 LW 5 (100 2 W) ? W 5 100W 2 W 2. So A has its maximum when W 5 50 m, since the vertex is at (50, 50). Thus L 5 50 m and the maximum area is 2500 m2. b. The same kind of argument applies. P 5 2L 1 2W; thus L 5 (P/2) 2 W and A 5 LW 5 W(P/2 2 W) 5 (P/2) ? W 2 W2, and this has its vertex at (P/4, (P/4)2) and the maximum area is (P/4)2 m2. 21. a. True b. False
c. False d. True
e. False
23. a. R(n) 5 (1250 1 100n) ? (50 2 2n) b. There will be no apartments rented when n 5 25. Thus the domain is 0 # n # 25. c. The graph is given in the accompanying diagram. From inspection, the practical maximum occurs when n 5 6. At that value R 5 $70,300.
(12, 6) y = (3/4)x 2
y = (1/24)x 2
x 15
0
–5
b. y 5 2(1/24)x2 –15
0
5
Rent in dollars
–5
80,000
x
d. y 5 16x2 0
5
x 150
R(n) = (1250 + 100n)(50 – 2n)
n 26
0
y
Rental units (3, 144)
y = –(1/24)x 2 (–12, –6)
Section 8.2 y = 16x –10 y
7. a. y 5 (1/16)x2 b. y 5 2(1/32)x2 9. a. 28 and 12
b. 10 and 14
–5
0
2
Algebra Aerobics 8.2a 5
x
c. y 5 4x2 d. y 5 26x2 c. 2 and 214
11. a. y 5 (1/6)x2 b. 5 5 (1/6)x2 implies x 5 6 "30 and thus the reflector is 2 "30 < 10.95 inches wide. 13. Assuming that we rotate the parabolic model to become concave up: a. The focus is at (0, 1.25). b. y 5 0.2x2 c. 2.5 5 0.2x2 gives x 5 6 "12.5; thus the reflector should be 2 "12.5 < 7.07 inches wide. 15. Student answers will vary. The general trend is that the smaller the focal length, the bigger the value of a and thus
1. s(x) 5 2x2 1 2 is narrower than r(x) 5 x2 1 2 because, looking at the coefficient of x2, 2 . 1. Both have a vertical intercept at (0, 2) since c 5 2 in both equations. 2. h(t) 5 t 2 1 5 is concave up since a (51) is positive; k(t) 5 2t 2 1 5 is concave down since a (5 21) is negative. They have the same shape because k a k 5 1 in both. They both cross the vertical axis at 5 since c 5 5 in both equations. 3. Both are concave down because both have a , 0. Both have a vertex at (0, 0) since b and c 5 0 in both equations. g(z) is flatter than f(z) because k 20.5 k , k 25 k . 4. They have the same shape and are concave up since both have a 5 1. g(x) is six units higher than f(x) because 8 is 6 more than 2. 5. They have the same shape and are concave down since a (5 23) is negative in both equations. g(t) 5 23t2 1 t 2 2 is three units higher than f(t) 5 23t2 1 t 2 5 since c 5 22 is three units vertically up from c 5 25.
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CH. 8 Algebra Aerobics Solutions for Section 8.2
b. y2 5 13x 2 2 2
6. a. y1 5 3x2 1 5
CH. 8 Algebra Aerobics Solutions for Section 8.2
c. y3 522x2 1 4
y 10
y1
y2 x 10
–10
4. y 5 a(x 2 h)2 1 k. The value of h 5 2 in (b), (c), and (d), so the vertices of those parabolas are two units to the right of the vertex of (a), where h 5 0. All are concave up with the same shape since a 5 1 in all four equations. (c) is four units above (a) and (b); (d) is three units below (a) and (b) since k 5 0 in (a) and (b), k 5 4 in (c), and k 5 23 in (d). y
y3
5 y y y y
–10 x 4
–2
7. a. g(x) 5 3x b. h(x) 5 25x2
c. j(x) 5 1/2x d. k(x) 5 2x2
8. a. y 5 5x2 2 2 b. y 5 x2 1 3
c. y 5 20.5x2 2 4.7 d. y 5 x2 2 71
2
665
= = = =
x2 (x – 2)2 (x – 2)2 + 4 (x – 2)2 – 3
2
–5
Algebra Aerobics 8.2b 1. a. vertex is (0, 0) b. vertex is (23, 0) c. vertex is (2, 0)
5. y 5 a(x 2 h) 1 k. All are concave down with the same shape since a 5 21 in all four equations. (b), (c), and (d) have vertices three units to the left of vertex of (a), since h 5 23 in those equations, and h 5 0 in (a). k 5 0 in (a) and (b), but k 5 21 in (c), so (c) is one unit below (a) and (b). k 5 4 in (d), so (d) is four units above (a) and (b). y 5 x –6
y = (x –
60
y = (x +3)2
y = x2
x –5
2. a. vertex is (0, 0) b. vertex is (1, 0)
y y y y
= = = =
–x 2 –(x + 3)2 –(x + 3)2 + 4 –(x + 3)2 – 1
–15
y 2)2
2
g(x) h(x) j(x) k(x)
7. a. b. c. d.
The vertex at (3, 24) is a minimum. The vertex at (21, 5) is a maximum. The vertex at (4, 0) is a maximum. The vertex at (0, 27) is a minimum.
5
–10
c. vertex is (24, 0)
All vertices lie on the x-axis. The vertex of (b) is one unit to the right of the vertex of (a). The vertex of (c) is four units to the left of vertex of (a). All have the same shape and are concave up. y 60 j(x) = 0.5(x – 1)2 j(x) = 0.5(x +4)2 f(x) = 0.5x2
5 3(x 2 2)2 2 1, the vertex is (2, 21) 5 22(x 1 3)2 1 5, the vertex is (23, 5) 5 15 (x 1 4) 2 2 3.5, the vertex is (24, 23.5) 5 2(x 2 1)2 1 4, the vertex is (1, 4)
6. a. b. c. d.
Algebra Aerobics 8.2c 1. a. vertex is (0, 24). b. vertex is (0, 6). c. vertex is (0, 1). 2. a. vertex is (0, 3) b. vertex is (0, 3)
x –5
3. a. vertex is (0, 0) b. vertex is (21.2, 0)
5
–10
y 5
y = x2 + 3
c. vertex is (0.9, 0) x –5
All of the vertices lie on the horizontal axis. The vertex of (b) is 1.2 units to the left of the vertex of (a). The vertex of (c) is 0.9 units to the right of the vertex of (a). All have the same shape and are concave down. r
5 –5
3. a. Vertex is (0, 0). Parabola is concave down since a (5 23) is negative. There is one x-intercept, (0, 0), since vertex is on the x-axis.
10
y
t –5
10
5
x 10 f(x) = –3x2
–10
r = –2(t – 0.9)2
–60
r = –2t 2 r = –2(t + 1.2)2
y = –x 2 + 3
–50
Ch. 8
All vertices lie on the x-axis. Vertex of (b) is three units to the left of the vertex of (a). Vertex of (c) is two units to the right of vertex of (a). All have same shape and are concave up.
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CH. 8 Algebra Aerobics Solutions for Section 8.2
b. Vertex is (0, 25), which is below x-axis. Parabola is concave down since a (5 22) is negative. So there are no x-intercepts.
CH. 8 Algebra Aerobics Solutions for Section 8.2
214 2 424 1 2 5 214, So vertex is A2112, 214B . Vertical intercept is 2. y
y 3
10
y = x 2 + 3x + 2 x –10
2
10 –10
2 f(x) = –2x – 5
1
–20
In parts (c) and (d), use the formula h 5 2b 2a as the horizontal coordinate of the vertex (where f(x) 5 ax2 1 bx 1 c). 24 c. a 5 1, b 5 4 1 h 5 2b 2a 5 2s1d 5 22, ƒ(22) 5 (22)2 1 4(22) 2 7 5 4 2 8 2 7 5 211 Vertex is (22, 211). Parabola is concave up since a (5 1) is positive. There are two x-intercepts. They are approximately (25, 0) and (1, 0). 10
y
x –3
1 –1
b. a 5 2, b 5 24, so horizontal coordinate of vertex is 2s24d 2b 2a 5 2s2d 5 1. ƒ(1) 5 2 2 4 1 5 5 3, so vertex is (1, 3). Since vertex is above x-axis, and a (5 2) is positive; the parabola is concave up, it does not cross x-axis, so no horizontal intercepts. Vertical intercept is 5. y
2 f (x) = x + 4x – 7
20 f (x) = 2x 2 – 4x + 5
x
10
Ch. 8
–10
–3
x 3
0
–10
2s24d
d. ƒ(x) 5 4 2 x 2 2x2 5 22x2 2 x 1 4; so a 5 22, 2(21) 1 1 b 5 21; h 5 2b 2a 5 2(22) 5 24 5 24
1 ƒ A214B 5 4 2 A214B 2 2 A214B 2 5 4 1 14 2 2 A16 B5 1 1 1 1 1 4 1 4 2 8 5 48 . So vertex is A24, 48B . Parabola is concave down since a (5 22) is negative. There are two x-intercepts. They are approximately (22, 0) and (1, 0).
4 c. a 5 21, b 5 24; 1 2b 2a 5 2s21d 5 22 5 22 g(22) 5 2(22)2 2 4(22) 2 7 5 24 1 8 2 7 5 23, so vertex is (22, 23). Since vertex is below t-axis and a (5 21) is negative, the parabola is concave down; it does not cross t-axis, so no horizontal intercepts. Vertical intercept 5 27. 3 t 5
–5
y 10
g(x) = –x 2 – 4x – 7 x
–10 –10
–8
10 f(x) = 4 – x – 2x 2
–20
4. y 5 ax2 1 bx 1 c compared with y 5 x2 (where a 5 1) a. y 5 2x2 2 5 1 a 5 2 i. a . 0 1 minimum at the vertex ii. k a k . 1, so parabola is narrower than y 5 x2. b. y 5 0.5x2 1 2x 2 10 1 a 5 0.5 i. a . 0 1 minimum at the vertex ii. k a k , 1, so parabola is flatter than y 5 x2. c. y 5 3 1 x 2 4x2 1 a 5 24 i. a , 0 1 maximum at the vertex ii. k a k . 1, so parabola is narrower than y 5 x2. d. y 5 20.2x2 1 11x 1 8 1 a 5 20.2 i. a , 0 1 maximum at the vertex ii. k a k , 1, so parabola is flatter than y 5 x2.
6. a. Vertex is (25, 211). y-intercept at x 5 0 1 y 5 0.1(5)2 2 11 5 0.1(25) 2 11 5 2.5 2 11 5 28.5. So y-intercept is 28.5. y 5 –20
–15
b. Vertex is (1, 4). y-intercept at x 5 0 1 y 5 22(21)2 1 4 5 22 1 4 5 2. So y-intercept is 2. y 4
–1
3 1 5 23 2 5 212 . If x 5 22 , 1
y5
2 A232B
1
3 A232B
125
9 4
2
9 2
125
1
2
x 3 y = –2(x – 1)2 + 4
–4
5. a. a 5 1, b 5 3; so horizontal coordinate of vertex is: 2b 2a
y = 0.1(x + 5)2 – 11 x 10
7. a. (x 1 3)2 2 9 b. (x 2 5)2 2 25
c. (x 2 15)2 2 225 2 d. Ax 1 12B 2 14
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CH. 8 Algebra Aerobics Solutions for Section 8.2
CH. 8 Exercises Solutions for Section 8.2
8. a. ƒ(x) 5 x2 1 2x 2 1 5 x2 1 2x 1 (1 2 1) 2 1 5 (x2 1 2x 1 1) 1 (21 2 1) 1 ƒ(x) 5 (x 1 1)2 2 2 vertex (21, 22); stretch factor 1 b. j(z) 5 4z2 2 8z 2 6 5 4(z2 2 2z) 2 6 5 4(z2 2 2z 1 1) 2 4(1) 2 6 1 j(z) 5 4(z 2 1)2 2 10 vertex (1, 210); stretch factor 4 c. h(x) 5 23x2 2 12x 5 23(x2 1 4x) 5 23(x2 1 4x 1 4) 1 3(4) 1 h(x) 5 23(x 1 2)2 1 12 vertex (22, 12); stretch factor 23 d. h(t) 5 216(t2 2 6t) 1 h(t) 5 216(t2 2 6t 1 9) 1 16(9) 1 h(t) 5 216(t 2 3)2 1 144 vertex (3, 144); stretch factor 216 e. h(t) 5 24.9(t2 1 20t) 1 200 1 h(t) 5 24.9(t2 1 20t 1 102) 1 200 1 4.9(102) 1 h(t) 5 24.9(t 1 10)2 1 690 vertex (210, 690); stretch factor 24.9 9. a. y 5 2 Ax 2
1 2 2B
1 5 5 2 Ax 2
5 2 Ax 2 x 1 2
2
1 2B
15
1 5 5 2x 2 2x 1 12 1 5 1 2
y 5 2x2 2 2x 1 vertex (1/2, 5); stretch factor 2 b. y 5 213 sx 1 2d 2 1 4 5 213 sx 1 2dsx 1 2d 1 4
12. a. ƒ(p) 5 21875p2 1 4500p 2 2400 1 a 5 21875, b 5 4500, c 5 22400 1 24500 h 5 2b 2a 5 23750 5 1.2 ƒ(1.2) 5 21875(1.2)2 1 4500(1.2) 2 2400 1 ƒ(1.2) 5 300 So the vertex is at ($1.20, $300). b. The maximum daily profit is $300 at a price of $1.20 per pretzel. 13. a. The vertex is (21, 4) 1 h 5 21, k 5 4 1 y 5 a(x 2 (21))2 1 4 1 y 5 a(x 1 1)2 1 4 Passing through the point (0, 2) 1 2 5 a(0 1 1)2 1 4 1 a 5 22 1 the equation is y 5 22(x 1 1)2 1 4. b. The vertex is (1, 23) 1 h 5 1, k 5 23 1 y 5 a(x 2 1)2 2 3. Passing through the point (22, 0) 1 0 5 a(22 2 1)2 2 3 1 3 5 9a 1 the equation is y 5 13 sx 2 1d 2 2 3.
Exercises for Section 8.2 1. a. b. c. d.
downward; the coefficient of x2 is negative. upward; the coefficient of t2 is positive. upward; the coefficient of x2 is positive. downward; the coefficient of x2 is negative.
3. The graphs of the four functions are given with their labels in the diagram below.
5 213 sx2 1 4x 1 4d 1 4
y 10
5 213x2 2 43x 2 43 1 4 1 y 5 213x2 2 43x 1 223 vertex (22, 4); stretch factor 21/3 c. y 5 10(x2 2 10x 1 25) 1 12 1 y 5 10x2 2 100x 1 250 1 12 1 y 5 10x2 2 100x 1 262 vertex (5, 12); stretch factor 10 d. y 5 0.1(x2 1 0.4x 1 0.04) 1 3.8 1 y 5 0.1x2 1 0.04x 1 3.804 vertex (20.2, 3.8); stretch factor 0.1
y = 2x2
–5
x
Top to bottom y = –2x2 + 3 y = –2x2 y = –2x2 – 3 –10
10. a. y 5 x2 1 6x 1 7 5 x2 1 6x 1 9 2 9 1 7 1 y 5 (x 1 3)2 2 2 b. y 5 2x2 1 4x 2 11 5 2(x2 1 2x) 2 11 5 2(x2 1 2x 1 1) 2 2(1) 2 11 1 y 5 2(x 1 1)2 2 13 11. a. “Completing the square,” y 5 x2 1 8x 1 11 1 y 5 x2 1 8x 1 16 2 16 1 11 1 y 5 (x 1 4)2 2 5 So vertex is (24, 25). b. Using the formula, y 5 3x2 1 4x 2 2 1 a 5 3, b 5 4, c 5 22 1 24 2 h 5 2b 2a 5 6 5 23 When x 5 22 3 , y 5 3 A223B 2 1 4 A22 3 B 2 21 8 4 y 5 3 A9B 2 3 2 2 5 43 2 83 2 2 5 243 2 2 5 2113 2 2 5 2313, so vertex In vertex form, y 3 Ax 1 23B 2 2 313
5
5. a. The compression factor is 0.3, the vertex is at (1, 8). b. The expansion or stretch factor is 30 and the vertex is at (0, 211). c. The compression factor is 0.01 and the vertex is at (220, 0). d. The expansion or stretch factor is 26 and the vertex is at (1, 6). c. (24, 27); maximum d. (2, 26); minimum
7. a. (3, 5); maximum b. (21, 8); minimum 9. a. b. c. d.
y 5 1(x 2 2)2 2 4 5 x2 2 4x y 5 2(x 2 4)2 1 3 5 2x2 1 8x 2 13 y 5 22(x 1 3)2 1 1 5 22x2 2 12x 2 17 y 5 0.5(x 1 4)2 1 6 5 0.5x2 1 4x 1 14
11. a. h(x) 5 23(x 2 4)2 1 5 13. a. 9; 3 b. 16; 4 is at A223, 2313B .
b. (4, 5)
c. 4; 2 d. 2.25; 1.5
15. a. y 5 (x 1 3)2 1 4 b. f (x) 5 (x 2 2.5)2 2 11.25
c. (0, 243) e. 1; 1 f. 0.25; 0.5
Ch. 8
1 4B 512
1 2 B Ax
667
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CH. 8 Exercises Solutions for Section 8.2
CH. 8 Exercises Solutions for Section 8.2
c. g(x) 5 (x 2 1.5)2 1 3.75 d. p(r) 5 23(r 2 3)2 1 18 e. m(z) 5 2(z 1 2)2 2 13
29. The maximum occurs when x 5 248/(26) 5 8 computers and the revenue from selling 8 will be 192 million dollars.
2
17. The larger in absolute value the coefficient of the x term, the narrower the opening. Thus the order from narrow to broad is: d, f, a, b, c, and finally e. Technology confirms the principle. 19. a. The diagram exhibits the graph of f, its horizontal and vertical intercepts, and its vertex. y 2 (4, 1) –1
(3, 0)
9
(5, 0)
x
f(x) = –x 2+ 8x – 15
31. a. 4 ft 5 how high off the ground the baseball was at the instant at which it was hit. b. t(sec) h(feet) 0 1 2 3 4
4 38 40 10 252
The ball hits the ground between t 5 3 and t 5 4 sec, since the height at t 5 3 is positive and the height at t 5 4 is negative. c. The accompanying diagram gives the graph and confirms the estimate given in part (b). h
(0, –15)
b. f (x) 5 2(x 2 4)2 1 1, and thus its vertex is at (4, 1). 21. a. y 5 a(x 2 2)2 1 4 and 7 5 a(1 2 2)2 1 4 gives a 5 3 and thus y 5 3(x 2 2)2 1 4 5 3x2 2 12x 1 16. b. If a . 0, the graph of y 5 a(x 2 2)2 2 3 is concave up; it is concave down if a , 0. c. The axis of symmetry is the line y 5 (22 1 4)/2 5 1; since the parabola is concave downward, we have a , 0; also we have y 5 a(x 2 1)2 1 k with 5 5 a(22 2 1)2 1 k and 5 5 a(4 2 1)2 1 k. Thus we have 9a 1 k 5 5 or k 5 5 2 9a. One can choose a 5 21/9 and then k 5 6, or one can choose a 5 22/9 and then k 5 7. Thus two examples are y 5 (21/9)(x 2 1)2 1 6 or y 5 (22/9)(x 2 1)2 1 7. 23. a. (4, 25); no
h = 4 + 50t – 16t 2
(3.20, 0) t
0
4
–20 t in seconds
d. Negative values for h make no sense since this would mean that the ball is below ground. e. Similarly, using technology we can determine that the ball reaches its maximum height of approximately 43.96 ft at t < 1.56 sec. 33. We are given that 2W 1 L 5 1 cowhide length. Thus L 5 1 2 2W and thus the area formula is A 5 W(1 2 2W) 5 W 2 2W 2. Note this is at its maximum when W is at the vertex, i.e., when W 5 21/[2 ? (22)] 5 1/4. At this value of W we have that L 5 1/2. Thus, Dido was correct.
c. 22
b. 0.5
(1.56, 43.06)
25. Derivation: from the data we have: y 5 a(x 2 2)2 1 3; and 21 5 a(4 2 2)2 1 3 This implies that a 5 21. Thus y 5 2(x 2 2)2 1 3 is its equation. Check: 2(4 2 2)2 1 3 5 24 1 3 5 21 This is confirmed in the accompanying graph.
35. a. The graph is given in the accompanying diagram, with x and y marked off in miles, and with a sample point (7.2, 9.6) marked on the highway along with the straight line from the origin to that point.
y 3
y 20
(2, 3)
15
N
Miles
Ch. 8
–16
h in feet above the ground
50
(7.2, 9.6)
y = –(x – 2)2+3 5
–1
y = 15 – 0.75x d = 12
x
x 20
0 (4, –1)
2
27. The maximum profit occurs when x 5 220/[2 ? (20.5)] 5 20; the maximum profit is 430 thousand dollars.
Miles
b. The highway goes through the points (0, 15) and (20, 0) and thus has the equation y 5 20.75x 1 15. c, d. The distance squared: d 2 5 x2 1 y2 5 x2 1 (15 2 0.75x)2 5 1.5625x2 2 22.5x 1 225
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CH. 8 Exercises Solutions for Section 8.2
CH. 8 Algebra Aerobics Solutions for Section 8.3
e. Letting D 5 d 2, we have D 5 1.5625x2 2 22.5x 1 225; the minimum occurs at the vertex, which is at x 5 22.5/(2 ? 1.5625) 5 7.2. The minimum for D is 144 and thus the minimum distance, d, is "144 or 12 miles. f. The coordinates of the point of this shortest distance from (0, 0) are (7.2, 9.6). [See the graph in part (a).]
669
Its graph is given in the accompanying diagram. Eyeballing gives the maximum mpg at the same M of approximately 73 mph. This is expected since maximum 5 1/minimum. Graph for y 5 M/G y 70 Miles per gallon
37. a, b. The graph is shown below. The minimum gas consumption rate suggested by the graph occurs when M is about 32 mph, and it is approximately 0.85 gph. (Computation on a calculator gives 32.5 for M, and the corresponding gas consumption rate is 0.86 (when rounded off).)
(73.14, 61.51)
y = M/(0.0002M 2 – 0.012M + 1.07)
M 200
0 Miles per hour
Gallons per hour
G 2
1 (32.5, 0.86) G = 0.0002M2 – 0.013M + 1.07 M 80
0
39. a. Time of release is at x 5 0. The height then is 2 meters. b. At x 5 4 m, y 5 5.2 m; at x 5 16 m, y 5 5.2 m. c, d. The graph in the accompanying diagram shows the highest point, namely when x 5 10 m and y 5 7 m, and the point where the shot put hits the ground, namely when x is approximately 22 m.
Miles per hour
0 10 20 30 40 50 60 70 80
1.07 0.96 0.89 0.86 0.87 0.92 1.01 1.14 1.31
— 0.09600 0.04450 0.02867 0.02175 0.01840 0.01683 0.01629 0.01638
0.0 10.4 22.5 34.9 46.0 54.3 59.4 61.4 61.1
For (f) we have: G/M 5 (0.0002M2 2 0.013M 1 1.07)/M. Its graph is in the accompanying diagram. Eyeballing gives the minimum gpm at M < 73 mph. Graph for y 5 G/M y Gallons per mile
0.20 y = (0.0002M 2 – 0.013M + 1.07)/M
(73.14, 0.02) M 200
0 Miles per hour
For (g) we have: M/G 5 M/(0.0002M2 2 0.013M 1 1.07).
Ch. 8
y
c. In 2 hours 1.72 gallons will be used and you will have traveled 65 miles. d. If M 5 60 mph, then G 5 1.01 gph. It takes 1 hour and 5 minutes to travel 65 miles at 60 mph and one will have used approximately 1.094 gallons. e. Clearly, traveling at the speed that supposedly minimizes the gas consumption rate does not conserve fuel if the trip lasts only 2 hours. f, g. M (mph) G (gph) G/M (gpm) M/G (mpg)
8 (10, 7)
Meters
6
4
2
0
5
10
15 Meters
20
x 25
Section 8.3 Algebra Aerobics 8.3a 1. a. y 5 22t(8t 2 25) 5 0 1 22t 5 0 or 8t 2 25 5 0 1 25 t 5 0 or t 5 25 8 1 horizontal intercepts at (0, 0) and A 8 , 0B b. y 5 (t 2 5)(t 1 5) 5 0 1 t 5 5 or t 5 25 1 horizontal intercepts at (5, 0) and (25, 0) c. h(z) 5 (z 2 4)(z 1 1) 5 0 1 z 2 4 5 0 or z 1 1 5 0 1 z 5 4 or z 5 21 1 horizontal intercepts at (4, 0) and (21, 0) d. g(x) 5 (2x 2 3)(2x 1 3) 5 0 1 2x 2 3 5 0 or 2x 1 3 5 0 1 x 5 3/2 or x 5 23/2 1 horizontal intercepts at (3/2, 0) and (23/2, 0) e. y 5 (5 2 x)(3 2 x) 5 0 1 5 2 x 5 0 or 3 2 x 5 0 1 x 5 5 or x 5 3 1 horizontal intercepts at (5, 0) and (3, 0) f. v(x) 5 (x 1 1)2 5 0 1 x 1 1 5 0 1 x 5 21, or one horizontal intercept at (21, 0) g. p(q) 5 (q 2 3)(q 2 3) 5 0 1 q 2 3 5 0 1 q 5 3 1 one horizontal intercept at (3, 0) 2. a. ƒ(x) 5 (5 1 4x)(1 2 x) 5 0 1 5 1 4x 5 0 or 1 2 x 5 0 1 x 5 25/4 or x 5 1 1 horizontal intercepts at (25/4, 0) and (1, 0)
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CH. 8 Algebra Aerobics Solutions for Section 8.3
Ch. 8
b. h(t) 5 (8 2 3t)(8 1 3t) 5 0 1 8 2 3t 5 0 or 8 1 3t 5 0 1 t 5 83 or t 5 283 1 horizontal intercepts at A83, 0B and A283, 0B c. y 5 (5 1 t)(2 2 3t) 5 0 1 5 1 t 5 0, 2 2 3t 5 0 1 t 5 25 or t 5 2/3 1 horizontal intercepts at (25, 0) and (2/3, 0) d. z 5 (2w 2 5)(2w 2 5) 5 0 1 2w 5 5 1 w 5 5/2 1 one horizontal intercept (5/2, 0) e. y 5 (2x 2 5)(x 1 1) 5 0 1 2x 2 5 5 0 or x 1 1 5 0 1 x 5 52 or x 5 21 1 horizontal intercepts at A52, 0B and (21, 0) f. Q(t) 5 (3t 2 2)(2t 1 5) 5 0 1 3t 2 2 5 0 or 2t 1 5 5 0 1 t 5 23 or t 5 252 1 horizontal intercepts at A23, 0B and A252, 0B 3. Product of a sum and difference: a. y 5 x2 2 9 5 (x 1 3)(x 2 3) d. y 5 9x2 2 25 5 (3x 1 5)(3x 2 5) f. y 5 16 2 25x2 5 (4 1 5x)(4 2 5x) Square of sum or difference: b. y 5 x2 1 4x 1 4 5 (x 1 2)2 e. y 5 x2 2 8x 1 16 5 (x 2 4)2 Neither: c. y 5 x2 1 5x 1 25 g. y 5 4 1 16x2 4. a. The vertical intercept is (0, 0), which means at time t 5 0, the object is on the ground. The horizontal intercepts: h(t) 5 216t(t 2 4) 5 0 1 t 5 0 or t 5 4 1 (0, 0) and (4, 0), which means the object left the ground at time t 5 0 seconds and returns to the ground at time t 5 4 seconds. b. The horizontal intercepts: P(q) 5 2(q2 2 60q 1 800) 1 P(q) 5 2(q 2 20)(q 2 40) 5 0 1 q 5 20 or q 5 40. This means that if either 20 or 40 units are sold, the profit is $0 (or at breakeven). From the graph, P(q) . 0 if 20 , q , 40. The vertical intercept is (0, 2$800), which means that if no items are sold, there is a loss of $800. 5. y1 is Graph C because (0, 0) and (2, 0) are horizontal intercepts, and graph is concave down. y2 is Graph A because (2, 0) and (21, 0) are horizontal intercepts, and graph is concave up. y3 is Graph B because (24, 0) and (21, 0) are horizontal intercepts, and graph is concave up. 6. a. ƒ(x) 5 (x 2 5)(x 1 6) b. (5, 0) and (26, 0) c.
y 20 f(x) x 10
–10
–35
d. ƒ(x) 5 0 1 (x 2 5)(x 1 6) 5 0 1 x 2 5 5 0 or x 1 6 5 0 1 x 5 5 or x 5 26. So ƒ(x) has two horizontal intercepts, at (5, 0) and (26, 0).
CH. 8 Algebra Aerobics Solutions for Section 8.3
Algebra Aerobics 8.3b 1. a. 4x 1 7 5 0 1 x 5 27/4 b. 4x2 2 7 5 0 1 4x2 5 7 1 x2 5 74 1 x 5 6 #74 1 x 5 6 "7 2 c. 4x 2 2 7x 5 0 1 xs4x 2 7d 5 0 1 x 5 0 or x 5 74 d. 2x 6 x2 1 0 5 x 2 2 2x 2 6 1 2s22d 6 "s22d 2 2 4s26d 1 2 2 6 "28 2 6 2 "7 x5 1x 5 1 2 2 2 A1 6 "7 B x5 1 x 5 1 6 "7 2
x5
e. s2x 2 11d 2 5 0 1 2x 2 11 5 0 1 x 5 11 2 f. sx 1 1d 2 5 81 1 x 1 1 5 6 "81 1 x 5 21 6 9 1 x 5 210 or x 5 8 g. 0 5 x 2 2 x 2 5 1 x5
2s21d 6 "s21d 2 2 4s25d 2
1 x 5 1 6 2"21
2. a. a 5 2, b 5 3, c 5 21 1 D 5 (3)2 2 4(2)(21) 5 17 1 two real, unequal zeros 1 two horizontal intercepts b. a 5 1, b 5 7, c 5 2 1 D 5 (7)2 2 4(1)(2) 5 41 1 two real, unequal zeros 1 two horizontal intercepts c. a 5 4, b 5 4, c 5 1 1 D 5 (4)2 2 4(4)(1) 5 0 1 one real zero (also known as a “double zero”) 1 one horizontal intercept d. a 5 2, b 5 1, c 5 5 1 D 5 (1)2 2 4(2)(5) 5 239 1 no real zeros (two imaginary zeros) 1 no horizontal intercepts 3. a. h 5 24.9t2 1 50t 1 80. The vertical intercept is the initial height (at 0 seconds), which is 80 m; coordinates are (0, 80). a 5 24.9, b 5 50, c 5 80. So when h 5 0 the horizontal intercepts are: t5
250 6 "2500 2 4(24.9) (80) 2(24.9)
1 1568 "4068 5 250 6 "2500 5 250 6 1 29.8 29.8 250 1 63.8 250 2 63.8 t5 or t 5 29.8 29.8 13.8 5 29.8
5 2113.8 29.8
5 21.41 5 11.61 seconds Negative values of t have no meaning in a height equation, so the horizontal intercept at (11.61, 0) means that the object hits the ground after 11.61 seconds. b. h 5 150 2 80t 2 490t2. The vertical intercept is (0, 150), which means that the initial height is 150 cm. a 5 2490, b 5 280, c 5 150 1 when h 5 0 the horizontal intercepts are at: t5
80 6 "6400 2 4(2490) (150) 2(2490)
5
80 6 "6400 1 294,000 2980
5
80 6 "300,400 6 548 5 802980 1 2980 80 1 548 80 2 548 2980 ort 5 2980 628 5 2468 2980 2980
t5 5
5 20.64
5 0.48 seconds.
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CH. 8 Algebra Aerobics Solutions for Section 8.3
CH. 8 Exercises Solutions for Section 8.3
18 at x 5 1 1 y 5 22 5 s1 1 2ds1 2 4d 1 y 5 5 1 the vertex 18 is at A1, 5 B .
Discard negative solution. The object hits the ground after 0.48 seconds. c. The vertical intercept is the height in feet at t 5 0 seconds 1 h 5 3 feet. If a 5 216, b 5 64, c 5 3 1 the horizontal intercepts are:
y
y=
5
t5
264 6
2 4s216d s3d 2s216d
"4288 1 t 5 264 6 232
6 65.5 1 t 5 264232 1 t < 4.05, which means the object hits the ground after about 4.05 seconds or t < 20.05 (which is meaningless in this context).
264 6 "(64) 2 2 4(216) (0) 2(216)
y
x –5
or
2 9 x 5 1210
10 5 210
28 5 210
5 21
5 45
So the x-intercepts are (21, 0) and (4/5, 0). b. a 5 4, b 5 228, c 5 49 1 discriminant 5 784 2 4(4)(49) 5 784 2 784 5 0 1 one x-intercept. "0 5 0, so root is rational. The x-intercept is where x 5 28 86 0 5 72; at A312, 0B . c. a 5 2, b 5 5, c 5 4 1 discriminant 5 25 2 4(2)(4) 5 25 2 32 5 27, which is negative 1 no x-intercepts. The function has two imaginary zeros at x 5 25 64 "27 5 25 64 "7 i d. a 5 2, b 5 23, c 5 21 1 discriminant 5 (23)2 2 4(2) (21) 5 9 1 8 5 17 1 two real zeros 2s23d 6 "17 1 x 5 3 6 4"17 1 2s2d x-intercepts are A3 1 4"17, 0B and A3 2 4"17,
x5
0B
e. a 5 23, b 5 0, c 5 2 1 discriminant 5 (0)2 2 4(23)(2) 5 24 1 two real zeros 6 "24 1 x 5 6 6"24 1 x 5 6 26 "6 1 x 5 202(23) "6 "6 x 5 7 "6 3 1 x-intercepts are Q2 3 , 0R, Q1 3 , 0R
5. a. y 5 a(x 1 2)(x 2 4). The point (3, 2) is on the parabola 1 2 5 a(3 1 2)(3 2 4) 1 a 5 225 1 y 5 225 (x 1 2) sx 2 4d . The vertex is on the line of symmetry, which lies halfway between the x-intercepts or
The vertex is on the line of symmetry, which lies halfway between the x-intercepts or at x 5 5 1 y 5 5 245 245 8 (5 2 2)(5 2 8) 1 y 5 8 and the vertex is A5, 8 B . 6. a. Vertex is (1, 5), above x-axis; a 5 3 is positive, so it opens up; so there are no x-intercepts. b. Vertex is (24, 21), below x-axis; a 5 22 is negative, so it opens down; so there are no x-intercepts. c. Vertex is (23, 0), on x-axis; a 5 25 is negative, so it opens down; so there is one x-intercept. d. Vertex is (1, 22), below x-axis; a 5 3 is positive, so it opens up; so there are two x-intercepts. 7. Answers will vary for different values of a and will be of the form: a. ƒ(x) 5 a (x 2 2)(x 1 3) If a 5 1, then ƒ(x) 5 (x22)(x 1 3) b. ƒ(x) 5 ax(x 1 5) If a 5 2, then ƒ(x) 5 2x(x 1 5) c. ƒ(x) 5 a (x 2 8)2 If a 5 22, then ƒ(x) 5 22(x 2 8)2 8. Graph A: two real, unequal zeros 1 discriminant is positive Graph B: one real zero 1 discriminant is equal to zero Graph C: no real zeros 1 discriminant is negative
Exercises for Section 8.3 0 5 (x 2 3) (x 1 3); thus x 5 3 or 23 0 5 x(4 2 x); thus x 5 0 or 4 0 5 x(3x 2 25); thus x 5 0 or 25/3 0 5 (x 1 5)(x 2 4); thus x 5 25 or 4 0 5 (2x 1 3)2; thus x 5 23/2 twice 0 5 (3x 1 2)(x 2 5); thus x 5 22/3 or 5 x2 1 4x 1 3 5 21 1 x2 1 4x 1 4 5 0 1 (x 1 2)2 5 0; thus x 5 22 (a double zero) h. x2 1 2x 5 3x2 2 3x 2 3 1 2x2 2 5x 2 3 5 0 1 (2x 1 1)(x 2 3) 5 0; thus x 5 21/2 or 3
1. a. b. c. d. e. f. g.
Ch. 8
1 9 x 5 1210
5 –5
horizontal intercepts are t 5 0 and t 5 4 1 the object hits the ground after exactly 4 seconds.
6 9 x 5 1210 1
4
5
6 64 1 t 5 264232 1
4. Discriminant is b2 2 4ac, from ax2 1 bx 1 c 5 y a. a 5 25, b 5 21, c 5 4 1 discriminant 5 1 2 4(25)(4) 5 81 . 0 1 two x-intercepts. "81 5 9, so roots are rational. The function has two real zeros where:
(x + 2)(x – 4)
b. y 5 a(x 2 2)(x 2 8). The vertical intercept is (0, 10) 1 10 5 a(0 2 2)(0 2 8) 1 a 5 58, so y 5 58 sx 2 2d sx 2 8d
d. The vertical intercept is the height in feet at t 5 0 seconds 1 h 5 64(0) 2 16(0)2 5 0 feet. a 5 216, b 5 64, c 5 0 1 horizontal intercepts are t5
–2/5
x –4
"s64d 2
671
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CH. 8 Exercises Solutions for Section 8.3
CH. 8 Exercises Solutions for Section 8.3
3. a. y 5 (x 1 4)(x 1 2)
f. g(x) 5 (3x 2 4)(x 1 1) 30 y 30
g(x) = 3x2 – x –4
2
y = x + 6x + 8
(–1, 0)
(4/3, 0)
–5 (–4, 0) (–2, 0) –5
–10
4
x 5
x
–10
5. a. y 5 x2 2 5x 1 6 has zeros at 2 and 3, as is shown in the graph below.
b. z 5 3(x 1 1)(x 2 3)
y 20 z 15 z = 3x2 – 6x – 9 y = x2 –5x + 6
(–1, 0)
–3
(3, 0)
x 7 x –1
(3, 0)
(2, 0)
6
–5
Ch. 8
–15
b. y 5 3x 2 2x 1 5 has no real zeros, as is shown in the graph below. 2
c. f (x) 5 (x 2 5)(x 1 2)
y 10 y 10 y = 3x2 – 2x + 5
(5, 0)
(–2, 0) –4
8
x
f(x) = (x – 5)(x + 2)
x –1
0
2
–15
c. y 5 3x2 2 12x 1 12 has a “double zero” at x 5 2 since y 5 3(x 2 2)2. See the graph below.
d. w 5 (t 2 5)(t 1 5)
y 40
12
w
w = t 2 – 25 y = 3x2 – 12x + 12 (–5, 0)
(5, 0)
–8
8
t
x 0
–30
4
(2, 0)
d. y 5 23x2 2 12x 1 15 has zeros at x 5 25 and 1; see the graph below.
e. r 5 4(s 2 5)(s 1 5) 200
y
r
30 y = –3x 2–12x + 15
r = 4s s – 100
(–5, 0) –8
(5, 0) 8
–100
s
–6
(–5, 0)
(1, 0) 4
–20
x
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CH. 8 Exercises Solutions for Section 8.3
CH. 8 Exercises Solutions for Section 8.3
e. y 5 0.05x2 1 1.1x has zeros at x 5 222 and 0, as the graph below shows. y
(–22, 0)
b. c. d. e.
673
a 5 2, b 5 3, c 5 22; b2 2 4ac 5 9 1 16 5 25 a 5 2, b 5 3, c 5 4; b2 2 4ac 5 9 2 32 5 223 a 5 21, b 5 2, c 5 21; b2 2 4ac 5 4 2 4 5 0 same as in (b) above.
Graph for (a)
(0, 0)
x y 5
y = –0.05x2 + 1.1x y = 2x2 + 3x + 1
f. y 5 22x2 2 x 1 3 has roots at x 5 21.5 and 1, as the graph below shows.
–5
5
x
–2
y 5
Graph for (b), (e) (1, 0)
(–1.5, 0)
3
–3
x
y 30
y = –2x2 – x + 3
Ch. 8
–10
7. a. t 5 5
7 6 "49 2 4 ? 6 ? s25d 2 ? 6 7 6 13 5 1 5 12 3 or 22
5 7 6 12"169
y = 2x2 + 3x – 2
x
2 4 ? 9 ? 4 5 12 618 "0 5 23 b. x 5 12 6 "144 2 ? 9 1 6 "1 2 4 ? 3 ? s29d 2 ? 3
c. z 5 or 1.907
5
1 6 "109 6
< 21.573
–4
3 –5
Graph for (c)
2 4 ? 1 ? 7 d. x 5 26 6 "36 5 26 62 "8 5 23 6 "2 2 ? 1 < –1.586 or –4.414
e. s 5 5
217 6 "172 2 4 ? 6 ? s210d 2 ? 6 217 6 23 1 5 210 12 3 or 2 3 6 "9 2 4 ? 2 ? s29d 2 ? 2
f. t 5 or 21.5
5 217 612 "529
9 5 3 6 4"81 5 3 6 53 4
11 6 "121 2 4 ? 4 ? s28d 2 ? 4
g. x 5 or 20.597
y 18
y = 2x2 + 3x + 4
5 11 6 8"249 < 3.347 x
2 4 ? 4 ? 2 h. x 5 12 6 "144 5 12 6 8"112 2 ? 4
5 12 684 "7 5 3 62 "7 < 2.823 or 0.177 9. a. y-intercept is 21; y 5 (3x 2 1)(x 1 1) and thus x-intercepts are 1/3 and 21. b. y-intercept is 11; the x-intercepts are 6 63 "3 < 2.58 and 1.42. c. y-intercept is 15; x-intercepts are 5/2 and –3/5. d. The vertical intercept is ƒ(0) 5 25 and the x-intercepts are 6 "5. 11. Student choices for values of a, b and c will vary. Here are some choices and the accompanying graphs. The equations in the form y 5 ax2 1 bx 1 c are in the graph diagrams. a. a 5 2, b 5 3, c 5 1; b2 2 4ac 5 9 2 8 5 1
–4
0
2
Graph for (d) y 2
x 3
–3
y = –x2 + 2x – 1
–4
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CH. 8 Exercises Solutions for Section 8.3
CH. 8 Algebra Aerobics Solutions for Section 8.4
13. a. Q(t) 5 24(t 1 1)2 b. No, since it must be of the form Q(t) 5 a(t 1 1)2 if 1 is to be a double root and a must be 24 if Q(0) 5 24. c. The axis of symmetry is the line whose equation is t 5 21. The vertex is at (21, 0).
b.
y 20
y = 7x2 – 5x – 9 (2, 9) (–1, 3)
15. For Graph A: f (x) 5 (x 1 5)(x 2 2) 5 x2 1 3x 2 10 For Graph B: g(x) 5 20.5(x 1 5)(x 2 2) 5 20.5x2 2 1.5x 1 5 17. a. 21 1 10i
c. 21 1 3i
b. 1
d. 25 1 11i y = –2x2 + 4x + 9)
19. f (x) 5 (x 2(1 1 i))(x 2 (1 2 i)) 5 x2 2 (1 2 i)x 2 (1 1 i)x 1 (1 1 i)(1 2 i) 5 x2 2 2x 1 (1 2 i2 ) 5 x2 2 2x 1 2 21. x 5 5
4 6 "16 2 4 ? 1 ? 13 2 ? 1 4 6 6i 5 2 6 3i. 2
5
–20
4 6 "236 2
Thus ƒ(x) 5 (x – 2 – 3i)(x – 2 1 3i). Its roots are not real and thus there are no x-intercepts.
Ch. 8
x 4
–4
23. a. Factoring, y 5 (x 1 4)(x – 2) and thus the roots are 24 and 2 and their average is –1. Thus h 5 21 and a 5 1; therefore y 5 (x 1 1)2 1 k 5 x2 1 2x 1 1 1 k 5 x2 1 2x 2 8 and thus k 1 1 5 28 or k 5 29. Thus the equation is y 5 (x 1 1)2 2 9. b. Factoring gives y 5 2(x 1 4)(x – 1) and thus its roots are 24 and 1 and thus h 5 21.5 and a 5 21; therefore y 5 2(x 1 1.5)2 1 k 5 2x2 2 3x 2 2.25 1 k 5 2x2 2 3x 1 8 and thus 4 5 k 2 2.25 or k 5 6.25. Thus the equation is y 5 2(x 1 1.5)2 1 6.25.
29. a. 1500 5 4W 1 2L and thus L 5 (1500 2 4W)/2 5 750 2 2W b. A(W) 5 W(750 2 2W) 5 2W(375 2 W) c. Domain for A(W) is 0 , W , 375 d. The area is largest when the W value is at the vertex of the parabola graph of A(W), namely, when W 5 187.5 ft. At that point L 5 375 ft. and thus the area of the largest rectangle is 70,312.5 sq. ft. See the accompanying graph for verification. 80000 (187.5, 70312.5)
A(W) = W(750 – 2W) 40000
25. y 5 4(x – 2)(x – 3). If the quadratic is to go through (2, 0) and (3, 0), then it must be of the form y 5 a(x – 2)(x 2 3); and if it is to stretch the graph of y 5 x2 by a factor of 4, then a 5 4 must hold. But this function’s graph does more. It shifts the vertex of the graph of y 5 x2 to (2.5, 21). Its graph is in the accompanying diagram. y
W 400
200
0
Section 8.4
25
Algebra Aerobics 8.4 1. y = 4(x – 2)(x – 3)
(2, 0)
(3, 0)
0 (2.5, –1)
x 5
–10
27. a. Setting the two formulas for y equal gives: 7x2 2 5x 2 9 5 22x2 1 4x 1 9 1 0 5 9x2 2 9x – 18 5 9(x2 2 x 2 2) 5 9(x 2 2)(x 1 1) and thus the intersection points are where x 5 21 and x 5 2. If x 5 21, then y 5 3 and if x 5 2, then y 5 9 (by substitution into either original equation). The graph confirming this information is given in the accompanying diagram.
x 21 0
y 4 5
1
4
2
1
3
24
4
211
Average Rate of Change n.a. 5 2 4 0 2 (21) 5 1 4 2 5 1 2 0
Average Rate of Change of Average Rate of Change n.a. n.a.
5 21
1 2 4 2 2 1 5 23 s24d21 3 2 2 5 25 s211d 2 s24d 5 4 2 3
27
s21d21 1 2 0 s23d 2 2 2 s25d 2 3 2 s27d 2 4 2
5 22 s21d 1 s23d 2 s25d 3
5 22 5 22 5 22
b. The third column tells us that the average rate of change of y with respect to x is decreasing at a constant rate, so the relationship is linear. The fourth column tells us that the average rate of change of the average rate of change is constant at 22.
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CH. 8 Exercises Solutions for Section 8.4
CH. 8 Algebra Aerobics Solutions for Section 8.5
2. a. The function is quadratic since the average rate of change is linear, that is, the average rate of change is increasing at a constant rate. b. The function is linear since the average rate of change is constant. c. The function is exponential since both the average rate of change and the average rate of change of the average rate of change are exponential, that is, are multiplied by a factor of 2 or increasing at a constant percentage. 3. a. The slope of the average rate of change 2a 5 2, giving y 5 2t 1 b. The vertical intercept b 5 1, so y 5 2t 11 is the equation of the average rate of change. b. The slope of the average rate of change 2a 5 6, giving y 5 6x 1 b. The vertical intercept b 5 5, so y 5 6x 1 5 is the equation of the average rate of change. c. The slope of the average rate of change is 10, giving y 5 10x 1 b. The vertical intercept b 5 2, so y 5 10x 1 2 is the equation of the average rate of change.
Exercises for Section 8.4 1. a. linear
b. positive, negative
b. The third column indicates that the average rate of change of Q is linear. For each increase of 1 in t it goes up by 4. 7. Graph A goes with Graph F Graph C goes with Graph D 9. a. F(t) 5 6t 1 1 11. a. quadratic
Q
Average Rate of Change
23 22 21 0 1 2 3
16 7 2 1 4 11 22
n.a. 29 25 21 3 7 11
c. exponential
Algebra Aerobics 8.5a 1. a. Degree is 5: ƒ(21) 5 11(21)5 1 4(21)3 2 11 5 11(21) 1 4(21) 2 11 5 211 2 4 2 11 5 226 b. Degree is 4: When x 5 21, y 5 1 1 7(21)4 2 5(21)3 5 1 1 7(1) 2 5(21) 5 1 1 7 1 5 5 13 c. Degree is 4: g(21) 5 22(21)4 2 20 5 22 2 20 5 222 d. Degree is 2: When x 5 21, z 5 3(21) 2 4 2 2(21)2 5 23 2 4 2 2 5 29 2. a. degree n 5 5 1 3 1 2 5 10 b. degree n 5 2 1 12 5 14 3. a. b. c. d.
degree 5; a quintic polynomial function The leading term is 22t5. The constant term is 0.5. ƒ(0) 5 0.5 2 2(0)5 1 4(0)3 2 6(0)2 2 (0) 5 0.5 ƒ(0.5) 5 0.5 2 2(0.5)5 1 4(0.5)3 2 6(0.5)2 2 (0.5) 5 21.0625 ƒ(21) 5 0.5 2 2(21)5 1 4(21)3 2 6(21)2 2 (21) 5 26.5
1.
y 100
x 5
–5
f(x) = –2(x – 3)2 (x + 4) –100
Q 30
Q = 2t2 + t + 1
t 0
b. linear
c. H(z) 5 2
Algebra Aerobics 8.5b
a. The function Q is quadratic. Its graph is given in the accompanying diagram.
–5
b. G(x) 5 210x 1 0.4
Section 8.5
5. The table and graph are given below: t
Graph B goes with Graph E
Ch. 8
3. a. y 5 3; y 5 22; y 5 a. It is horizontal line with slope 0 going through the point (0, a). b. It is a straight line with slope 5 2a and y-intercept b. No, the slope of linear function is constant. The slope of the quadratic is a linear function. c. Guesses will vary. You may guess by analogy from the answers to parts (a) and (b) that its function is the quadratic y 5 3ax2 1 2bx 1 c.
675
4
a. b. c. d. e.
degree 3 two turning points as x S 1`, y S 2` ; and as x S 2`, y S 1` y 5 22x3 The horizontal intercepts are (3, 0), where the graph “touches” the x-axis and (24, 0), where the graph crosses the x-axis. f. y 5 272; the vertical intercept 2. Graph A: Minimum degree 5 3 because two “bumps”; positive leading coefficient Graph B: Minimum degree 5 4 because three “bumps”; negative leading coefficient Graph C: Minimum degree 5 5 because four “bumps”; negative leading coefficient
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CH. 8 Algebra Aerobics Solutions for Section 8.5
3. a. The y-intercept is at 23. The graph crosses the x-axis only once. It happens at about x 5 1.3.
CH. 8 Exercises Solutions for Section 8.5
(ii) h(x) 5 2x2(x 1 3)(x 2 5)(x 2 7) 1000
y 4
h (x)
500
–2
–1
x 2
1
x
–4 –2 0
2
4
6
8
–500 –4
–1000
b. The y-intercept is at 3. The graph does not intersect the x-axis.
(iii) g(x) 5 220x2(x 1 3)(x 2 5)(x 2 7)
y g(x) 8000
10
x 10
–10
g (x)
4000
–10
x
–4 –2 0
2
4
6
8
–4000
4. a. Degree is 1. x-intercept is 22.
–8000
y
Ch. 8
8
Exercises for Section 8.5 x –3
3 –2
b. Degree is 2. x-intercepts are 24 and 1. y 4
6
–8
c. Degree is 3. x-intercepts are 25, 3, and
2212.
100
x 10
–10
d. not polynomial e. polynomial, 5 f. polynomial, 3
3. a. 81, 218 b. 12, 212
c. 212, 12 d. 232, 32
5. a. goes with Graph A b. goes with Graph C c. goes with Graph B
x –6
1. a. polynomial, 1 b. polynomial, 3 c. not polynomial
7. For Graph A: i. 2 For Graph B: i. 3 For Graph C: i. 4
ii. 2 ii. 2 ii. 5
iii. minus iii. minus iii. plus
iv. 3 iv. 4 iv. 5
9. (a) and (e) are cubics; (b), (d), and (f) are quartics; (c) is by itself, since it is the only quintic. 11. a. Always negative; could have up to three turning points; has exactly one turning point (see accompanying graph).
–100
5. There are infinitely many examples of such functions. To have exactly those four x-intercepts, the functions are of the form: ƒ(x) 5 axn(x 1 3)m(x 2 5) p(x 2 7)r, where a is a real number, and n, m, p and r are positive integers. (i) ƒ(x) 5 x(x 1 3)(x 2 5)(x 2 7)
y 6
x –4
5
150 f(x)
100
y = –2x2 + 4x + 3
50 x –4 –2
2 4 –50
6
8
–100 –150
–16
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CH. 8 Exercises Solutions for Section 8.5
CH. 8 Exercises Solutions for Section 8.5
b. Always positive; in general a quartic has at most three turning points; here exactly one turning point (see accompanying graph).
677
y 5 y = –2x2 + 4x + 3
y 5
(–0.6, 0)
(2.6, 0)
x 3
–3
y = (t2 + 1)(t2 – 1) –5
t –2
2
b. A quartic at most four horizontal intercepts; from the accompanying graph we see that it has two: at t 5 21 and 1.
–2
y 3
y = (t2 – 1)(t2 + 1)
c. Negative if x is negative; positive if x is positive; at most two turning points; here none (see the accompanying graph).
Ch. 8
(–1, 0)
–3
y
t 3
(1, 0)
2
–3
–2
y = x3 + x + 1
x 2
–2
d. Positive if x is positive; negative if x is negative; in general a quintic has at most four turning points; here there are exactly four (see the accompanying graph).
c. A cubic has at most three horizontal intercepts; from the accompanying graph we see that there is only one, at x < 0.682. d. A quintic has at most five horizontal intercepts; from the accompanying graph we see that there are five, at x 5 22, 21, 0, 1, and 5; in fact we have that y 5 (x 1 2)(x 1 1)x(x 2 1)(x 2 5).
Graph for (c) 3
y
Graph for (d) y
y
100
25
–2 –1 0 1 –6
x 6
–4
5
x 6
3
y=x +x +1 (–0.682, 0) 3
–3
x y = x 5 – 3x 4 – 11x 3 + 3x 2 + 10x
y = x5 – 3x4 – 11x3 + 3x2 + 10x –3
–400
13. a. A quadratic has at most two horizontal intercepts; from the accompanying graph we see that there are two, at x < 20.6 and 2.6.
–500
15. a. y 5 3x3 2 2x2 2 3 has only one real zero at x < 1.28. (See the accompanying graph.) b. y 5 x2 1 5x 1 3 has two real zeros: x 5 22.5 6 0.5 "13 < 24.302 and 20.697. (See the accompanying graph.)
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CH. 8 Exercises Solutions for Section 8.5
Graph for (a)
CH. 8 Algebra Aerobics Solutions for Section 8.6
Graph for (b)
y
y
5
5
y = 3x 3 – 2x 2 – 3
–5
(1.28,0)
5
x
–5
(–0.697, 0) (–4.302, 0)
1
x
f(x) = x 2 + 5x + 3 –5
–5
17. a. The y-values of the polynomial change sign when going from very large negative values of x to very large positive values of x and thus the graph must cross the x-axis. b. y 5 x2 1 1 and y 5 x4 1 1 are such polynomials, and their graphs are both in the accompanying graph. y 3
y = x2 + 1
Ch. 8
x 0
Section 8.6 Algebra Aerobics 8.6 1. a. 2ƒ(x) matches the graph of g(x) in Graph B because it is a reflection of ƒ across the x-axis. b. ƒ(2x) matches the graph of h(x) in Graph C because it is a reflection of ƒ across the y-axis. c. 2ƒ(2x) matches the graph of j(x) in Graph A because it is a double reflection of ƒ across both the x- and y-axes. 2. ƒ(x) 5 2x 2 3 a. ƒ(x 1 2) 5 2(x 1 2) 2 3 1 ƒ(x 1 2) 5 2x 1 1
y = x4 + 1
–2
c. The number written in base 2 as 11001 evaluates to 25 when written in base 10 notation, since (1 ? 24) 1 (1 ? 23) 1 (0 ? 22) 1 (0 ? 21) 1 (1 ? 20) 5 25. d. Here is one way to find the base two equivalent: find the highest power of 2 in 35. This is 2 5 5 32. Subtracting that from 35 leaves 3, which is easily written as 2 1 1. Thus 35, in base 10, can be written as (1 ? 25) 1 (0 ? 24) 1 (0 ? 23) 1 (0 ? 22) 1 (1 ? 21) 1 (1 ? 20) and this is 100011 in base 2.
2
c. i. The polynomial y 5 3x3 – 2x2 2 3 has exactly one real zero and its graph is given in the Solution to Exercise 15(a). ii. The polynomial y 5 (x – 1)x(x 1 1) has three real zeros at x 5 21, 0, and 1, as can be seen in the following graph on the left. d. The polynomial y 5 (t2 – 1)(t2 1 1) has exactly two real zeros and its graph is shown on the right.
b. c. d. e.
5 12 s2x 2 3d 1 12 ƒsxd 5 x 2 32 2ƒ(x) 5 2(2x 2 3) 1 2ƒ(x) 5 22x 1 3 ƒ(2x) 5 2(2x) 2 3 1 ƒ(2x) 5 22x 2 3 2ƒ(2x) 5 2(2(2x) 2 3) 1 2ƒ(2x) 5 2(22x 2 3) 5 2x 1 3
1 2 ƒsxd
ƒ(x) 5 1.5x a. ƒ(x 1 2) 5 1.5x12 1 ƒ(x 1 2) 5 1.52 ? 1.5x or 2.25(1.5)x b. 12 ƒsxd 5 12 s1.5x d c. 2ƒ(x) 5 2(1.5x) d. ƒs2xd 5 1.52x 1 ƒs2xd 5 1 x 1.5
e. 2ƒs2xd 5 2s1.52x d 1 2ƒs2xd 5 21x 2
y
3
1.5
y
3. Graph B is symmetric across the vertical axis; Graph A is symmetric across the horizontal axis; Graph C is symmetric about the origin.
y = (x – 1)x(x + 1)
x –2
–1
0
1
2
–3
1
–1
t 3
y = (t2 – 1)(t2 + 1) –2
–3
4. a. hst 2 2d 5 e t22 b. 2hst 2 2d 5 2e t22 c. 2hst 2 2d 2 1 5 2e t22 2 1 5. a. Qst 1 2d 5 2 ? 1.06 t12
19. a. b. c. d. e.
True; it is a function and 0 is in its domain. True; it is a cubic. True. False; it is d and d can be positive, negative, or 0. True, since the cubic term dominates for large positive or negative values of x. f. False, the origin is a point on the graph if and only if d 5 0.
21. a. 8701 5 8n3 1 7n2 1 0n 1 1n0 b. 239 5 2n2 1 3n 1 9n0
b. Qst 1 2d 2 1 5 2 ? 1.06 t12 2 1 c. 2sQst 1 2d 2 1d 5 2s2 ? 1.06 t12 2 1d 1 2(Q(t 1 2) 2 1) 5 22 ? 1.06 t12 1 1 6. a. g(x) is a reflection of ƒ(x) across the x-axis, since g(x) 5 x2 2 5 5 2(5 2 x2) 1 g(x) 5 2ƒ(x). b. g(x) is a reflection of ƒ(x) across the y-axis, since ƒ(x) 5 3 ? 2x and ƒ(2x) 5 3 ? 22x 1 ƒ(2x) 5 g(x). 7. a. g(x) is a compression of ƒ(x) by a factor of 13 since 1 1 gsxd 5 3x 12 6 5 3sx 12 2d 5 13 ? x 2 2 5 3 f(x).
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CH. 8 Exercises Solutions for Section 8.6
CH. 8 Exercises Solutions for Section 8.6
b. g(x) is a vertical shift up of ƒ(x) by ln 3 units since g(x) 5 ln 3x 5 ln 3 1 ln x 1 g(x) 5 ln 3 1 ƒ(x).
Exercises for Section 8.6 1. a. g’s graph is a reflection of ƒ’s graph across the x-axis followed by a stretching of the graph by a factor of 2. Thus gsxd 5 22 ? fsxd 5 22 "x. b. g’s graph is the graph of ƒ(x) shifted two units to the right. Thus g(x) 5 ƒ(x – 2) 5 ex–2. c. g’s graph is the graph of ƒ(x) shifted three units to the left. Thus g(x) 5 ƒ(x 1 3) 5 ln(x 1 3). 3. a.
Graph for (i) y
5
y
f(x) = x 3
f(x) = x 3
5
–5 g(x) = –x
x
5
–5
x
3
g(x) = –2x 3–1 –5
9. a. y 5 20(0.5)x12 2 5 b. y 5 4(x 1 2)3 2 5 11. a. y 5
1 2 s2 2 xd
b. y 5 0.5 1
s2td
c. y 5 d.
2
21 5 0.52 t
1 2 s "s 2 3d
1 sx 1 hd 2
5 s 21 3
2 12 5 x
c. y 5 log(x 1 2)1/3 2 5
x2 x 2 sx 1 hd 2
2
sx 1 hd 2 x 2 sx 1 hd 2
2 h2 5 22hx 2 2 x sx 1 hd
13. a. kss 2 2d 5 s 21 2 b. 13 kss 2 2d 5 3ss 12 2d c. 213 kss 2 2d 5 3ss21 2 2d d. j(s) 5 213 k(s 2 2) 1 4 1 j(s) 5 3(s21 2 2) 1 4
Graph for (ii)
5
iii. h(2x) 5 (2x)4 1 (2x)3 5 x4 2 x3 2 h(x) and 2 2h(x), and so h(x) is neither even nor odd. iv. g(2x) 5 10 ? 32x 2 g(x) and 2 2g(x), and so g(x) is neither even nor odd. d. The graphs of even functions are symmetric across the y-axis and the graphs of odd functions are symmetric about the origin, as the graphs of the functions in (c)(i) and show.
–5
15. ƒ(x) 5 ln x a. ƒ(x 1 2) 5 ln(x 1 2) b. 12 ƒ(x) 5 12 ln x 1 12 ƒ(x) 5 ln x1/2 1 12 ƒsxd 5 ln "x c. 2ƒsxd 5 2ln x 1 2ƒsxd 5 ln x21 1 2ƒsxd 5 ln 1x d. ƒ(2x) 5 ln(2x) e. 2ƒ(2x) 5 2ln(2x) 1 2ƒ(2x) 5 ln(2x21) 1 2ƒ(2x) 5 ln A21x B 1 x3 a. ƒsx 1 2d 5
ƒ(x) 5 Graph for (iii) 5 g(x) = (x + 2)
3
Graph for (iv)
y
5 f(x) = x
1 sx 1 2d 3
b. 12 ƒsxd 5 1 3 2x
3
c. 2ƒsxd 5 21 x3
f(x) = x 3 = –(–x)3 = g(x)
5
–5
y
x
5
–5
x
1 1 d. ƒ(2x) 5 (2x) 3 5 2 3 x 1 1 e. 2ƒ(2x) 5 2Q (2x) 3 R 5 x3
–5
–5
5. a. symmetric about the origin. b. symmetric across the x-axis c. symmetric across the y-axis 7. a. If ƒ(x) 5 a ? x2k, then ƒ(2x) 5 a(2x)2k 5 ax2k 5 ƒ(x). b. If ƒ(x) 5 a ? x2k11, then ƒ(2x) 5 a(2x)2k11 5 2ax2k 5 2ƒ(x). c. i. ƒ(2x) 5 (2x)4 1 (2x)2 5 x4 1 x2 5 ƒ(x), so this is an even function. ii. u(2x) 5 (2x)5 1 (2x)3 5 (2x5) 1 (2x3) 5 2(x5 1 x3) 5 2u(x), so this is an odd function.
17. a. The “parent” function is f(t) 5 t. So g(t) is f(t) 5 t shifted to the right by 1 and compressed by a factor of 1/2 to get g(t) 5 12 f(t 2 1). b. The “parent” function is f(t) 5 (1/2)t. So g(t) is f(t) 5 A12B t shifted left by 4 and stretched by a factor of 3 to get g(t) 5 3f(t 1 4). c. The “parent” function is f(t) 5 1/t. So g(t) is f(t) 5 1t shifted to the right by 5, stretched by a factor of 7, reflected across the x-axis, and shifted down by 2 to get g(t) 5 27f(t 2 5) 2 2. 19. a. g(x) 5 3 ln(x 1 2) 2 4, so the graph of f(x) 5 ln x was shifted to the left by 2, with the result stretched by a factor of 3 and then shifted down by 4.
Ch. 8
i. f(2x) 5 2x3; the original graph has been reflected across the y-axis. ii. 22ƒ(x) 2 1 5 22x3 2 1; the original graph has been first stretched by a factor of 2, then reflected across the x-axis and then lowered one unit in the y direction. iii. ƒ(x 1 2) 5 (x 1 2)3; the graph has been shifted two units to the left along the x-axis. iv. 2ƒ(2x) 5 x3; the graph was reflected across both axes, but the original graph is symmetric with respect to the origin and, effectively, no visual change has occurred. The graphs are in the diagrams below, each with the graph of the original ƒ.
679
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CH. 8 Algebra Aerobics Solutions for Section 8.7
21. a. If we let t 5 time (in hours) since the corpse was discovered and T 5 temperature of the corpse, then since the ambient temperature is 608 according to Newton’s Law of Cooling, T 5 60 1 Ce2kt for some constants k and C. When t 5 0, T 5 80, so we have 80 5 60 1 C 1 C 5 20. So the equation becomes T 5 60 1 20e2kt. When t 5 2, T 5 75, so we have 75 5 60 1 20e22k 1 (15/20) 5 e22k 1 (3/4) 5 e22k 1 ln(3/4) 5 ln(e22k) 1 20.288 < 22k 1 k < 0.144. So the full equation is T 5 60 1 20e20.144t. b. If we assume that the normal body temperature is 98.68, then to find the time of death we need to solve 98.6 5 60 1 20e20.144t for t 1 (38.6/20) 5 e20.144t 1 ln(38.6/20) 5 ln(e20.144t) 1 0.658 < 20.144t 1 t < 24.6 hours. So the person died about 4.6 hours before the corpse was discovered.
c. ƒ(t) ? h(t) 5 (3 2 2t)(t2 2 1) 5 22t3 1 3t2 1 2t 2 3 2 hstd d. ƒstd 5 3t 22 2t1
4. a. ƒsx 1 1d 5 sx 1 1d 2 1 2sx 1 1d 2 3 5 x2 1 4x 2 2 b. ƒsxd 1 1 5 x 1 2x 2 3 1 1 5 x 1 2x 2 2 1 1 c. gsx 1 1d 5 sx 1 1d 2 1 5 x x 2 1 x 1 1 d. g(x) 1 1 5 x 2 1 1 1 5 x 2 1 1 x 2 1 5 x 2 1
5. a. horizontal intercepts at x 5 1 and x 5 25; vertical asymptote at x 5 23 y 10
Person alive [–5, –4.5]
x 10
–10
–10
b. horizontal intercept at x 5 22/3 and vertical asymptotes at x 5 21 and x 5 3 y 5
x
Temperature of Body Over Time Temperature of body (Fahrenheit)
Ch. 8
b. To find the vertical intercept, set x 5 0 to get g(0) 5 3 ln (0 1 2) 2 4 < 21.92. So the vertical intercept is at approximately (0, 21.92). To find any horizontal intercepts, set g(x) 5 0, to get 0 5 3 ln(x 1 2) 2 4 1 4/3 5 ln(x 1 2) 1 e4/3 5 x 1 2 1 x 5 e4/3 2 2 < 1.79. So there is a single horizontal intercept at approximately (1.79, 0).
CH. 8 Algebra Aerobics Solutions for Section 8.7
–5
5
120 –5
100
60
(2, 75°)
40 20
–5 –4 –3 –2 –1 0 1 2 3 4 Time since corpse discovered (hrs)
(r 1 2) (r 2 6)
2 12 6. a. f(r) 5 rr2 22 4r 5 (r 2 1) (r 2 3) 4r 1 3 To find any horizontal intercepts, set the numerator 5 0 to get (r 1 2)(r 2 6) 5 0 1 r 5 22 or r 5 6. So the horizontal intercepts are at (22, 0), (6, 0). To find any vertical asymptotes, set the denominator 5 0 to get (r 2 1)(r 2 3) 5 0 1 r 5 1 or r 5 3. So f(r) is not defined at r 5 1 and r 5 3. So there are two vertical asymptotes at the lines r 5 1 and r 5 3. The graph of f(r) follows. 2
(0, 80°) 80
5
30
Section 8.7
f(r)
Algebra Aerobics 8.7
20
1. a. ƒ(2) 5 23 5 8 b. g(2) 5 2(2) 2 1 5 3 c. hs2d 5 12 d. sh ? gds2d 5 hs2d ? gs2d 5 12 ? 3 5 32 e. (ƒ 1 g)(2) 5 ƒ(2) 1 g(2) 5 8 1 3 5 11
10
r
hs2d
1 f. AhgB s2d 5 gs2d 5 1/2 3 5 6
10
–10
2. a. Q(1) 1 P(1) 5 7 1 3 5 10 b. Q(2) 2 P(2) 5 9 2 24 5 215 c. P(21) ? Q(21) 5 (23 ? 3) 5 29 Q(3)
d. P(3) 5 11 81 3. a. ƒ(t) 2 h(t) 5 (3 2 2t) 2 (t2 2 1) 5 2t2 2 2t 1 4 b. ƒ(t) 1 h(t) 5 (3 2 2t) 1 (t2 2 1) 5 t2 2 2t 1 2
–10
(r 2 1) (r 2 3)
4r 1 3 b. g(r) 5 rr2 22 4r 5 (r 1 2) (r 2 6) 2 12 To find any horizontal intercepts, set the numerator 5 0 to get (r 2 1)(r 2 3) 5 0 1 r 5 1 or r 5 3. So the horizontal intercepts are at (1, 0), (3, 0). 2
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CH. 8 Algebra Aerobics Solutions for Section 8.7
CH. 8 Exercises Solutions for Section 8.7
To find any vertical asymptotes, set the denominator 5 0 to get (r 1 2)(r 2 6) 5 0 1 r 5 22 or r 5 6. So g(r) is not defined at r 5 22 and r 5 6. So there are two vertical asymptotes at the lines r 5 22 and r 5 6. The graph of g(r) follows. 10 g(r)
r 10
–10
681
2x 2 3 10. g(x) 5 x 1 3x 2 1 i. To find horizontal intercept(s): set g(x) 5 0, which is equivalent to setting x2 1 2x 2 3 5 0 1 (x 2 1)(x 1 3)5 0 1 x 5 1, x 5 23. So there are horizontal intercepts at (1, 0) and (23, 0). To find vertical intercept: evaluate g(0) 5 3. So the vertical intercept is at (0, 3). ii. To find vertical asymptote: set 3x 2 1 5 0 1 x 5 1/3. So the vertical asymptote is the vertical line at x 5 1/3. x2 5 13x. So g(x) iii. The end behavior as x S 6` is g(x) 1 3x 1 is asymptotic to the line y 5 3x (called an oblique asymptote). iv. Graph of g(x) 2
15 –10
g(x)
10
(0, 3)
7. a. g(x) 5
2 (x 11 3) 2
5
11
(–3, 0)
b. g(x) 5 2 (x 11 3) 2 1 1 ?
(x 1 3) 2 (x 1 3) 2
–15 –10
5
2 5 (x 21 1 x (x1 16x3)12 9 1 3) 2
15
(1, 0)
Ch. 8
–10
x2 1 6x 1 9 5 21 x1 2 1 6x 1 9
–15
2 p(x) 6x 1 8 5 xx2 1 5 q(x) 1 6x 1 9
c. To find any horizontal intercepts, let the numerator p(x) 5 0 to get x2 1 6x 1 8 5 (x 1 4)(x 1 2) 5 0 1 x 5 24 or x 5 22. So the horizontal intercepts are at (24, 0) and (22, 0). To find any vertical asymptotes, let the denominator q(x) 5 0 to get (x 1 3)2 5 0 1 x 5 23. So g(x) is not defined when x 5 23. So there is one vertical asymptote at the line x 5 23. 8. a. Horizontal intercepts b. Vertical intercepts
c. Horizontal intercept d. Vertical asymptote
1 6 9. f(x) 5 2x x 2 3 i. To find horizontal intercept(s): set f(x) 5 0, which is equivalent to setting the numerator 2x 1 6 5 0 1 x 5 23. So there is one horizontal intercept at (23, 0). To find vertical intercept: evaluate f(0) 5 6/23 5 22. So the vertical intercept is at (0, 22). ii. To find vertical asymptote: set x 2 3 5 0 1 x 5 3. So the vertical asymptote is the line at x 5 3. iii. The end behavior as x S 6` is f(x) S 2x x 5 2. So there is a horizontal asymptote at the line y 5 2. iv. Graph of f (x) 15
(–3, 0) 5
5
–5
10
–5 (0, –2) –10 –15
Exercises for Section 8.7 1. a. 3t2 1 10t 2 4 b. 23t2 1 2t 1 6
c. 18t3 1 27t2 – 26t 2 5 2 1 4t 2 5 d. 3t 6t 1 1
3. a. j(x) 5 3x5 1 x2 1 x – 1; k(x) 5 3x5 – x2 1 x 1 1; l(x) 5 3x7 2 3x5 1 x3 – x b. j(2) 5 101; k(3) 5 724; l(21) 5 0 5. a. R(n) 5 25n b. R(n) 5 25n 2 500 c. R(n) 5 25(n 2 30) 2 500 5 25n 2 1250 7. If a worker works t hours a week (where t $ 40), then the worker’s weekly paycheck, P (in dollars), is the sum of two terms: regular pay 1 overtime pay. The regular pay is 20 ? 40 5 $800 a week. The overtime pay 5 30 ? (t 2 40), where t 5 total number of hours worked. So the weekly paycheck is P 5 800 1 30(t 2 40), where t $ 40. 9. x h(x) j(x) k(x)
0 26 0 9
1 26 24 5
2 26 216 255
3 26 236 2315
4 26 264 21015
5 26 2100 22491
11. f (x)
10
–15 –10
10
–5
15
x 23 22 f(x) 9 4 g(x) 24 23 f(x) 1 g(x) 5 1 f(x) 2 g(x) 13 7 f(x) ? g(x) 236 212 g(x)/f(x) 24/9 23/4
21 1 22 21 3 22 22
0 0 21 21 1 0 undefined
1 1 0 1 1 0 0
2 4 1 5 3 4 1/4
3 9 2 11 7 18 2/9
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CH. 8 Exercises Solutions for Section 8.7
CH. 8 Algebra Aerobics Solutions for Section 8.8
13. a. C(n) 5 500 1 40n 40n b. P(n) 5 500 1 n c. P(25) 5 60; P(100) 5 45. As the number of people attending increases, the cost per person decreases. If only 25 attend (the minimum size), the cost would be $60 per person. If 100 attend (the maximum size), the cost would be $45 per person.
Algebra Aerobics 8.8a 1. ƒ(x) 5 2x 1 3, g(x) 5 x2 2 4 a. ƒsgs2dd 5 ƒs0d 5 2s0d 1 3 5 3 b. gsƒs2dd 5 gs7d 5 s7d 2 2 4 5 49 2 4 5 45
15. Graph A, f(x) asymptotes: horizontal, y 5 23; vertical, x 5 22 Graph B, g(x) asymptotes: horizontal, y 5 4; vertical, x 5 2 Graph C, h(x) asymptotes: horizontal, y 5 22; vertical, x 5 22 and x 5 3
c. ƒsgs3dd 5 ƒs5d 5 2s5d 1 3 5 10 1 3 5 13
17. g(x) has no horizontal intercepts and a vertical asymptote at the line x 5 23. The graph of g(x) verifies this.
f. (g + ƒ)(x) 5 g(ƒ(x)) 5 g(2x 1 3) 5 (2x 1 3) 2 2 4 5 4x2 1 12x 1 9 2 4 5 4x2 1 12x 1 5
y
g(x)
1 d. sP + Qdstd 5 PsQstdd 5 Ps3t 2 5d 5 3t 2 5 e. sQ + Pdstd 5 QsPstdd 5 Q A1t B 5 3 A1t B 2 5 5 3t 2 5
–10
19. a. The domain is all real numbers. b. Yes, there is one horizontal intercept at x 5 0, the origin. There are no vertical asymptotes since x is defined for all real values (so there are no singularities). x c. As x S 6`, S(x) 5 x2 1 < xx2 5 1x . So as x S 6`, 1 S(x) S 0. So S(x) has a horizontal asymptote at the x-axis. d. To some the graph looks like a slithering serpent. y S(x)
–1
21. There are infinitely many possibilities. If a and b are nonzero real numbers, then one set of answers would be of the form a(x 1 3) (x 2 4) f(x) 5 b(x 2 1) (x 1 5) . (Note: You could create additional rational functions by, say, squaring or cubing one or more terms.) If we let a 5 b 5 1, then the graph of f(x) is as follows: 10 8
f(x) =
2 3. Fsxd 5 x 2 1 , Gsxd 5 3x 2 5 a. (F + G)(x) 5 F(G(x)) 5 F(3x 2 5) 5 (3x 2 25) 2 1 5 3x 22 6 2 2 b. (G + F)(x) 5 G(F(x)) 5 G Ax 2 1 B 5 3 Ax 2 1 B 2 5 6 5x2125 6 c. (F + G)(x) 5 3x 22 6 2 (G + F)(x) 5 x 2 1 2 5 5 6 2 5(x 2 1) 11 2 5x 5 x21 x 2 1
4. a. b. c. d.
x 10
–10
6
e. (ƒ + g)(x) 5 ƒ(g(x)) 5 ƒ(x2 2 4) 5 2(x2 2 4) 1 3 5 2x2 2 8 1 3 5 2x2 2 5
b. sQ + Pds2d 5 QsPs2dd 5 Q A12B 5 3 A12B 2 5 5 27 2 c. (Q + Q)(3) 5 Q(Q(3)) 5 Q(3(3) 2 5) 5 Q(4) 5 3(4) 2 5 5 7
x 10
1
d. ƒsƒs3dd 5 ƒs9d 5 2s9d 1 3 5 18 1 3 5 21
2. a. (P + Q)(2) 5 P(Q(2)) 5 P(3(2) 2 5) 5 P(1) 5 11 5 1
5
–10
Ch. 8
Section 8.8
(x + 3)(x – 4) (x – 1)(x + 5)
f(22) 5 2 g(22) 5 0 f(0) 5 22 g(0) 5 1
–10 –8 –6
–2 –2
(g (f (g (f
s s s s
f )(22) 5 g(2) 5 2 g)(22) 5 f(0) 5 22 f )(0) 5 g(22) 5 0 g)(0) 5 f(1) 5 21
5. a. f(x) 5 x2 2 2 b. g(x) 5 12 x 1 1 c. (g f )(x) 5 g(x2 2 2) 5 12 (x2 2 2) 1 1 5 12x2 2 1 1 1 5 12x2 s
d. (f g)(x) 5 f A12 x 1 1B 5 A12 x 1 1B 2 2 2 s
5 A 14 x2 1 x 1 1B 2 2 5 14 x2 1 x 2 1
4 2
e. f. g. h.
6. a. (g f )(22) 5 12 (22) 2 5 2 and ( f g)(22) 5 14 (22) 2 1 (22) 2 1 5 22. s
2
4
6
8
x 10
–4 –6 –8 –10
23. f(x) matches Graph C; g(x) matches Graph B; h(x) matches Graph A.
s
So both answers agree with the answers in Problem 4, parts (e) and (f). 7. a. (h f g)(4) 5 h( f(g(4))) 5 h( f(0)) 5 h(3) 5 4 b. (f h g)(1) 5 f(h(g(1))) 5 f (h(3)) 5 f(4) 5 5 s
s
s
s
8. a. (h f g)(3) 5 h( f(g(3))) 5 h( f (2)) 5 h(21/3) 5 3 b. (f g h)(100) 5 f(g(h(100))) 5 f(g(3)) 5 f(2) 5 21/3 s
s
s
s
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CH. 8 Algebra Aerobics Solutions for Section 8.8
CH. 8 Exercises Solutions for Section 8.8
Algebra Aerobics 8.8b 1. a.
t
g(t)
t
h(t)
0 1 2 3
5 3 1 21
21 1 3 5
3 2 1 0
d. Dividing x by 5 and then adding 3 e. Subtracting 2 from z and then dividing the result by 23 9. The functions in Graphs A and D are 1-1 since they pass the horizontal line test. The functions in Graphs B and C are not 1-1, since they fail that test.
b. sg + hds3d 5 gshs3dd 5 gs1d 5 3; sh + gds3d 5 hsgs3dd 5 hs21d 5 3. t c. Yes, since sg + hdstd 5 gshstdd 5 g A5 2 2 B 5 5 2 t 5 2 2 A 2 B 5 5 2 s5 2 td 5 t and sh + gdstd5
10. a. f is not one-to-one on the domain of all real numbers (since it fails the horizontal line test), so it cannot have an inverse. 5
5 2 s5 2 2td
1 2. ƒsxd 5 2x 1 1,gsxd 5 x 2 2 1 x 2 1 (ƒ + g)(x) 5 ƒ(g(x)) 5 ƒ Ax 2 2 B 5 2A 2 B 1 1 2(x 2 1) 2
x 5
–5
hsgstdd 5 hs5 2 2td 5 5 t and the domains 2 and ranges of both g and h are all real numbers. d. They are inverse functions.
5
683
–5
b. If we restrict the domain of f to x $ 22, then f has an inverse on this new domain (since it now passes the vertical test).
115x21115x
(g + ƒ)(x) 5 g(ƒ(x)) 5 g(2x 1 1) 5
5
(2x 1 1) 2 1 2
x 5
–5
3 3. ƒsxd 5 "x 1 1,gsxd 5 x3 2 1
–5
3 (ƒ + g)(x) 5 ƒ(g(x)) 5 ƒ(x3 2 1) 5 " (x3 2 1) 1 1
c. Letting f(x) 5 y, we have
3 3 5 " x 5 (x3 ) 1/3 5 x1 5 x
y 5 (x 1 2) 2 1 x 5 1 "y 2 2 1 f21 (x) 5 "x 2 2 (using the convention of designating x as the input variable). The graph of f 21(x) is shown below.
(g + ƒ)(x) 5 g(ƒ(x)) 5 g( "x 1 1) 3
3 5 A" x 1 1 B 2 1 5 [(x 1 1) 1/3]3 2 1
3
5 (x 1 1) 2 1 5 x
5
1
1
x 511x 4. ƒ(ƒ21 (x)) 5 ƒ A1 1 x B 5 1 1 x 21 2 xx x x 1 5 (1/x) 5x x 2 1 1 11x2 1 1 1 1 ƒ21 (ƒ(x)) 5 ƒ21 Ax 2 5x21 1 x21 1B 5 1
5
x x 2 1 1 x 2 1
x 2 1
5
x x 2 1
?
x 2 1 1
x 2 1
5x
So ƒsƒ21 sxdd 5 x and ƒ21 sƒsxdd 5 x. 5. Letting f (x) 5 y, we have y 5 3x 1 5 1 y 2 5 5 3x 1 3 3 21 x5y2 5 1 f (x) 5 x 2 5 (using the convention of designating x as the input variable). 6. Letting g(x) 5 y, we have y 5 (x 2 2) 3/2 1 x 5 y 2/3 1 2 1 g21 (x) 5 x2/3 1 2 (using the convention of designating x as the input variable). 7. Letting h(x) 5 y, we have y 5 5x3 2 4 1 y 1 4 5 5x3 1 y 1 4
4 1/3 x 5 A 5 B 1/3 1 h21 (x) 5 Ax 1 (using the convention 5 B of designating x as the input variable).
8. a. Saying “no” b. Taking the bus from home, then going to class. c. Turning off the light, leaving the room, closing the door, and then locking the door.
x 5
–5
–5
11. a. The radius R(t) as a function of time t is R(t) 5 10t. b. R(2) 5 10 ? 2 5 20, so area A(20) 5 p(20)2 < 1257 square feet. c. sA + Rdstd 5 AsRstdd 1 As10td 5 ps10td 2 5 100pt2
Exercises for Section 8.8 1. a. ƒ(g(1)) 5 ƒ(0) 5 2 b. g(ƒ(1)) 5 g(1) 5 0 c. ƒ(g(0)) 5 ƒ(1) 5 1
d. g(ƒ(0)) 5 g(2) 5 3 e. ƒ(ƒ(2)) 5 ƒ(3) 5 0
3. a. g(ƒ(2)) 5 g(0) 5 1 b. ƒ(g(21)) 5 ƒ(2) 5 0
c. g(ƒ(0)) 5 g(4) 5 23 d. g(ƒ(1)) 5 g(3) 5 22
5. a. b. c. d.
F(G(1)) 5 F(0) 5 1 G(F(22)) 5 G(23) 5 4 F(G(2)) 5 F(0.25) 5 1.5 F(F(0)) 5 F(1) 5 3
Ch. 8
5 2x 2 5 x
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CH. 8 Exercises Solutions for Section 8.8
CH. 8 Check Your Understanding
1 e. (F G)(x) 5 2 Axx 2 1 2B 1 1
letter equivalent. The domain of the inverse function would be the integers 1 through 26. b. “MATH RULES”
s
5
(2x 2 2) 1 (x 1 2) x 1 2
5 x 3x 1 2
(2x 1 1) 2 1
f. (G F)(x) 5 (2x 1 1) 1 2 5 2x 2x 1 3 s
7. a. A(r) 5 pr2, where r is measured in feet and A(r) is measured in square feet. b. r 5 R(t) 5 5t, where t is measured in minutes and R(t) is measured in feet. c. A(R(t)) 5 p25t2, where t is measured in minutes and A is measured in square feet. d. A(R(10)) 5 p ? 25 ? 102 5 2500p < 7854 sq. ft and A(R(60)) 5 p ? 25 ? 602 < 282,743 sq. ft. 9. r(t) 5 13t and thus A(r(t)) 5 p(13t)2 5 169pt 2 11. a. T 5 32 – 5s b. If the road is 40 feet wide, then k 5 20 and thus x 2 x2 Ssxd 5 S 1 2 12 ? A20 B T Sd 5 S 1 2 800 T Sd.
27. a. Yes, f (x) has an inverse since its graph passes the horizontal line test. b. The domain of f (x) is the interval [24, `]. The range of f(x) is the interval [0, `]. c. f(24) 5 0, f(0) 5 2, and f(5) 5 3. This means that the points (24, 0), (0, 2), and (5, 3) all lie on the graph of f(x). d. Given the results in part (c), the points (0, 24), (2, 0), and (3, 5) all lie on the graph of f 21(x). So f 21(0) 5 24, f 21(2) 5 0, f 21(3) 5 5. 15 29. Q21 (x) 5 3x 1 ,Q(3) 5 23,Q21 (3) 5 12 2 3 21 31. Q21 (x) 5 x 2 1 ,Q(3) 5 2,Q (3) 5 3/2
33. a.
c. At the middle of the 40-ft road x 5 0 and therefore 0 Ss0d 5 S 1 2 800 T Sd 5 Sd. At the edge of the 40-ft road, 2
Ch. 8
20 1 x 5 20, so Ss20d 5 S 1 2 800 T Sd 5 2 Sd.
d. T(S(x)) 5 5
x2 T A S 1 2 800 T SdB 5 32 x2 32 2 5Sd 1 160 Sd
2 5S1 2
13. a. M(x) 5 (L J K)(x) 5 L(J(K(x))) 5 L(J(log x)) 5 L((log x)3) 5 1/(log x)3 b. Take the log of x, cube the result, and then place it in the denominator, with 1 as the numerator. s
15. If f (x) 5 4x, g(x) 5 ex, and h(x) 5 x 2 1, then f (g(h(x))) 5 f(g(x 2 1)) 5 f(ex21) 5 4ex21 5 j(x) 17. f(g(x)) 5 f(x2 1 1) 5 "(x2 1 1) 2 1 5 "x2 5 x since x . 0 g(f(x)) 5 g A !x 2 1 B 5 A !x 2 1B 2 1 1 5 (x 2 1) 1 1 5 x x3 2 5 4 B
34 A 3
3 g(f(x)) 5 g A ! 4x 1 5 B 5
5
(4x 1 5) 2 5 4
4
5x 1/3
G(F(t)) 5 G(e ) 5 ln (e ) 3t
23.
x
f 21(x)
5 1 2 4
22 21 0 1
3t 1/ 3
)
32 8
x (quarts) g(x) (gallons)
2 0.5
4 1
8 2
16 4
s
s
s
35. a. Wmen(h) 5 50 1 2.3(h 2 60), where a reasonable domain might be 600 # h # 780; Wwomen(h) 5 45.5 1 2.3(h 2 60), where a reasonable domain might be 600 # h # 740. b. Wmen(70) 5 50 1 2.3(70 2 60) 5 73 kg, so 73 kg is the “ideal” weight of a 59100 man. Wwomen(66) 5 45.5 1 2.3(66 2 60) 5 59.3 kg, so 59.3 kg is the “ideal” weight of a 5960 woman. c. W21men(77.6) means 77.6 5 50 1 2.3(h 2 60) 1 h 5 72 inches. A man with a IBW of 77.6 kg should be 72 inches or 6 ft tall. 50 1 2.3(h 2 60) d. Wnewman(h) 5 ; 0.4356
5 eln(t) 5 t (where t . 0)
5 ln(e ) 5 t t
25. a. Yes, this is a 1-1 function since each letter is associated with a unique number. The inverse function would just consist of matching each number between 1 and 26 wih its
45.5 1 2.3(h 2 60) . 0.4356 45.5 1 2.3(h 2 60) 21 W newwomen(125) means 125 5 0.4356
Wnewwomen(h) 5
3
21. F(G(t)) 5 F(ln (t ) 5 e3 ln(t
16 4
i. (g f)(8) 5 g(2) 5 0.5 gal ii. g21(2) 5 8 qt iii. (f 21 g21)(1) 5 f 21(4) 5 16 cups iv. (f 21 g21)(2) 5 f 21(8) 5 32 cups c. (f 21 g21)(x) is a function that converts gallons to cups.
3 3 155 " x 5x
A "(4x 1 5) B 3 2 5
1/3
8 2
s
2
3 5 19. f(g(x)) 5 f Ax 2 4 B 5
4 1
b. x2 800 T Sd
20 Sd 5 e. T(S(0)) 5 32 2 5Sd and T(S(20)) 5 32 2 5Sd 1 160 32 2 5Sd 1 2.5Sd 5 32 2 2.5Sd s
x (cups) f(x) (quarts)
e.
1 h < 63.89 < 64 inches. So 125 lb is the IBW for a woman about 5940 in height.
37. F(G(x)) 5 F(loga (x)) 5 aloga(x) 5 x G(F(x)) 5 G(ax) 5 loga(a x) 5 x
Ch. 8: Check Your Understanding 1. False
6. True
11. False
16. False
2. True
7. True
12. False
17. False
3. True
8. False
13. False
18. True
4. False
9. True
14. True
19. True
5. False
10. True
15. False
20. True
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CH. 8 Review: Putting It All Together
CH. 8 Review: Putting It All Together
21. True
e.
22. Possible answer: y 5 x 1 x 1 1 2
23. y 5 20.25x 2 24. Possible answer: ƒ(x) 5 (x 1 1)(x 2 3)(x 2 4) 25. Possible answers: h(x) 5 2(x 1 1)(x 2 3)(x 2 4), g(x) 5 23(x 1 1)(x 2 3)(x 2 4) 26. Possible answer: y 5 2(x 2 1)2 1 3 27. Possible answer: y 5 2(x 2 3)2 2 5 28. y 5 21 2 sx 2 2dsx 1 2d
Cost of Making One Tile Cost per tile ($)
4
685
12 10 8 6 4 2 0
0
5 10 x (in inches)
15
f. Estimating from the graph, when C(x) 5 $7, then x < 7.50, so the width (and length) of the whole tile would be about 9.50. To keep the cost/tile at $7 or below, the dimensions of a tile must be at most 9.50 by 9.50. 5. g(x) 5 2x2 and h(x) 5 20.5x2
29. r 5 s 2 s 1 5 2
7. a. Vertex for F(x) is (0, 0), vertex for G(x) 5 (0, 5), vertex for H(x) is (22, 0), vertex for J(x) is (1, 25)
30. G(x) 5 (x 1 2) 1 2(x 1 2) 2
b.
31. 2h(t) 5 2(t 2 2)2
y 20
32. Possible answer: y 5 x and y 5 2(x 2 1) 1 1 2
G(x) H(x)
2
33. Possible answer: y 5 (x 1 4)2
F(x) 4
–4
x
J(x)
35. Possible answer: ƒ(x) 5 x3 1 2x2, g(x) 5 5x 2 2 36. H(t) 5 3t 1 1, Qstd 5 "t
–20
x 2 2 37. Possible answer: ƒsxd 5 xsx 1 3d
38. False
44. False
50. True
56. True
39. True
45. True
51. False
57. False
40. False
46. False
52. True
58. False
41. True
47. False
53. False
42. True
48. True
54. True
43. False
49. True
55. False
Ch. 8 Review: Putting It All Together 1. In Graph A, the parabola is concave up with an estimated minimum at (2, –4). Hence the axis of symmetry is the line x 5 2 and there are two horizontal intercepts, at x 5 0 and x 5 4. In Graph B, the parabola is concave down with an estimated maximum at (0, 23). Hence the axis of symmetry is the vertical axis (the line t 5 0) and there are no horizontal intercepts. 3. a. Area of interior square 5 x2 square inches; area of each of the maple strips 5 (x 1 1) ? 1 5 x 1 1 square inches. b. Cost of white oak: ($2.39/ft2) ? (1 ft2/144 in2) < $0.02/in2; cost of maple: ($4.49/ft2) ? (1 ft2/144 in2) < $0.03/in2 c. For white oak: 0.02x2 (in dollars); for all four maple strips: 4(0.03)(x 1 1) 5 0.12x 1 0.12 (in dollars) d. C(x) 5 0.02x2 1 0.12x 1 5.12, a quadratic function
c. The graph of G(x) is the graph of F(x) shifted up five units. The graph of H(x) is the graph of F(x) shifted left two units. The graph of J(x) is the graph of F(x) shifted right one unit, flipped over the x-axis, and shifted down five units. 9. a. One possibility is Q(t) 5 (t 2 4)(t 1 2) 5 t2 2 2t 2 8. The vertex of Q(t) is at (1, 29). b. One possibility is M(t) 5 3Q(t) 5 3(t 2 4)(t 1 2) 5 3t2 2 6t 2 24. The vertex of M(t) is at (1, 227). So the vertices are not the same, though they share the same t-coordinate. c. One possibility is P(t) 5 Q(t) 1 10 5 t2 2 2t 1 2. P(t) has no horizontal intercepts since the discriminant 5 (22)2 2 (4 ? 1 ? 2) 5 4 2 8 5 24, which is negative. 11. Since we have set the vertex at the origin, then the equation is of the form y 5 ax2 (where a , 0). We know two points on the parabola, (d, 232) and (2(100 2 d), 272) 5 (d 2 100, 272). Substituting each set of points into the equation y 5 ax2, we get the two equations 232 5 ad 2
and
272 5 a(d 2 100)2
Solving both equations for a, we get a 5 232/d 2
and
a 5 272/(d 2 100)2
Setting both expressions for a equal and solving for d gives us 232/(d 2) 5 272/(d 2 100)2 cross-multiply 232(d 2 100)2 5 272d2 simplify 232(d 2 100)2 1 72d 2 5 0 2 232(d 2 200d 1 10,000) 1 72d 2 5 0 40d 2 1 6400d 2 320,000 5 0 divide by 40 d 2 1 160d 2 8000 5 0
Ch. 8
34. hstd 5 14 sx 1 2dsx 1 1dsx 2 2dsx 2 3d
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CH. 8 Review: Putting It All Together
CH. 8 Review: Putting It All Together
Now we can solve for d either using the quadratic formula or factoring. Quadratic formula (letting a 5 1, b 5 160, c 5 28000) gives:
17. a. Behavior looks like a cubic polynomial. (Note: The graph doesn’t look linear on semi-log plot, so data are not exponential.) The best-fit cubic is P(t) 5 1.26x3 2 6.94x2 1 9.22x 1 50.21
2160 6 "(160) 2 2 4(1)(28000) 2?1 2160 6 !25,600 1 32,000 5 2 2160 6 !57,600 2160 6 240 5 5 5 280 6 120 2 2 So d 5 40 feet or 2200 feet. Only d 5 40 feet makes sense. d5
[where P(t) 5 price per hive in t years after 2000]. P(t) is plotted on the accompanying graph along with the data. Average Price Paid for a Bee Hive for California Farmers Average price per hive ($)
$140
Factoring d 1 160d 2 8000 5 0 gives (d 2 40) (d 1 200) 5 0, which confirms that either d 5 40 feet or 2200 feet, where d 5 40 feet is the only valid answer in this context. 2
Ch. 8
Substituting d 5 40 feet into the equation 232 5 ad 2, we get 232 5 a(40)2 1 a 5 232/1600 5 20.02. So the equation for the swimming pool parabolic roof is y 5 20.02x 2, where x and y are both in feet. 13. a. The highest point of her dive will be at the vertex of the height function (which is concave down). Letting a 5 216, b 5 12, and c 5 25, the t-coordinate of the vertex is at 212/(2 ? (216)) 5 0.375 seconds. Then H(0.375) 5 25 1 (12 ? 0.375) 2 16(0.375)2 5 27.25 feet above water will be the highest point of her dive. b. She will hit the water when H(t) 5 0 1 25 1 12t 2 16t2 5 0. Using the quadratic formula, letting a 5 216, b 5 12, and c 5 25, we have t5 5
$100 $80 $60 $40 $20 $0
0
1
2 3 4 5 Years since 2000
19.
300
y
g(x) 0
21 0 1 2 3 4 5
5 0 23 24 23 0 5
n.a. 25 23 21 1 3 5
x
–100
<20.93 seconds or 1.68 seconds. Only the positive value makes sense in this context. So about 1.68 seconds (a little under 2 seconds) after she starts her dive, she will hit the water.
y
5
h(x)
212 6 "(144 1 1600) 232
x
7
f(x)
212 6 "(12) 2 4(216)(25) 2 ? (216)
Average Rate of Change
6
b. P(10) < $708/hive c. (2.5 hives/acre) ? ($708/hive) 5 $1770/acre d. Since healthy beehives would be scarce, the price per beehive would probably go up.
2
212 6 "1744 212 6 41.8 5 < 232 232
15. a.
$120
Average Rate of Change of Average Rate of Change n.a. n.a.
23 2 (25) 1 2 0
The graph of g(x) is f(x) vertically stretched by a factor of 4; that is, for each value of x, g(x) is four times the value of f(x). The graph of h(x) is the graph of f(x) flipped over the x-axis and then vertically compressed by a factor of 0.5. 21. a. g(x) 5 (2/3)f(x 2 4) 2 1. So the graph of f(x) was shifted four units to the right, compressed by a factor of 2/3, then shifted down one unit to create the graph of g(x). 3(x 2 4) 1 14 b. g(x) 5 3(x 22 4) 2 1 5 3(x 22 4) 2 3(x 2 4) 5 23x 3x 2 12
c. The domain of g(x) is all real numbers except 4. 10
52
2 2 2 2
g(x) –10
10 x
–10
b. A linear function. c. The fourth column shows that the average rate of change (of the third column with respect to x) is constant, which means that the third column is a linear function of x.
d. If g(x) 0, then 0 5 (23x 1 14)/(3x212) 1 0 5 23x 1 14 1 x 14/3 423. So g(x) has a single horizontal intercept at x 423.
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CH. 8 Review: Putting It All Together
Exercises Solutions for EE on Mathematics of Motion
If x 0, then g(0) 14/(212) 5 2116. So g(x) has a vertical intercept at 2116. e. If the denominator 3x 2 12 0 1 x 4. So g(x) is not defined when x 4, but does have a vertical asymptote at the line x 4. 1 14 23x f. As x S 6`, g(x) 5 23x 3x 2 12 < 3x 5 21. So as, x S 6`, g(x) S 21 (but never reaches 21). So g(x) has a horizontal asymptote at the line y 5 21.
687
25. The function is 121 since if f(x1) 5 f(x2), then (x1 2 2)3 1 1 5 (x2 2 2)3 1 1 1 (x1 2 2)3 5 (x2 2 2)3 1 x1 2 2 5 x2 2 2 1 x1 5 x2. So on an appropriate domain f 21 exists. Letting y 5 f(x), we can solve for x in terms of y. Given Subtract 1 Take the cube root Add 2, switch sides
y 5 (x 2 2)3 1 1 y 2 1 5 (x 2 2)3 (y 2 1)1/3 5 x 2 2 x 5 (y 2 1)1/3 1 2
So f 21(y) 5 (y 2 1)1/3 1 2, or since the function is abstract, we can use any name for the input variable, in particular x, to get the conventional form, f21 (x) 5 (x 2 1)1/3 1 2.
23. a. S(t) 5000(1.04)t; R(t) 5000(1.10)t; T(t) 5 S(t) 1 R(t) 5000(1.04)t 1 5000(1.10)t where t 5 years since the start of the investments.
feet 1 hr 1 min i. 7611 miles ? 5280 hr 1 mile ? 60 min ? 60 sec < 1116 ft/sec. ii. D(t) 1116t, where t is in seconds and D(t) is in feet from the lightning strike. iii. Yes, the rule of thumb is reasonable since for each second after the strike, the sound thunder travels about 1116 feet. b. i. A(r) 5 pr2, where r is the radius (in feet) of the sound circle. ii. A(D(t)) 5 A(1116t) 5 p(1116t)2 ≈ 1,245,500pt2 < 3,913,000 t2, where t is in seconds and A(D(t)) is in square feet. A(D(4)) 3,913,000(4)2 < 63 million 1sq. mile square feet or (63 ? 106 sq. ft) ? < 2.3 2 5280 sq. ft square miles. iii. When the time doubles, the distance doubles, but the area increases by a factor of 4.
27. a. b.
Individual Return on $5000 at 4% (S(t)) and $5000 at 10% (R(t)), and Total Return T(t) = R(t) + S(t) $120,000 S(t) R(t) T(t) = S(t) + R(t)
$100,000 $80,000 $60,000 $40,000 $20,000 $0
0
10 20 30 Years since start of investment
Years Since Start of Investment
T(t) 5 S(t) 1 R(t)
U(t) 5 5000(1.14)t
0 5 10 15 20 25 30
$10,000 $14,136 $20,370 $29,891 $44,593 $67,503 $103,464
$5,000 $9,627 $18,536 $35,690 $68,717 $132,310 $254,751
Comparing the Return on Investing $5000 at 4% plus $5000 at 10% (T(t)) vs. Investing $5000 at 14% (U(t))
Exercises for EE on Mathematics of Motion 1.
Time (sec) 0.0000 0.0333 0.0667 0.1000 0.1333 0.1667
Distance (cm)
Avg. Vel. over Previous 1/30 sec. (cm/sec)
0.00 3.75 8.67 14.71 21.77 29.90
n.a. 113 147 181 212 243
The average velocity (over each 1/60 of a second) increases rapidly as time progresses. 3. For d 5 490t2 1 50t: a. 50 is measured in cm/sec; it is the initial velocity of the object falling; 490 is measured in (cm/sec)/sec and is half the acceleration due to gravity when measured in these units. b, c. Below is a small table of values and the graph of the equation with the table points marked on it.
$300,000 $250,000
60
$200,000 U(t) with 14% return $150,000 $100,000 $50,000 $0
T(t) = S(t) + R(t) 0
5 10 15 20 25 Years since start of investments
30
t
d
0.0 0.1 0.2 0.3
0.0 9.9 29.6 59.1
d (0.3, 59.1)
d = 490t 2 + 50t (0.2, 29.6)
(0.1, 9.9) 0 0
t 0.3
EE
c. She would have only S(30) 5000(1.04)30 < $16,217. d. Let U(t) 5000(1.14)t, where 5000 is the initial value invested at 14% per year. The following table and graph show that after 30 years the U(t) account contains almost 2.5 times as much as the T(t) account (or about 150% more).
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Exercises Solutions for EE on Mathematics of Motion
5. For d 5 4.9t2 1 1.7t: a. 1.7 is the initial velocity of the object falling; it is measured in meters per second; 4.9 is half the gravitational constant when it is measured in (meters/sec)/sec. b, c. Below is a small table of values, and next to it is the graph with the table points marked on it. 1
t
d
0.0 0.1 0.2 0.3
0.00 0.22 0.54 0.95
t
0
45
0
d = 4.9t 2 + 1.7t (0.20, 0.54)
(0.10, 0.22) t 0.3
m m 2 7. m 5 sec 2 ? sec 1 sec ? sec
9. a. d 5 490t2 1 50t and v 5 980t 1 50 b. At t 5 1, d 5 540 cm and v 5 1030 cm/sec. At t 5 2.5, d 5 3187.5 cm and v 5 2500 cm/sec.
EE
h
h = 200t – 4.9t 2
d. The results in this question are very similar to those in earlier parts of this chapter. The shape of the graph is that of a quadratic; the coefficient of t 2 is half the gravitational constant, and the coefficient of t is the initial velocity.
0 28 88 180 304 460
2,500
(0.30, 0.95)
0
0 1 2 3 4 5
b. The graph of h over t is given in the following diagram.
d
0
11. a. d 5 16t2 1 12t b. t d
Exercises Solutions for EE on Mathematics of Motion
Height
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c. The object reaches a maximum height of approximately 2000 meters after 20 seconds. It reaches the ground after approximately 40 seconds of flight. 23. For h 5 85 2 490t2: a. 85 is the height in centimeters of the falling object at the start; 2490 is half the gravitational constant when measured in (cm/sec)/sec; it is negative in value since h measures height above the ground and the gravitational constant is connected with pulling objects down. This will mean subtraction from the starting height of 85 cm. b. The initial velocity is 0 cm/sec. c. Below is a table of values for this function
d 500
t
h
0.0 0.1 0.2 0.4
85.0 80.1 65.4 6.6
d = 16t2 + 12t
d. The following diagram is the graph of the function with the table entries marked on it. h 90 0
t 0
5
c. The graph and table of the function are given above. h = 85 – 490t 2
13. The distance is measured in meters if the time is measured in seconds. The use of 4.9 for half of the gravity constant is the indicator of these units. 15. a. Student answers will vary considerably. b. Since the velocity is changing at a constant rate, a straight line should be a good fit. The graph of this line is a representation of average velocity. 17. a. The coefficient of t2 is one-half the gravity constant. Since the coefficient of t2 is approximately 490, distance is measured in centimeters and time in seconds, and 490 is measured in cm/sec2. The coefficient of t is an initial velocity of 7.6 cm/sec. b. When t 5 0.05 sec, d 5 1.59 cm; when t 5 0.10 sec, d 5 5.62 cm, and when t 5 0.30 sec, d 5 45.99 cm. 19. It represents an initial velocity of the object measured in meters per second. 21. a. At t 5 0 sec, h 5 0 m; at t 5 1 sec, h 5 195.1 m; at t 5 2 sec, h 5 380.4 m; at t 5 10 sec, h 51510 m.
0
0
t 0.6
25. a. The initial velocity is positive since we are measuring height above ground and the object is going up at the start. b. The equation of motion is h 5 50 1 10t 2 16t2, where height is measured in feet and t in seconds. 27. a. 980 cm/sec2 since we are measuring in cm and in sec. b. t v 0 1 2 3 4
266 21046 22026 23006 23986
c. The graph is given below. The object is traveling faster and faster toward the ground. The increase in downward
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velocity is at a constant rate, as we can see from the constant slope of the graph. This constant acceleration, of course, is due to gravity.
0
v 0
4 (0, –66)
t
(1, –1046) (2, –2026) v = –980t – 66
(3, –3036)
–4,000 (4, –3986)
d. Ordinarily, if t 5 0 corresponds to the actual start of the flight, then the initial condition given would indicate that the object was thrown downward at a speed of 66 meters per second. This interpretation comes from the negative sign given to the initial velocity. But this would contradict the statement in the problem to the effect that it is a “freely falling body.” In this context, another interpretation is suggested by the laboratory experiment, namely that the object started being timed at a point along its downward fall.
v0 1 sv0 1 atd and 2 2v0 t 1 at 2 5 v0t 1 12at 2 2
31. Forming dt 5 d5
solving for d, we get
689
This is very similar in form to the falling-body formula. The acceleration factor increases the velocity in a manner proportional to the square of the time traveled, and the initial velocity increases the distance in a manner proportional to the time. 33. a. After 5 seconds its velocity is 110 cm/sec; after 1 minute (or 60 seconds) its velocity is 660 cm/sec; after t seconds, its velocity is v(t) 5 60 1 10t cm/sec. b. After 5 seconds its average velocity is (60 1 110)/2 5 85 cm/sec. 35. a. v(t) 5 200 1 60t meters/sec b. d(t) 5 200t 1 30t2 meters 37. a. Using units of feet and seconds, the equation governing the water spout is h 5 216t 2 1 v0 t, where h is measured in feet and t, time, in seconds and where v0 is the soughtafter initial velocity. We are given that the maximum height reached is 120 ft. The maximum height is achieved v
at the vertex, i.e., when t 5 2v0 /(232) 5 320 . Substituting for t and h gives us 120 5 216 Q
v0 2 32 R
v
v2
v2
2
1 v0 Q 0R 5 2640 1 2 0 5 v0 32 64 64
Thus v02 5 7680 or v0 < 87.6 ft per sec b. t 5 v0 /32 5 87.6/32 < 2.74 sec 39. a. dc 5 vc t 1 ac t2/2; dp 5 ap t2/2. One wants to solve for the 2 2 t at which dc 5 dp, i.e., when v0t 1 ac t2 5 ap t2 1 vc t 1 ac t2/2 2 ap t2/2 5 0 1 t (vc 1 [ac /2 2 ap /2]t) 5 0. This occurs when t 5 0 (when the criminal passes by the police car) and again when t 5 2vc /(ap 2 ac), (when the police catch up to the criminal). b. Now vc 5 ac t 1 vc and vp 5 ap t. One wants to solve for the t at which vc 5 vp, i.e., when ac t 1 vc 5 ap t or for t 5 (ap 2 ac)/vc. This does not mean that the police have caught up to the criminal, but rather that the police are at that moment going as fast as the criminal is and that they are starting to go faster than the criminal.
EE
29. a. Its velocity starts out negative and continues to be so since the object is falling; h is measured in cm above the ground; t is measured in seconds. b. h 5 150 2 25t 2 490t2; for 0 # t # 0.528 (the second value being the approximate time in seconds it takes for the object to hit the ground). c. The average velocity is the slope, i.e., (15 2 150)/0.5 5 2270 cm/sec; the initial velocity is 225 cm/sec. The average velocity is 10.8 times as great in magnitude as the initial velocity.
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INDEX A page number followed by f indicates a figure; a number followed by t indicates a table; a number followed by n indicates a footnote.
absolute value function, 115–117, 132 abstract variables, relationship between, 16–18 Acadia National Park, Maine, night temperatures at, 93 acceleration average, 566, 567f, 567t, 576 defined, 566 of falling body, 559, 566–567 and force relationship, 410, 411 gravity as cause of, 566, 567, 569, 575t and mass relationship, 410, 411 and time relationship, 411 and velocity relationship, 411 acidic or alkaline classification, 246–247 acidic solution (defined), 255, 357 acidity, measuring, 357–359 acid rain, 246, 358 acre (defined), 233, 257 adding functions, 521–522, 548 addition of ordinates, 529 additive versus log scales. See linear versus log scales aerobic exercise heart rate, target, achieving through, 207 level, recommended, 200–201 pulse rate, minimum and maximum during, 191 aerospace industry, net loss and profit of, 70 Africa, elephant population in, 70 age height as function of, 102 human versus dog, 206 pulse rate, minimum and maximum based on, 191 age distribution calculating, 8t for Ghana, 12f for Tanzania, 5, 5f, 6 age for U.S. population distribution of, 4, 4f, 5f, 6, 11, 12f median age, 13–14, 13f, 68 ages of presidents, 11 aggravated assault, juvenile arrests for, 74t AIDS cases in U.S., 19f, 86f airborne toxins, simulated, release of, 532 airplane collisions, preventing, 181–182 air temperature-water temperature relationship, 81 air traffic computers and controllers, 181–182 alcohol consumption versus education level, 81f alkaline solution (defined), 357 allometry, 434–437 Alpha Centauri (star), 232 American Lung Association, 125 ammonia, hydrogen ion concentration of, 358 ammonia, pH of, 358t ampicillin dosage, 29
AM radio waves, 247 Andromeda galaxy, 215, 255 angstrom (defined), 217, 232, 234 annual percentage rate (APR), 341, 379 annual percentage yield (APY), 341, 342, 379 anthrax spores, 232, 246 antifreeze for car radiators, 107 ant, length of, 246 apartments in Manhattan, purchase price of, 7 Apple Computer, 134 area, maximizing, 459 arguments, selecting, 14 Armenia, earthquakes in, 245 aspirin levels in bloodstream, 377f asteroids, 229–30 astronomical unit (defined), 262 asymptotes, 423, 526. See also horizontal asymptotes; vertical asymptotes AT&T’s One Rate plan, 101 atmospheric pressure at sea level, 340, 419 atomic weapons, 302, 303 atoms first, age of, 261t radius of, 246, 261t size of, 213t, 215–216, 316 Australia rabbit population, increase in, 368 automobile insurance, average cost of, 69 average (word origin), 6n average acceleration, 566, 567f, 567t, 576 average loan rates, 45f average rate of change calculating and interpreting, 65, 67–69, 76–79, 131 change in, 71–73, 72f, 72t limitations of, 68–69 for linear and exponential functions, 273 for quadratic functions, 493–495, 496, 547 as slope, 76–79 in time-distance relationship, 561–562, 562t, 572 of velocity (see acceleration) average velocity, 562–563, 562f, 565, 566 Avogadro’s number, 217, 229 axes, labeled, 4 axis of symmetry, 454, 457, 458, 547 “baby boom echo,” 73 baby names, 54f bacteria, 225, 266. See also under specific type of bacteria, e.g.: E. coli bacteria Baja California, population growth of, 296 ballast water, bacteria contained in, 225 ball thrown upward, height of, 549, 570f, 576 ball traveling distance down inclined plane, 396 Bank of America, 379 bar charts, 2–3, 2f, 5, 8f. See also histograms barometric pressure, calculating, 367
base defined, 218, 219, 255 effect of, 284–285 equalizing, 336 natural base e, 340–341, 342–344, 346, 363–366 baseball players, major league, average salaries for, 135f basic solution (defined), 255, 357 bathtub, capacity of, 445 bathtub, water level in, 61f battleship gun range, quadratic model for, 481 Bayer aspirin bottle, contents of, 445 bee hives, average price for, 553t beer, pH of, 379 Bejan, Adrian, Constructal Theory of, 447 Bell, Alexander Graham, 359 best-fit distance versus time function, 563, 563f best-fit exponential function, 268, 269f, 298, 304f best-fit line. See also regression line for children’s heights, 123 on log-log plot, 432, 435–436 overview of, 123, 132 on semi-log plot, 371 smokers, percentage of, 125, 126f best-fit linear function, 565f Betelgeuse (star), 233 bicycle use statistics, 51f Big Bang, 216–217 binary numbers, polynomial expressions for, 509 biological half-life, 325 bird and egg masses, 440t birth rate data, 20f, 73 population size, impact on, 182 per second, 258 for unmarried mothers, 270 bismuth-214, decay of, 340 blood alcohol concentration (BAC) average values for, 107t calculating, 102–103, 103t, 108, 297 legal limits for, 191 blood, hydrogen ion concentration of, 359 blood lead levels, children’s, reducing, 377f blood, pH of, 247, 359 blood type, U.S. population by, 53f body mass oxygen consumption versus, 441f population density versus, 441f surface area versus, 434–436, 435f body mass index (BMI), 203, 203f, 234 body temperature, surface area to volume ratio impact on, 387 boiling temperature of water, 163t, 246 bone marrow transplant, white blood cell count growth following, 303–304
691
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books shipped by publishers, number of, 120f Boston Marathon, women’s winning times for, 162f Boston radio stations, transmitting frequency of, 247 boundary lines, 183, 187 Bourton-on-the-Water, England, scale models of, 325 bowling ball floating in water, 396 Boyle’s Law for Gases, 410–411, 412–413, 418–419, 420 Braille code, 108 brakes, reaction time before hitting, 521 braking distance, 521–522 Brazil, inflation rate in, 306 breakeven point, 187–189, 188f breast cancer risk of, 19–20, 19f, 130f survival rates for, 546 tumor growth, modeling, 540–542, 546 brick, mass and volume of, 258 British pounds versus euros, 107 British thermal unit (BTU) (defined), 257 BSA (body surface area) as drug dosage determining factor, 259 Bush, George W., body mass index (BMI) of, 234 caffeine levels, measuring, 280–281, 291–292 calculators log function of, 249, 250 scientific notation display by, 216 speed of, 214n2 square root function of, 236 California, tomato production in, 551 calories burned from exercising, 55t, 162t, 206t Canadian-U.S. exchange rate, 135, 544 cancer trends, projecting, 22 candy corn production, 229 carbon-12, 312 carbon-14, decay of, 312 carbon dioxide, percent of greenhouse gas emissions, 138 carrying capacity (defined), 268 car sales, domestic, new versus Japanese, 70 Carthage, new, perimeter of, 479 cart, motion of, 478f car, trade-in value of, decrease in, 297 categorial (qualitative) data, 2 causation versus correlation, 145, 153–154 CD (compact disk), storage capacity of, 214, 222 ceiling function, 137 cell, living, radius of, 243 cell phones, U.S. households with, 130f cell phones, U.S. subscriptions to, 135 cells living, components of, manipulating, 246 marking, 349 size of, 213t chain letters, 315–316 chemotherapy, white blood cell count growth following, 270 child-bearing versus childless women, breast cancer risk of, 130f childhood obesity, 20, 20f children out of school, global breakdown of, 56f
children’s blood lead levels, reducing, 377f China area in square miles, 225, 378 oil production and consumption in, 205f population, 22f, 225, 312 U.S. trade with, 376f, 376t chip computing power, increase in, 368, 368t chlorine level in swimming pools, 296 cholesterol, measuring, 191 cigarettes, U.S. consumption and exports of, 71t circles area of, 390, 454 circumference of, 390 circumference to area ratio for, 387 radius of, 390, 532, 554 civil disturbances in U.S., 78–79, 79f, 79t closed interval (defined), 34 clothing sizes, international differences in, 134 Coca-Cola beverage consumption, 225, 233 Classic, 232 hydrogen ion concentration of, 358 pH of, 358t cocaine-related emergency room episodes, number of, 128t coefficient. See also correlation coefficient of friction, 396 in power functions, 400–402, 401f, 423, 424, 426–429, 442–444, 451–452 in scientific notation, 215, 216, 255 in time-distance relationship, 564, 566, 569 coffee cooling, 42, 42f, 517 hydrogen ion concentration of, 358 pH of, 358t, 363 college graduates, U.S., number of, 127f colleges, mean cost of 4-year, public versus private, 161f collinear points, 80 collision of objects and force exerted on, 513 Colorado River, length of, 257 common factors, 482 common logarithms calculating, 211, 248–249, 249t defined, 248, 256, 374 exponential functions as inverse of, 355 overview of, 349 representing, 352, 353f, 353t rules for, 329, 331–335, 350, 355, 374 comparisons of numbers, 211, 217, 220–221 completing the square (strategy), 473 complex number (defined), 487, 547 complex number expressions, multiplying, 492 complex zeros, 488 compound calculations, 305–306 compound inequality, 183–184, 187, 201 compound interest calculations based on, 343 overview of, 341–342, 374–375 rates, 304–306, 315 compression, function transformation through, 453, 468, 548 compression, vertical of graph, 400–401, 402, 442, 464, 465, 510 computer expenditures for universities, 528 computers in schools, statistics on, 70 computer storage, measurement units for, 217, 234
computer worms, 325 concave down (defined), 40, 50f concave up (defined), 40, 50f concert numbers versus attendance, 160f cone, volume of, 546 constant of proportionality, 108, 390, 410, 412, 417, 431, 442, 443 of vertical and horizontal intercepts, 111 constant dollars, 306, 326 constant functions, 498t constant percent change, 290–294 constant percent growth rates, 301–302 constant rate of change, 65, 87–89, 109 Constructal Theory, 447 continuous compounding, 329, 341–344, 343t, 366, 375 continuous growth or decay rate, 344, 345–346, 364, 365, 366, 375 continuous role in deflection, 397 contracting expression, 334, 350–351 conversion factors, 230, 231 cooking devices, designing, 455–456 corpse, temperature of, 520 correlation coefficient, 151–152, 153, 156, 160 correlation versus causation, 145, 153–154 cosmic chronology, 247 cost comparisons of heating systems, 167–168, 167f cost versus revenue, 188–189 cotton and wool fabrics, shrinking of, 192 counterarguments, examining, 14 countries, square areas of, 257t credit card debt, increase in, 298–299 crimes in U.S., 19t, 79f crude oil prices, 207f Cruithne (asteroid), 230 cryptology, 545 cube duplication of, 449 scaling up, 384–386, 388 surface area of, 384–385, 389, 454 volume of, 385–386, 389 cubic functions, 498, 498t, 501f Current Population Survey (CPS), 146, 147t, 157 cylinder plastic in aspirin bottle, 445 surface area of, 388 volume of, 388, 397, 413–414 dark energy, 214n3 dark matter, 214n3 data defined, 2 dictionary, 146, 147t presentation and interpretation, biases in, 82–84 table, exponential equation solving with, 330t daylight, hours of, seasonal, 520f death, causes of, international breakdown of, 53t death rate. See also under cause of death, e.g.: large intestine cancer, U.S. death rate from population size, impact on, 182 per second, 258 from smoking, 319, 319f
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decay factor annual, 301 daily, 500f defined, 279, 280, 284, 364 determining, 281, 282, 300, 364, 499 hourly, 281 representing, 292, 321 decay phenomena, modeling, 265, 280 decay rate, 292, 293, 300, 302–333, 321, 364, 499 decay rate, determining through, 293 decay, time required for, 337 decibel scale, 359–361, 360t, 381 decimal form, decay rate in, 292 decimal form, growth rate in, 291, 292, 299 deep space, 212–213 deep time, 216–217 deflection, 394, 396–397 6 notation, 76 demand curve, 176, 176f, 177 Deneb (star), 220, 232, 233 density, formula for determining, 258 dependent variables, 24, 29, 30, 31, 35, 50 depreciation, rate of, calculating, 89, 89f, 89t Dido, queen of Carthage, 479 digital information, generation of, 258 direct proportionality of cube surface area to square of edge length, 385 for linear functions, 108–109, 120, 132, 390 with more than one variable, 393–394, 397 overview of, 383 for power functions, 390–394, 396, 410, 442 properties of, 390–393 and weight-length relationship, 390 disco dancing, calories burned from, 55t discriminant, 484, 485, 486, 490, 547 distance-gravity relationship, 411 distance-time relationship average rate of change in, 561–562, 562t, 572 for falling bodies, 559, 561, 561f, 561t, 563–564, 565–566, 569, 573 for vehicles, 577 distributive law, 488, 525 dividing functions, 521, 548 division property in logarithms, 335 DNA double helix, 213 DNA molecule, 213 DNA strand, unwound, length of, 243 dog versus human life years, 206 dollar, value of, inflation impact on, 305–306 domain, 33–37, 50, 124, 538 “dotcom bubble,” 42 doubling time calculating, 301–302 defined, 299, 321 for E. coli bacteria, 329 for investment, 336 for lung cancer, 325 and Moore’s law, 368 for vibration frequency, 307 Dow Jones Industrial Average, 19f, 371–372, 371f Dragons of Eden, The (Sagan), 247 drug dosages, BSA (body surface area) as determining factor, 259
drug elimination from body, 312 duckling species, weights of, 271 duplicate lines, linear equation system represented by, 171f, 174, 175–176, 176f e (natural base). See natural base e as constant percent, 290–291 plotting, 268, 269f predicting, 266–267, 266f, 266t, 267t, 339 semi-log plot of, 316–17, 316f time calculations for, 329, 330, 336 Earth age of, 216t, 218f, 252, 252f, 253f, 256f, 261t circumference of, 229, 449, 538 companion asteroids of, 230 core of, 392 land, square miles on, 223 mass of, 258 moon distance from, 217 orbit, radius of, 233 organisms, living, first, 261t radius of, 212t, 221, 236, 246, 258, 261t, 388, 392, 538 stars’ distances from, 233 sun distance from, 232, 233, 268 surface area of, 236 volume of, 221, 392 earthquakes, energy magnitude of, 379 earthquakes, magnitude of, 243–244, 243t, 245, 359, 379 E. coli bacteria, growth of economics, linear systems in, 176–177 education as variable, 2–3, 2f education-health relationship, 81f education level college, 127f and earnings, relationship between, 145–154, 148f, 150f, 156, 157, 159f high school, 70, 70t, 73t effective interest rate, 341, 342, 344, 350, 375, 379 egg masses, 440t electrical resistance as wire length function, 120, 419 electrical resistance relationship to voltage and wattage, 446 electromagnetic radiation, 234, 247 electron, mass of, 218 electron, weight of, 246 elephant population, 69, 70 elephant seals, 257 elimination method of solving system of linear equations, 172, 173–174 ellipse, 17f, 18 embryo, average growth of, 480t energy calculating formula for, 397 comparisons, 259t consumption, projections for, 324 units of, 258 English versus metric unit conversion, 211, 212t, 218, 231 Environmental Protection Agency, 138 equations describing change with, 16–18 in function form, 31–33, 172–173, 175 function representation through, 23 in two variables, 50
693
equilibrium number (defined), 182 equilibrium point, 176, 176f, 177, 182 equivalent equations, 171f, 175–176 Eratosthenes, 449 Eros (asteroid), 230 estimates of solutions to problems, 17–18, 168–69, 223, 242, 330–331, 459–460 ethnicity, classifying, 2 eukaryotes, first, age and size of, 261t euros versus British pounds, 107 euros versus U.S. dollars, 107, 110, 545 evidence, selecting, 14 exchange rates, 107, 110, 135, 544, 545 excretion, drug elimination through, 312 exercise, pulse rate, minimum and maximum during, 191 expanding expression, 334, 350 experimentation, 560 exponential decay constant percent decrease, 291–293 graph of, 285, 287, 320, 320f versus growth, 280, 364, 375 overview of, 279–82 exponential decay function, 279–281, 285, 287, 292, 300, 320, 345 exponential equations estimating solutions for, 330–331 interpreting, 268 versus linear equations, 272 solving with logarithms, 329, 330, 331–335, 336–337, 374 exponential functions average rate of change for, 273 base e transformation of, 344–347, 363–366 compound interest rates, 304–306 constant percent change, 290–294 constructing, 268, 299, 300, 371–372 defined, 317 equations for, 329, 405t height-weight relationships expressed as, 434 identifying, 275–276 input, finding for, 329 inverses of, 540 logarithmic functions, relationship to, 329, 354–356 overview of, 43, 265 versus power functions, 390, 405–407 properties of, 321 representing, 265, 284, 292, 320, 321 selecting for set of data, 383 semi-log plots of, 265, 316–317, 316f, 321, 369–372, 375 shift of, 515–516, 517 visualizing, 284–287, 317, 405–407, 429, 430 exponential growth base effect on, 284–285, 287 by constant percent increase, 290–291 versus decay, 280, 364, 375 E. coli bacteria example, 266–268, 269f function, 267–268, 271–276, 272f, 272t, 279, 293–294, 300, 320 versus linear growth (see linear growth versus exponential growth) versus logarithmic growth, 354 variables affecting, 320 visualizing, 320f
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exponential models for continuous compounding, 329 exponential models of data, 274–275, 369–372 exponents base 10, 212, 213 defined, 255 errors involving, 221–222 fractional, 235–238, 239–240 logarithm relationship to, 248, 249 log scale labeling with, 253 negative, 213, 240 negative integer, 219, 226–228 positive integer (non-base 10), 218–222 as rational numbers, 219 rules for, 211, 219–221, 223, 226–228, 235, 238, 255, 332, 333 expressions, simplification of using exponents, 211 extrapolation, 126, 127f factoring for polynomials, 505 factoring for quadratics, 480, 481, 482–483 Factor Theorem, 488, 505, 547 falling object distance-time relationship for, 559, 561, 561f, 561t, 563–564, 565–566, 573 distance traveled by, 397, 551, 555–556 experimentation on, 560–562, 568f, 568t, 570–573 height, equation for, 568–569 family tree as fractal tree, 310 FAM1000 Census Graphs, 146, 149, 159 farmable land, people per square mile of, 223 farmable land, portion being farmed, 224 farm output, gross, 128f father functions, 531 federal debt, 47, 47f, 220–221, 315t federal debt per person, 523–524 federal funds rate, 44, 45f, 119 Federal Reserve, 44, 119 federal surplus or deficit, 14–15, 14f, 22, 47 female clothing sizes, international differences in, 134 Fidelity Fund investment, growth of, 319–320, 320f film exposure relationship to luminance, 323 first-class stamp prices, U.S., 122t fish and shellfish production, U.S., 340f fish caught, metric tons of, 53t flat versus graduated income tax, 193–198, 195f, 197t, 198f, 209 floor function, 137 flying bodies, cruising speed of, 447 FM radio waves, 247 focal length (defined), 458, 463 focal point of parabola, 454, 455, 456, 458, 547 FOIL technique, 456n, 488, 492 Food and Drug Administration, 191 food commodities, per capita consumption of, 283t food production increase, predicting, 308–309, 308f foods, pH of, 357, 358t, 363 football field, dimensions of, 233, 257 football players, salaries of, 257 force-acceleration relationship, 410, 411 force as function of time, 513
force-distance relationship, 410, 411 fractal tree, 309–310, 309f fractional exponents, 219, 235–238, 239–240, 255 free fall experiment, 560–562, 568f, 568t, 570–573 freezing point as function of salinity, 107t frequency count, 2, 3 distribution tables, 8 radio (defined), 247 frictional force between rope and pole, 242 “Friends and Family” long-distance service program, 315 functions algebra of, 521, 548 combining, 453, 521–527 composing, 531–533, 548 defined, 1, 22–23, 50 equations for, 538 families of, 43 finding from graph, 469 identifying, 26–27, 26f, 33, 35 increasing or decreasing, 40, 50f language of, 29–32, 33–37, 50 models, selecting, 383, 429–442 overview of, 22–26 product of, 531 properties of, 1 quotient of, 523–524 representing, 1, 23–24 requirements of, 536–537, 539 transforming, 453, 510–517, 513f, 533, 548 (see also quadratic functions: transforming) undefined, 35, 36, 37 visualizing, 39–43, 50f, 61 galaxies, recession rate from one another, 226 Galileo, 559, 563, 564, 565, 573, 660 gamma rays, wavelengths of, 234 gas consumption rate, speed impact on, 479 Gas Guzzler Tax, 131t gas prices international, 205t in Maine, 56t U.S. averages, 135t genealogy functions, 531, 536, 537 girls’ weight, median, constant rate of change for, 87–89, 88f, 88t global warming, 326, 553 gold price increases, 84f government trust fund, growth of, 314 graduated versus flat income tax, 193–198, 195f, 197t, 198f, 209 grain production and consumption, worldwide, 204f graphing calculators, 158 graphs of absolute value function, 115–116, 116f, 117f average rates of change represented through, 76–79 of cube surface areas, 385f of cube surface area/volume ratios, 386f of cube volumes, 385f data interpretation, role in, 83–84
describing and creating, 1 direct proportionality represented by, 108, 108f, 109f domain and range, finding from, 34–36 elements of, 4 of equations, 16–18, 32, 50 exponential equation solving with, 330t of exponential functions, 265, 285f, 286f, 287, 317, 320, 345f, 406, 406f, 407f, 443 of functions, 23, 26, 26f, 31f, 35, 36f, 37, 39f, 40, 529 of inequalities, 184f, 185f of inverse functions, 355–356, 375 of linear equation systems, 167, 167f, 171, 171f of linear functions, 90, 94, 97, 99–100, 100f, 101, 101f, 131–132, 406, 406f, 443 of logarithmic functions, 353–354, 514, 515f of piecewise linear functions, 114f, 115f, 193f, 195–196, 195f, 202f of polynomial functions, 453, 500–506, 501f, 507, 515, 547–548 of power functions, 383, 397–405, 406, 407f, 410f, 420–429, 431, 442, 443, 511, 512 of projectile’s paths, 460f of quadratic functions, 453, 454, 457, 458, 468, 469, 547 of rational functions, 527, 527f of revenue versus cost, 188f of step functions, 118f time as variable in, 13–14, 13f transforming, 469 zooming in on, 504–505 gravitational constant, 564 gravity acceleration due to, 566, 567, 569, 575t force between objects, 411, 415 impact on falling objects, 555–556 greatest integer function, 122, 137 greenhouse gas emissions, carbon dioxide, percent in, 138 gross domestic product (GDP), health care costs, U.S. as percentage of, 130t growth factor for bacteria, 268, 290–291 for cancerous tumors, 540, 541 defined, 267, 284, 364 determining, 282, 299, 371 growth rate used in determining, 291 for populations, 274 representing, 292, 293, 321 for white blood cells, 304 growth patterns, 41–42 growth phenomena, modeling, 265 growth rate, converting to growth factor, 291, 321 growth rate, determining, 293, 299, 304, 364, 371 hair conditioners, acidic, 357n Haldane, J. B. S., 387 HALE (health-adjusted life expectancy) versus health spending per person, 441f half-life biological, 325 of bismuth-214, 340 defined, 299
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of drugs, 312 exponential applications, 321 of radioactive substances, 300–301, 302–303, 311, 313, 348 half-open intervals, 34 Harvard Bridge, length of, 232 HDL (good cholesterol), 191 health care costs, U.S., 130t, 374t health care expenditures, increase in, 270, 324 health care expenditures, projected, 324 health-education relationship, 81f health spending per person, HALE (healthadjusted life expectancy) versus, 441f health spending per person, international breakdown of, 52f hearing, threshold of, 359, 360t, 363 heart attack, survival chances after, 297 heart rate of small versus large animals, 383, 384, 438f heart rate, target, aerobic exercise aimed at, 207 heart, restarting, timing impact on survival chances, 297 heating systems, solar versus conventional, 165, 166–169, 167f heat wave (defined), 552 hectare (defined), 233 hectometer, square (defined), 233 height age, function of, 102, 200t children’s mean, 123–125, 123f, 123t children’s versus weight, 431–434, 431f, 431t fathers’ versus sons’, 160 of first Homo sapiens, 261t in metric units, 213, 213t, 220, 246 paper thicknesses versus human, 268 hertz (defined), 247 hidden variable, 154, 157 high school completion, 70, 70t, 73t Hispanics, 2, 8f, 10f histograms, 4–5 Holdeen, Jonathan, 314 holding breath, avoidance of, 383, 412–413 home owners, U.S. percentage of, 70 home prices, increase in Medford, 278 home, single-family, median price for, 134 Homo sapiens, first, age of, 261t horizontal asymptotes, 287, 320, 420, 420f, 422, 424, 442, 526, 527 horizontal intercept of parabola, 457, 458, 459–460, 480–483, 484–490 of polynomial, 500, 502, 503–505, 503f, 506, 507, 525, 526, 547 of quadratic function, 490, 547 of rational function, 525, 548 of vertical line, 111 horizontal lines, 110, 111, 132 horizontal line test, 539–540, 548 hot water heater, diameter and volume of, 397 household income, median black versus white, 86f increase in, 324f inflation-adjusted, 44f, 306 household net worth, 15, 15f Hubble constant, 226 Hubble, Edwin, 216 Hubble’s Law, 226
human body temperature, normal, 122 human height. See height human life, origin of, 216t, 217 humidity impact on perceived temperature, 552 hydrogen atom, radius of, 215–216, 232, 243 hydrogen bomb, 312 hydrogen ion concentration, 246, 254, 357, 357f, 358, 359 icing temperature, salt impact on, 544 ideal body weight (IBW), formulas for, 546 imaginary numbers (defined), 487 immigration rate per second, 258 immigration rate, rise in, 73 income age impact on, 155 disposable, growth projection for, 291 and education level, relationship between, 159f gender impact on, 155–156, 160 household median, 44f, 86f, 306, 324f U.S. inequality of, 514, 514f incomes in United Kingdom (UK), 53 independent variable defined, 24, 29, 30, 50 graphical representation of, 31 reinitializing, 125–126 time as, 35 India area in square miles, 378 population, 378 wireless data market, growth of, 293 individual cell phone number mapping, 536t inequalities, manipulating, 184–189 infant mortality rates, U.S., decline in, 297, 439f infant mortality rates, U.S., male–female breakdown of, 129f infant mortality rates versus literacy rates, 129f inflation impact on money worth, 305–306 influenza-related deaths, 20–21, 21f information, relevant, collecting, 14 infrared light as electromagnetic radiation, 247 initial value, 267, 285–286 input in data tables, 31t doubling, 412 as independent variable, 24, 29, 30, 31, 50 insanity versus radio ownership, 154 instantaneous growth or decay rate, 344, 345–346, 364, 365, 366, 375 integer exponents, 218–222, 226–228, 238, 239 intercepts. See also horizontal intercept; vertical intercept characteristics of, 458 determining, 453, 457 estimating, 459–460 interest rate, 44, 119 Internet access, percent of U.S. adults with, 127f backbone bandwidth, 326f teenager use of, 8, 9f Internet users, increase in, 312, 324 Internet users, projections for, 324 interpolation, 126, 127f
695
intersection points in linear equation system, 171, 171f, 172, 174–176, 182, 195–196 interval notation, 34, 36 inverse functions, 355–356, 375, 453, 534–540, 546, 548 inverse proportionality (for power functions), 383, 411–422, 442 inverse square laws, 415–417, 446 investments comparing, 305, 305t doubling, 302, 336 growth of, 319–320, 320f, 515–516, 539f growth, targeted, interest required for, 504–505 mixed portfolio of, 553 iodine-131, decay of, 279, 280, 325 iPod sales, 134 Iraq war, U.S. opinion on, 54f Japan area in square miles, 225 Japan population, 57f, 225, 378 jet airplanes, noise level of, 361 job type and income, 156 jointly proportional variable (defined), 396 Jordan, Hamilton, 22 joule (defined), 258 joule measurement, applications of, 379 jump, height and velocity of, 576 Jupiter falling object, distance traveled on, 551 radius of, 224 sun distance from, 232 juvenile arrests, 56f, 74t Kalama, Egypt study on children’s heights, 123–125 Kenya, elephant population in, 69 Kepler, Johannes, 262 key ideas, identifying, 14 kilograms versus pounds, 128t kilometers to miles conversion, 231 King, Martin Luther, assassination of, 79 Kleiber’s Law, 436–437 K-12 education, federal spending on, 85f labor force, participation by gender, 170f land, measuring, 233 land on Earth, square miles of, 223 large intestine cancer, U.S. death rate from, 439f Las Vegas, NV population trends, 170f LDL (bad cholesterol), 191 lead-206, 311 lead-210, 311, 368 lemon juice, pH of, 246 Libby, Willard, 312 life expectancy increase in, 326f U.S. by race and sex, 71t U.S. increase in, 137f world versus Botswana, 55f light Andromeda galaxy, time to travel to, 255 intensity of, 415, 418 speed of, 230, 231, 233, 234, 554 visible, wavelengths of, 234 lightning and thunder, delay between, 554 light year, 220, 225, 232
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linear correlation, 153 linear equations applications of, 122–123, 132 constructing, 65, 90, 100–103, 124 duplicate lines represented by, 171f, 174, 175–176, 176f versus exponential equations, 272 systems of, 165, 166–169, 171–173, 201 linear factors, 482 linear functions average rates of change for, 273, 493–495, 496 constructing, 371 data approximation through, 149 direct proportionality for, 108–109, 120, 132, 390 equations for, 90–91, 131, 405t identifying, 275–276, 369–370 inverses of, 540 overview of, 43, 87–91, 265 piecewise, 114–118, 132, 165, 193–198, 202 polynomials, other, compared to, 498t, 499 selecting for set of data, 383 visualizing, 94–97, 131–132, 405–406, 429 linear growth versus exponential growth functions, 271–276, 272f, 272t, 293–294 overview of, 265, 320, 320f population growth versus food production, 308–309, 308f linear inequalities, 165, 183–189, 201 linear models of data, 65, 122–127, 150, 156, 274–275, 429 linear regression (defined), 160 linear relationship, 65, 109, 122–123, 150 linear versus log scales, 244, 244f, 245, 245f, 252, 316 literacy rates, 129f Little Rock, Arkansas, earthquakes in, 245 Litvinenko, Alexander, 378 logarithmic functions and exponential functions, relationship between, 329, 354–356 inverses of, 540 power functions translated into, 430 visualized, 515f logarithmic growth, 354 logarithms applications of, 357–361 base e (see natural logarithms) base 10 (see common logarithms) calculating, 250–251 contracting expressions with, 334, 350–351 defined, 248, 374 errors, common in using, 335 expanding expressions with, 334, 350 exponential equation solving with, 329, 330, 331–335, 336–337, 374 functions of, 43, 329, 352–361, 375 inverse functions and, 538, 546 numbers, calculating based on, 251 properties of, 353, 514 rules for, 331–335, 337, 374 simplifying expressions with, 334 solving equations containing, 335 log-linear plots. See semi-log plots log-log plots of height versus weight data, 432, 432f
power functions on, 431, 432–433 of surface area versus body mass, 435–436, 435f visualizing, 429, 429f, 430, 443 log scales. See also semi-log plots advantages of, 369 equivalent variations of, 370, 370f as function model selection aid, 383 noise, measuring with, 359 overview of, 244–245 plotting numbers on, 251–253, 252f, 253f, 256 sample, 247f long-distance phone calls, cost of, 101 Los Angeles, earthquakes in, 245 luminance, film exposure relationship to, 323 lung cancer, doubling times for, 325 lung cancer risk, smoking-related, 323, 369, 369f lung tumor, growth of, 314 lye, pH of, 379 magnitude of slope, 95–97, 96f major league baseball players, average salaries for, 135f Malthusian dilemma, 308–309 manufacturing employment, growth projection for, 291 margarine, consumption of, decrease in, 70 marital status of U.S. population, 71t Mars, falling object, distance traveled on, 551 mass-acceleration relationship, 410, 411 Massachusetts flat versus graduated income tax debate, 196–198, 197t, 198f Massachusetts Registry of Motor Vehicles, 359 mass-gravity relationship, 411 mathematical model (defined), 16 Matterhorn, height of, 268 maximum value, 39, 50f MCI “Friends and Family” program, 315 mean calculating, 7–8, 7n defined, 6 limitations of, 68 overview of, 6–7 regression lines, 152 meat, pH of, 358t median, 6, 7, 7n median age, 13–14, 13f, 22, 68 median regression lines, 152 medical degrees awarded to women, 129f Medicare, 298, 366 medicine dosage, calculating, 29 men’s mile, world records for, 162f metabolic mass, body surface area as indicator of, 259 metabolic rate of small versus large animals, 383, 384, 436f metabolic rate versus body mass, 436–437 metabolism, drug elimination through, 312 methane gas, increase in, 326f metric system, 212, 214, 230–231 metric to English unit conversion, 211, 212t, 567 Mexico City, population decline of, 296 Mexico population, growth rate of, 302 microorganisms, global distribution of, ballast as factor in, 225
microwaves as electromagnetic radiation, 247 military reserve enlisted personnel, 48, 48f milk pH of, 363 supply and demand curves, 177 whole, per capita consumption of, 339 Milky Way age of, 261t limits of, 268 radius of, 212t, 232, 246, 261t minimum value, 39, 50f minimum wage, 117–118, 118f minimum wage workers, salaries of, 257 Mirex pesticide, decay of, 324 mirror images, graphs as, 355–356, 356f, 375, 401, 424, 442 mirror images of parabolas, 458 Missouri state tax, 200t mole (defined), 217 Monaco area in square miles, 225, 257t Monaco population, 225 money, spending breakdown/distribution of, 9f, 10f money, worth of, inflation impact on, 305–306 moon Earth distance from, 217 radius of, 246 sun distance from, 232 Moore’s law, 317, 368 mortality rate for children, male and female in Sweden, 137f, 138 mortgage interest rates, 30-year, 85f mother functions, 531, 536, 537 motion bodies in, 559 describing, 453 equation of, 564, 573 experiments on, 660 upward versus downward, 569 motor vehicle accidents, deaths in, 69t motor vehicle registrations, number of, 74t, 164f multilevel marketing, 316 multiplicative scales. See log scales multiplying functions, 521, 531, 548 multivariable analysis, 155 murder, juvenile arrests for, 56f musical pitch, 307 mute swans, population growth of, 278 myelodysplastic syndrome, 303 nails, size and weight of, 240 nanometer (defined), 234 nanosecond (defined), 233 “nanotweezers,” 246 National Assessment of Educational Progress (NAEP) reading scores, 85f National Association of Realtors, 134 natural base e base conversion to, 346, 365–366 continuous compounding using, 341, 342–344 defined, 340–341 exponential functions, transforming to, 344–347, 363–366 overview of, 340–341 natural disasters, 54f, 377f natural gas production, 40, 40f
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natural logarithms defined, 349, 374 inverse of, 355 overview of, 349–351 representing, 352, 353f, 353t rules for, 329, 350, 355, 374 NEAR (robot spacecraft), 229–230 NEC company, 214 negative integer exponents, 219, 226–228 negative integer powers, 398, 409–429, 442 negative logarithms, 249 neutral solution (defined), 255, 357 neutrons, number in nuclear reactor, 368 neutrons, radius of, 246 news communication in U.S., methods of, 76 Newton’s Law of Cooling, 369, 520 Newton’s Law of Universal Gravitation, 411 New York City taxi fares, 205 1918 flu epidemic, population figure impact of, 73 nitric oxide, concentration of, 397 Nobel prizes in science, 86f noise exposure, maximum safe limits for, 314t level, legal limits of, 359 measuring, 359–361 nominal interest rate, 341, 342t, 344, 350, 374–375, 379 nonagriculture employment, growth projection for, 291 noninvertible functions, 536, 537 Norman window, shape of, 479 notation. See also scientific notation 6, 76 of functions, 29–33, 50 -, 11 nth roots, 237–238 nuclear energy generation in U.S. versus United Kingdom, 225 nuclear reactor, neutrons, number in, 368 number in poverty, 42f, 43 object sizes comparing with ratios, 220–221 on linear scale, 244f on order-of-magnitude (logarithmic) scale, 244f objects thrown upward, height of, 478, 549, 569–570, 570f, 576 observation, use of, 560 ocean wave velocity, calculating, 419 octaves, function of, vibration frequency as, 307 odd versus even powers inverse functions and, 540 for polynomial functions, 502 for power functions, 399–400, 399f, 400f, 422–423, 422f, 442, 512 oil, barrel of, U.S. versus British, 233 Old Faithful geyser, 576 Olympic shot put, 24–26, 25t, 129f one-child-per-family policy, 378 one-to-one functions, 537, 538, 539–540, 545, 548 open interval (defined), 34 orange juice, hydrogen ion concentration of, 255 orbital period, 262, 262t, 263t
order of magnitude scales. See log scales orders of magnitude, 242–245, 256, 359 ordinates, addition of, 529 organism length, population density versus, 441f organs, surface areas of, 387 Orion constellation, 233 oscillation frequency, calculating, 418 outliers (defined), 7 output, 24, 29, 30, 31, 31t, 50 oxygen, concentration of, 397 oxygen consumption versus body mass, 441f Pangaea (land mass), 253, 261t paper recycled, 136 paper, thickness of, 268 parabola defined, 454, 547 equation for, 474, 489 intercepts of, 453, 457, 458, 459–460 properties of, 458 parabolic devices, 454–456, 550 parabolic roof, 552 parallel lines, linear equation system represented by, 171f, 174–175, 175f parallel lines, slopes of, 112, 113, 132 parental education, 154 parental socioeconomic status, 154, 157 part-time jobs, 156 patterns, searching for, 14 Pearson, Karl, 160 pedaling into the wind, 391–392 pediatric growth charts, 87, 102 pendulum, period of motion for, 242 Pensacola, Florida tides, 34–35, 35f percentage form, decay rate in, 292 percentage form, growth rate in, 291 perfect square (defined), 235 period of pendulum motion, 242 period of planetary orbit, 262, 262t, 263t perpendicular lines, slopes of, 112–113, 132 perspectives, changing on functions, 534–535 pH function, non-linearity of, 359 pH level, calculating, 358 phone bill, monthly, modeling, 101 phone tree as fractal tree, 310 phosphorus-32, decay of, 349 pH scale, 246–247, 254–255, 357–359, 357f, 357t, 359 piecewise functions (defined), 194–195, 202 piecewise linear functions, 114–118, 132, 165, 193–198, 202 pie charts, 5–6, 8f Pittsburgh, PA population trends, 170f Planck’s length, 234 Planck time, 234 planets. See also Earth falling objects on, 551 objects thrown upward on, 478 orbital periods of, 262, 262t, 263t radii of, 224 sun distance from, 232, 233, 262, 262t, 263t plank deflection, calculating, 394, 396–397 Pleiades (galactic cluster), age and size of, 261t plotting numbers on logarithmic scale, 251–253, 252f, 253f, 256 Pluto, height of objects thrown upward on, 478 plutonium, decay of, 302–303
697
plutonium-238, decay of, 284 pneumonia-related deaths, 20–21, 21f point spread in football game (defined), 9 poison ivy, treating, 499 poll results, range in values of, 116 polonium-210, 378 polonium-214, decay of, 311 polynomial functions constructing, 505–507 creating, 453 definition and terminology for, 498–500, 498t in factored form, 505, 507 global characteristics of, 500–502 overview of, 43, 507, 547–548 quotient of, 524–525, 548 ratio of, 453 visualizing, 500–502 polynomials applications of, 498, 499 binary numbers, expressing as, 509 of degree 0, 498t of degree 1 (see linear functions) of degree 2 (see quadratic functions) of degree 3, 498, 498t, 501f of degree 4, 498t, 499, 501f of degree 5, 498t, 499, 501, 501f positive integers, expressing as, 509 polystyrene molecules, components of, manipulating, 246 population density versus body mass, 441f population density versus organism length, 441f population growth predictions, 274–275, 275f, 275t, 308–309, 308f population growth rate, 268 population pyramids, 12, 12f, 57f population size, factors affecting, 182, 268 population statistics, international, 9f positive integer powers, 397–402, 442 positive integers, polynomial expressions for, 509 potassium-40, decay of, 284 pounds versus kilograms, 128t poverty rate, 42f, 43, 82–83, 82f, 82t, 84t power (defined), 219, 255 power functions adding, 453 defined, 389 of degree –1, 386 of degree 2, 384 equations for, 389, 405t, 409, 431, 432–433, 443 versus exponential functions, 390, 405–407 height as, 432–433 inverses of, 540 logarithmic functions, translating into, 430–431 with negative powers, 409–417, 442 overview of, 43, 442 with positive powers, 389–394, 397–402, 442 properties of, 383 quadratics as, 464 transformations of, 516 visualizing, 405–406, 420–426, 429, 430 power intensity, calculating, 416, 416f powers of 10, 212–214, 255 predator-prey model, 42, 42f prednisone level, changes in, 499, 500f
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Presley, Elvis, impersonators of, 312 prices, increase in, 258t primates, chest circumference and body weight of, 240 principal square root (defined), 235 profit, calculating, 522–523, 523f projectile, trajectory of, 459–460, 480, 481, 481f, 539f projectile, travel distance as function of launch angle, 62 protons, radius of, 246 pulse emission, optical communications laser, speed of, 214 pulse rate, minimum and maximum during aerobic exercise, 191 pyramid growth schemes, 316 pyramids, dimensions of, 449 quadratic formula, 480, 484–487, 490, 492, 547 quadratic functions applications of, 453 average rate of change for, 493–495, 496, 547 equations of, 458 factored form, 488–489, 490, 492, 547 falling object modeling with, 574 forms of, converting, 453, 472–474, 488–489, 490 for free fall data, 563 overview of, 454–456, 547 polynomials, other, compared to, 498t, 499, 501f properties of, 457–458 roots or solutions of, 485, 486 standard form, 456, 470–474, 475, 490, 547 terminology for, 456, 498t transforming, 463–466, 467–470, 471–475 vertex and intercepts of (see intercepts; vertex) vertex form, 468–470, 472–474, 475, 547 qualitative (categorial) data, 2 quantitative reports, 14 quantitative variables, 2 quantities as functions of time, 26f quantities, ratios of, 220–221 quartic functions, 498t, 499, 501f Quintana Roo (Mexico), population growth of, 296 quintic functions, 498t, 499, 501, 501f rabbit population, increase in, 368 race, classifying, 2 radiation, electromagnetic. See electromagnetic radiation radiation intensity as wavelength function, 26f radicals, rules for, 238 radioactive decay, 300–301, 311 radioactive substances, storage for, 313, 349 radio ownership versus insanity, 154 radio stations, transmitting frequency of, 247 radio waves, 247 rain, acid, 246, 358 raindrop, weight of, 246 rain, pH of, 246, 361 range, 33–37, 50, 122 range of error, 116, 116f rate of change. See also average rate of change constant, 65, 87–89, 109 varied by interval, 68–69, 69f, 69t
rational exponents (defined), 239 rational functions creating, 453, 548 overview of, 524–525 visualizing, 525–527 rational numbers, 219 ratios, estimating, 242 ratios, object sizes, comparing with, 220–221 reading scores, National Assessment of Educational Progress (NAEP), 85f real dollars (defined), 306 real-world situation, equation describing, 16 rebound height (defined), 340 rectangular area, maximizing, 459 recycled paper, 136 reflections function transformation through, 453 across horizontal and vertical axes, 511, 512, 512f, 514 across horizontal axis, 464–465, 511, 512, 548 across vertical axis, 511, 512, 548 region of profit (defined), 189 regions, terminology describing, 183–184, 189 regression analysis, 149 regression line defined, 149, 160 for earnings versus education, 149f, 150f, 156, 159 for fathers’ versus sons’ heights, 160 interpreting, 145, 153–154 limitations of, 157 mean versus median, 152 reindeer population growth, St. Paul Island, Alaska, 313 relative frequency, 2, 3 representations of data, bias in, 65 restrictions on domains and ranges, 35–37 retirement allowance, annual, determining, 510 revenue (defined), 463 revenue versus costs, 522–523, 523f Richmond, Virginia population growth patterns, 41–42, 41f Richter scale, 243–244, 243t, 245, 359, 379 Rigel (star), 233 rock band, noise level of, 360–361 roller coaster car, speed of, 41 roots as fractional exponents, 219, 235–238, 239–240, 255 roots of quadratic functions, 485, 486 rope and pole, frictional force between, 242 rope, breaking strenth of, 242 rounding off numbers, 223 rubber ducky race, 91 rule of 70, 301–303, 321, 336, 366 running, calories burned from, 206t Ryan, Nolan, 576 Sagan, Carl, 247 sales tax in Canada, 135 function, 534–535, 537 price, calculating based on, 534–535 in U.S., 23, 24, 93 salinity, freezing point as function of, 107t salt, icing temperature, impact on, 544 SAT scores, female, 70 scale, effect of, 434–437
scale models, 325 scaling factor, 448, 449 scatter plots analyzing, 148–149, 153f for education versus earnings, 148–149 for infant mortality versus literacy rates, 129f for linear models of data, 123 overview of, 13–14, 13f, 14f scientific method, 559, 660 scientific notation, 211, 212, 214–217, 214n2, 223, 252, 255 scuba diving, rule of, 383, 412–413 sea depth impact on species density, 368 sea ice, melting rate for, 292 sea level, global warming impact on, 553 sea water, pH of, 358t seismic energy, 379 self-similar growth, 433–434 self-similarity (defined), 325 semi-log plots defined, 316f exponential function equations, finding on, 329, 370 of exponential functions, 265, 316–317, 316f, 321, 369–372, 375 of height versus weight data, 431, 432f, 434 and other plots compared, 429, 429f, 430, 443 shampoo, alkaline nature of, 357n shifting function transformation through, 453, 510, 514 horizontal, 465, 466–468, 515–516, 548 vertical, 465–466, 467, 468, 517, 548 short-term interest rate, 119 shot put, 24–26, 25t, 129f, 480 - notation, 11 sigmoid (defined), 268 sigmoid curve, 269f, 271f sign of slope variable, 94–95, 95f, 96f, 97 single-family home, median price for, 134 single-variable data describing, 2–7 patterns in, 1 visualizing, 2–6, 49 singularity, 423, 525, 526 “60-second summary,” 1, 14–15, 40 size and shape, relationship between, 383, 386–387 skid distance, 393, 396, 521–522 slope average rate of change as, 561–562, 562f calculating, 76–78, 90, 94–97, 100, 110–113, 123, 131 function of, 132 in semi-log plot, 371, 375 undefined, 111 smallpox deaths in Sweden, 138 smokers, percentage of adult, 125–126, 125f, 126f, 127f, 161t smokers, state breakdown of, 100f smoking deaths due to, 319, 319f versus education level, 81f lung cancer due to, 323 Smoots (unit) (defined), 232 snow, pH of, 361
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snowplow speed versus snow depth, 101, 101f snow tree cricket, chirps per minute of, 106 soap, alkaline nature of, 357n soap, pH of, 358t SoBig (computer worm), 325 Social Security number function and inverse, 537f soda can, designing, 413–414 solar cooking devices, designing, 455–456 solar energy units consumed, number of, 128t solar system, age of, 261t solar system, radius of, 212t, 220, 233, 242, 246, 261t, 316 solar versus conventional heating systems, 165, 166–169, 167f solution to equation, estimating from graph, 17–18, 168–169 solution to equation for abstract variables, 16–18, 50 solution to system of linear equations, 168–169, 171–173, 174–177, 201 sound circle, radius of, 554 intensity of, 359–360, 363 perception of, 359, 360t, 361 reduction through walls, 447t speed of, 258, 554 sources of data in graphs, 4 Soviet Union, collapse of, 378 space and time, measuring, 212–217 species density, sea depth impact on, 368 speed calculation formula, 233 and calories per minute, relationship between, 162t skid proportionality to, 393 speed-engine efficiency relationship, 479 sphere circumference of, 538 radius of, 538 velocity in fluid, 397 volume of, 387, 388 sport utility vehicle (SUV) sales, 130f square roots, 235–236 standard plots, 429, 429f, 431 star magnitude, calculating, 363, 381 statistics, selecting and interpreting, 84 steepness, appearance versus magnitude of slope, 97 step functions, 117–118, 118f, 132 stocks and bonds, investing in, 174 stretching, function transformation through, 453, 468, 514, 548 stretching, vertical of graph, 400, 442, 464, 465, 510 stroke, neuron loss, average from, 161f strontium-90, decay of, 300–301, 301f, 301t, 337, 374f substitution method of solving system of linear equations, 172–173 subtracting functions, 521, 522–523, 548 sulfuric acid, pH of, 358 sun age of, 261t core, temperature of, 246 Earth distance from, 232, 233, 268 Jupiter distance from, 232 moon distance from, 232
planet distances from, 262–263, 262t, 263t radius of, 212t, 221, 243, 246, 261t Venus distance from, 233 volume of, 221 sun protection factor (SPF), 191 Super Bowl game (2004), 9 supply and demand, 176–177 supply curve, 176, 176f, 177 surface area to volume ratio for cubes, 384t, 386, 387, 389 surface area to volume ratio for threedimensional objects (general), 386–387 surface area versus body mass, 434–436, 435f Surtsey (volcanic island), 253, 261t swans, population growth of, 278 swimming, calories burned from, 206t swimming pool, chlorine level in, 296 swimming pool, parabolic roof of, 552 symmetry axis of, 454, 457, 458, 547 of graphs, 423, 424, 442, 512–513, 512f, 514, 548 of parabolas, 458, 547 tables exponential equation solving with, 330t frequency distribution, 8 function representation through, 23, 24–26, 25t, 31 linear relationships represented through, 102–103 tank emptying time, calculating, 418 tank filling time, calculating, 417 tap water, pH of, 246 taxpayers, mean income of, 43, 43f tax plans, 193–198, 209 tax rate schedule for single persons, 210t tax rebates, 9 TC (total cholesterol)/HDL (good cholesterol) ratio, 191 teenagers, Internet use by, 8, 9f temperature functions of, 26f, 28, 106 of object, ambient temperature impact on, 369, 520 perceived, humidity impact on, 552 rate of change in Acadia National Park, Maine, 93 speed of sound, impact on, 258 textbook prices, increase in, 325 thorium-230, decay of, 368 thorium-234, 311 3-D objects, representing, 448 thunder, delay following lightning, 554 tides, sun and moon effect on, 416–417 time and acceleration relationship, 411 functions of, 22, 26f, 66–67, 513, 540, 541, 551 and space, measuring, 212–217 tumor diameter growth over, 542, 542f time-distance relationship. See distance-time relationship time series, 125–126 time series graph, 13–14, 13f, 49, 49f time zones, 28 toilets, gallons per flush used by, 207 tomatoes, water requirements for, 551
699
tornadoes, U.S. fatalities due to, 79f total cholesterol (TC)/good cholesterol (HDL) ratio, 191 toxic plumes, 532 trade balance (defined), 319 trade deficit (defined), 319 traffic flow, measuring, 471–472, 472f traffic flow models, describing, 453 transistors per integrated circuit, increase projections for, 317 triangle area, calculating, 397 tritium, decay of, 312 trout, water temperature, optimal for, 119 tub, water level in, 61f tuition revenue, quadratic model for, 456 turning points on graph, 500, 502, 547 TV signal, speed of, 228, 229 two-variable data patterns in, 1 visualizing, 13–15, 49 Tyrannosaurus rex, age and size of, 261t ultraviolet light as electromagnetic radiation, 247 ultraviolet light, wavelengths of, 234 Under Armour sports clothing company, profit increase of, 296 unemployment rate, 81f, 156 union membership, U.S., 164t unit conversions in metric system, 230–231 unit conversions, metric to English, 211, 212t, 567 unit costs, 524 United Kingdom, motor vehicles in, 324 United Kingdom population growth projection (1798) for, 308–309 in 2002, 309 in 2006, 225 units of measurement in graphs, 4 universe age of, 216–217, 216t, 261t edge of, 268 object sizes in, 212, 212t, 244f, 261t, 316 radius of, 242, 261t University of Massachusetts parking fee increases, proposed, 278 unmarried mothers, births to, 270 upward velocity, 569–570 uranium-238, decay of, 311, 348 urban population, growth projection for, 291 U.S. area in square miles, 225 U.S. Army sleeping bags, 184–185 U.S. bicycle use, 51f U.S.–Canadian exchange rate, 135, 544 U.S. Census data, 145, 146, 147t, 157 U.S.–Chinese exchange rate, 55t U.S. dollars versus euros, 107, 110, 545 U.S. GDP, increase in, 326f U.S. government debt, 47, 47f, 220–221, 315t, 523–524 U.S. health care costs, 130t, 374t U.S. infant mortality rates, 129f, 297, 439f U.S. international trade, 69t, 319t, 376f, 376t U.S. life expectancy, 71t, 137f U.S. military sales to foreign governments, 47, 47f U.S. natural gas production, 40, 40f U.S. newspapers, number of, 76
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INDEX
U.S. population age of (see age for U.S. population) average rate of change in, 71–73, 72f, 72t, 75t change in, describing, 66–67, 66f–67f, 66t, 68–69, 83–84, 83f, 84f versus China population, 22f coastal inhabitants, percentage in, 136, 138 federal debt per person, 220–221 growth of, 258f, 296, 311f, 311t growth rate, constant, hypothetical, 87, 87f versus Japan population, 225 marital status of, 71t Pittsburgh, PA, and Las Vegas, NV, trends, 170f racial and ethnic composition of, 10f rural to urban shift, 78, 78f versus United Kingdom population, 225 versus world population, 256 U.S. TV stations, commercial, number of, 76 U.S. venture capital, 42, 42f U.S. wireless data service revenues, growth of, 293 variables defined, 2, 50 and functions, 22–23 relationships between, 13–15, 16–18, 24–26, 32–33, 148 velocity and acceleration relationship, 411 average, 562–563, 562f, 565, 566 of falling body, 559, 562–563, 565–566, 573 Venus, sun distance from, 233 verbal description of linear equation, 101–102 vertex defined, 454, 458, 547 determining, 453, 457, 463–466, 468, 470–472, 475, 489, 547 formula for, 470–472, 474 for parabolic device, 456 vertical asymptotes, 353–354, 420, 423, 442, 516, 525, 526, 548 vertical intercept as constant, 111 defined, 90 determining, 101, 124, 126 function of, 94, 97, 131 negative value for, 150 of parabola, 457, 458 of polynomial function, 502, 503f, 507 of regression line, 156 vertical lines, 111, 132 vertical line test, 26, 26f, 33, 35, 50, 149, 539
vibration frequency as function of octaves, 307 vibration frequency as function of string length, 419, 445–446 Vick, Michael, 257 Vietnam, U.S. troop withdrawal from, 79 vinegar, hydrogen ion concentration of, 357 vinegar, pH of, 358t virus, radius of, 218 visible light as electromagnetic radiation, 247 visible-light spectrum, 234 volume, calculating, 387. See also term volume under name of three-dimensional object, e.g.: cube: volume of volume–diameter relationship, 541 volume–pressure relationship, 410–411, 412–413, 413f, 413t, 420t voter turnout data, 137t walking speed and duration, recommended, 200–201 wall insulation cost, calculating, 396 water acidic-alkaline neutrality of, 246 boiling temperature of, 163t, 246 bottled, cost of, 346–347 bottled, per-capita consumption of, 270 density as temperature function, 26f density of, 258 depth–pressure relationship of, 419t hydrogen ion concentration of, 246 (pure), 254 jet of, shooting up, 576 pH of, 357 salinity, freezing point as function of, 107t tap water, pH of, 363 water temperature air temperature relationship to, 81 boiling, altitude relationship to, 163t optimal for trout, 119 wavelength, frequency, relationship to, 247 wavelength, functions of, 26f Weber-Fechner stimulus law, 359n weight, healthy based on height for men, 185–186, 185f, 191 weight, healthy based on height for women, 106, 191 weight impact on blood alcohol concentration, 107t weight–length relationship, 390 weightlessness in space, 415 weight of children age, function of, 200t versus height, 431–434, 431f, 431t standard growth curve for, 575
weight versus distance from Earth center, 419 wheat, barrel of, 233 wheelchair ramps, slope of, 81 white blood cell count growth, 270, 303–304 Wikipedia, number of articles posted on, 339 wildfire spread, predicting, 453, 544 wind force, 391–392, 392f speed, 447, 532, 544 velocity, 391–392, 392f wind chill effect, 28 wind chill temperature, estimating, 544 wind turbines, power generated by, 445 wine hydrogen ion concentration of, 357 pH of, 247, 358t U.S. consumption of, 128 wolf population, Yellowstone National Park, 50–51, 50f, 52f women long jump world records of, 163f medical degrees awarded to, 129f recommended weight formula for, 106 in U.S. military, 86t wording role in data interpretation, 83–84 work, amount of and income, 156 world energy use, 257 World Health Organization Report 2005, 52 world population average rate of change in, 73f, 73t current, 323 growth, 15, 15f, 313f historical and projected, 48, 48f predicting, 323 in 2004, 256 in 2005, 233, 313f in 2007, 223 World War I, population figure impact of, 73 World War II, water conservation in, 207 xenon gas (Xe-133), 325 X-rays as electromagnetic radiation, 247 X-rays, wavelengths of, 234 Yangtze River (China), length of, 257 Yellowstone National Park wolf population, 50–51, 50f, 52f Yemen, population of, 57f y-intercept. See vertical intercept Yugoslavia, inflation rate in, 314 zero as exponent, 226, 240 zero population growth, 268 zero product rule, 481 zeros of function, 485, 503, 507, 547
